This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "List of integrals of rational functions" – news · newspapers · books · scholar · JSTOR (February 2021) (Learn how and when to remove this template message)

The verifiability of the claims made in this article is disputed. Please help improve this article by verifying its references and removing any that are not reliable or do not support the article. Relevant discussion may be found on the talk page. (February 2021) (Learn how and when to remove this template message)

The following is a list of integrals (antiderivative functions) of rational functions. Any rational function can be integrated by partial fraction decomposition of the function into a sum of functions of the form:

{\frac {a}{(x-b)^{n))}

, and

{\frac {ax+b}{\left((x-c)^{2}+d^{2}\right)^{n))}.

which can then be integrated term by term.

For other types of functions, see lists of integrals.

Miscellaneous integrands

$\int {\frac {f'(x)}{f(x)))\,dx=\ln \left|f(x)\right|+C$
$\int {\frac {1}{x^{2}+a^{2))}\,dx={\frac {1}{a))\arctan {\frac {x}{a))\,\!+C$
$\int {\frac {1}{x^{2}-a^{2))}\,dx={\frac {1}{2a))\ln \left|{\frac {x-a}{x+a))\right|+C={\begin{cases}\displaystyle -{\frac {1}{a))\,\operatorname {artanh} {\frac {x}{a))+C={\frac {1}{2a))\ln {\frac {a-x}{a+x))+C&{\text{(for ))|x|<|a|{\mbox{)))\\[12pt]\displaystyle -{\frac {1}{a))\,\operatorname {arcoth} {\frac {x}{a))+C={\frac {1}{2a))\ln {\frac {x-a}{x+a))+C&{\text{(for ))|x|>|a|{\mbox{)))\end{cases))$
$\int {\frac {1}{a^{2}-x^{2))}\,dx={\frac {1}{2a))\ln \left|{\frac {a+x}{a-x))\right|+C={\begin{cases}\displaystyle {\frac {1}{a))\,\operatorname {artanh} {\frac {x}{a))+C={\frac {1}{2a))\ln {\frac {a+x}{a-x))+C&{\text{(for ))|x|<|a|{\mbox{)))\\[12pt]\displaystyle {\frac {1}{a))\,\operatorname {arcoth} {\frac {x}{a))+C={\frac {1}{2a))\ln {\frac {x+a}{x-a))+C&{\text{(for ))|x|>|a|{\mbox{)))\end{cases))$
$\int {\frac {dx}{x^{2^{n))+1))={\frac {1}{2^{n-1))}\sum _{k=1}^{2^{n-1))\sin \left({\frac {2k-1}{2^{n))}\pi \right)\arctan \left[\left(x-\cos \left({\frac {2k-1}{2^{n))}\pi \right)\right)\csc \left({\frac {2k-1}{2^{n))}\pi \right)\right]-{\frac {1}{2))\cos \left({\frac {2k-1}{2^{n))}\pi \right)\ln \left|x^{2}-2x\cos \left({\frac {2k-1}{2^{n))}\pi \right)+1\right|+C$

Integrands of the form x^m(a x + b)ⁿ

Many of the following antiderivatives have a term of the form ln |ax + b|. Because this is undefined when x = −b / a, the most general form of the antiderivative replaces the constant of integration with a locally constant function.^[1] However, it is conventional to omit this from the notation. For example,

\int {\frac {1}{ax+b))\,dx={\begin{cases}{\dfrac {1}{a))\ln(-(ax+b))+C^{-}&ax+b<0\\{\dfrac {1}{a))\ln(ax+b)+C^{+}&ax+b>0\end{cases))

is usually abbreviated as

\int {\frac {1}{ax+b))\,dx={\frac {1}{a))\ln \left|ax+b\right|+C,

where C is to be understood as notation for a locally constant function of x. This convention will be adhered to in the following.

$\int (ax+b)^{n}\,dx={\frac {(ax+b)^{n+1)){a(n+1)))+C\qquad {\text{(for ))n\neq -1{\mbox{)))$ (Cavalieri's quadrature formula)
$\int {\frac {x}{ax+b))\,dx={\frac {x}{a))-{\frac {b}{a^{2))}\ln \left|ax+b\right|+C$
$\int {\frac {mx+n}{ax+b))\,dx={\frac {m}{a))x+{\frac {an-bm}{a^{2))}\ln \left|ax+b\right|+C$
$\int {\frac {x}{(ax+b)^{2))}\,dx={\frac {b}{a^{2}(ax+b)))+{\frac {1}{a^{2))}\ln \left|ax+b\right|+C$
$\int {\frac {x}{(ax+b)^{n))}\,dx={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1))}+C\qquad {\text{(for ))n\not \in \{1,2\}{\mbox{)))$
$\int x(ax+b)^{n}\,dx={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)))(ax+b)^{n+1}+C\qquad {\text{(for ))n\not \in \{-1,-2\}{\mbox{)))$
$\int {\frac {x^{2)){ax+b))\,dx={\frac {b^{2}\ln(\left|ax+b\right|)}{a^{3))}+{\frac {ax^{2}-2bx}{2a^{2))}+C$
$\int {\frac {x^{2)){(ax+b)^{2))}\,dx={\frac {1}{a^{3))}\left(ax-2b\ln \left|ax+b\right|-{\frac {b^{2)){ax+b))\right)+C$
$\int {\frac {x^{2)){(ax+b)^{3))}\,dx={\frac {1}{a^{3))}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b))-{\frac {b^{2)){2(ax+b)^{2))}\right)+C$
$\int {\frac {x^{2)){(ax+b)^{n))}\,dx={\frac {1}{a^{3))}\left(-{\frac {(ax+b)^{3-n)){(n-3)))+{\frac {2b(ax+b)^{2-n)){(n-2)))-{\frac {b^{2}(ax+b)^{1-n)){(n-1)))\right)+C\qquad {\text{(for ))n\not \in \{1,2,3\}{\mbox{)))$
$\int {\frac {1}{x(ax+b)))\,dx=-{\frac {1}{b))\ln \left|{\frac {ax+b}{x))\right|+C$
$\int {\frac {1}{x^{2}(ax+b)))\,dx=-{\frac {1}{bx))+{\frac {a}{b^{2))}\ln \left|{\frac {ax+b}{x))\right|+C$
$\int {\frac {1}{x^{2}(ax+b)^{2))}\,dx=-a\left({\frac {1}{b^{2}(ax+b)))+{\frac {1}{ab^{2}x))-{\frac {2}{b^{3))}\ln \left|{\frac {ax+b}{x))\right|\right)+C$

Integrands of the form x^m / (a x² + b x + c)ⁿ

For $a\neq 0:$

$\int {\frac {1}{ax^{2}+bx+c))dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2))))\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2))))+C&{\text{(for ))4ac-b^{2}>0{\mbox{)))\\[12pt]\displaystyle {\frac {1}{\sqrt {b^{2}-4ac))}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac))}{2ax+b+{\sqrt {b^{2}-4ac))))\right|+C={\begin{cases}\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac))}\,\operatorname {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac))}+C&{\text{(for ))|2ax+b|<{\sqrt {b^{2}-4ac)){\mbox{)))\\[6pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac))}\,\operatorname {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac))}+C&{\text{(else)))\end{cases))&{\text{(for ))4ac-b^{2}<0{\mbox{)))\\[12pt]\displaystyle -{\frac {2}{2ax+b))+C&{\text{(for ))4ac-b^{2}=0{\mbox{)))\end{cases))$
$\int {\frac {x}{ax^{2}+bx+c))\,dx={\frac {1}{2a))\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a))\int {\frac {dx}{ax^{2}+bx+c))+C$
$\int {\frac {mx+n}{ax^{2}+bx+c))\,dx={\begin{cases}\displaystyle {\frac {m}{2a))\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2))))}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2))))+C&{\text{(for ))4ac-b^{2}>0{\mbox{)))\\[12pt]\displaystyle {\frac {m}{2a))\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{2a{\sqrt {b^{2}-4ac))))\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac))}{2ax+b+{\sqrt {b^{2}-4ac))))\right|+C={\begin{cases}\displaystyle {\frac {m}{2a))\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac))))\,\operatorname {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac))}+C&{\text{(for ))|2ax+b|<{\sqrt {b^{2}-4ac)){\mbox{)))\\[6pt]\displaystyle {\frac {m}{2a))\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac))))\,\operatorname {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac))}+C&{\text{(else)))\end{cases))&{\text{(for ))4ac-b^{2}<0{\mbox{)))\\[12pt]\displaystyle {\frac {m}{2a))\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)))+C={\frac {m}{a))\ln \left|x+{\frac {b}{2a))\right|-{\frac {2an-bm}{a(2ax+b)))+C&{\text{(for ))4ac-b^{2}=0{\mbox{)))\end{cases))$
$\int {\frac {1}{(ax^{2}+bx+c)^{n))}\,dx={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1))}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})))\int {\frac {1}{(ax^{2}+bx+c)^{n-1))}\,dx+C$
$\int {\frac {x}{(ax^{2}+bx+c)^{n))}\,dx=-{\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1))}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})))\int {\frac {1}{(ax^{2}+bx+c)^{n-1))}\,dx+C$
$\int {\frac {1}{x(ax^{2}+bx+c)))\,dx={\frac {1}{2c))\ln \left|{\frac {x^{2)){ax^{2}+bx+c))\right|-{\frac {b}{2c))\int {\frac {1}{ax^{2}+bx+c))\,dx+C$

Integrands of the form x^m (a + b xⁿ)^p

The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
These reduction formulas can be used for integrands having integer and/or fractional exponents.

$\int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p)){m+n\,p+1))\,+\,{\frac {a\,n\,p}{m+n\,p+1))\int x^{m}\left(a+b\,x^{n}\right)^{p-1}dx$
$\int x^{m}\left(a+b\,x^{n}\right)^{p}dx=-{\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1)){a\,n(p+1)))\,+\,{\frac {m+n(p+1)+1}{a\,n(p+1)))\int x^{m}\left(a+b\,x^{n}\right)^{p+1}dx$
$\int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p)){m+1))\,-\,{\frac {b\,n\,p}{m+1))\int x^{m+n}\left(a+b\,x^{n}\right)^{p-1}dx$
$\int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1)){b\,n(p+1)))\,-\,{\frac {m-n+1}{b\,n(p+1)))\int x^{m-n}\left(a+b\,x^{n}\right)^{p+1}dx$
$\int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1)){b(m+n\,p+1)))\,-\,{\frac {a(m-n+1)}{b(m+n\,p+1)))\int x^{m-n}\left(a+b\,x^{n}\right)^{p}dx$
$\int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1)){a(m+1)))\,-\,{\frac {b(m+n(p+1)+1)}{a(m+1)))\int x^{m+n}\left(a+b\,x^{n}\right)^{p}dx$

Integrands of the form (A + B x) (a + b x)^m (c + d x)ⁿ (e + f x)^p

The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, n and p toward 0.
These reduction formulas can be used for integrands having integer and/or fractional exponents.
Special cases of these reductions formulas can be used for integrands of the form $(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}$ by setting B to 0.

${\begin{aligned}&\int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx=-{\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p+1)){b(m+1)(a\,f-b\,e)))\,+\,{\frac {1}{b(m+1)(a\,f-b\,e)))\,\cdot \\&\qquad \int (b\,c(m+1)(A\,f-B\,e)+(A\,b-a\,B)(n\,d\,e+c\,f(p+1))+d(b(m+1)(A\,f-B\,e)+f(n+p+1)(A\,b-a\,B))x)(a+b\,x)^{m+1}(c+d\,x)^{n-1}(e+f\,x)^{p}dx\end{aligned))$
${\begin{aligned}&\int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {B(a+b\,x)^{m}(c+d\,x)^{n+1}(e+f\,x)^{p+1)){d\,f(m+n+p+2)))\,+\,{\frac {1}{d\,f(m+n+p+2)))\,\cdot \\&\qquad \int (A\,a\,d\,f(m+n+p+2)-B(b\,c\,e\,m+a(d\,e(n+1)+c\,f(p+1)))+(A\,b\,d\,f(m+n+p+2)+B(a\,d\,f\,m-b(d\,e(m+n+1)+c\,f(m+p+1))))x)(a+b\,x)^{m-1}(c+d\,x)^{n}(e+f\,x)^{p}dx\end{aligned))$
${\begin{aligned}&\int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n+1}(e+f\,x)^{p+1)){(m+1)(a\,d-b\,c)(a\,f-b\,e)))\,+\,{\frac {1}{(m+1)(a\,d-b\,c)(a\,f-b\,e)))\,\cdot \\&\qquad \int ((m+1)(A(a\,d\,f-b(c\,f+d\,e))+B\,b\,c\,e)-(A\,b-a\,B)(d\,e(n+1)+c\,f(p+1))-d\,f(m+n+p+3)(A\,b-a\,B)x)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p}dx\end{aligned))$

Integrands of the form x^m (A + B xⁿ) (a + b xⁿ)^p (c + d xⁿ)^q

The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, p and q toward 0.
These reduction formulas can be used for integrands having integer and/or fractional exponents.
Special cases of these reductions formulas can be used for integrands of the form $\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}$ and $x^{m}\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}$ by setting m and/or B to 0.

${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q)){a\,b\,n(p+1)))\,+\,{\frac {1}{a\,b\,n(p+1)))\,\cdot \\&\qquad \int x^{m}\left(c(A\,b\,n(p+1)+(A\,b-a\,B)(m+1))+d(A\,b\,n(p+1)+(A\,b-a\,B)(m+n\,q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q-1}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q)){b(m+n(p+q+1)+1)))\,+\,{\frac {1}{b(m+n(p+q+1)+1)))\,\cdot \\&\qquad \int x^{m}\left(c((A\,b-a\,B)(1+m)+A\,b\,n(1+p+q))+(d(A\,b-a\,B)(1+m)+B\,n\,q(b\,c-a\,d)+A\,b\,d\,n(1+p+q))\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1)){a\,n(b\,c-a\,d)(p+1)))\,+\,{\frac {1}{a\,n(b\,c-a\,d)(p+1)))\,\cdot \\&\qquad \int x^{m}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c-a\,d)(p+1)+d(A\,b-a\,B)(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1)){b\,d(m+n(p+q+1)+1)))\,-\,{\frac {1}{b\,d(m+n(p+q+1)+1)))\,\cdot \\&\qquad \int x^{m-n}\left(a\,B\,c(m-n+1)+(a\,B\,d(m+n\,q+1)-b(-B\,c(m+n\,p+1)+A\,d(m+n(p+q+1)+1)))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1)){a\,c(m+1)))\,+\,{\frac {1}{a\,c(m+1)))\,\cdot \\&\qquad \int x^{m+n}\left(a\,B\,c(m+1)-A(b\,c+a\,d)(m+n+1)-A\,n(b\,c\,p+a\,d\,q)-A\,b\,d(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q)){a(m+1)))\,-\,{\frac {1}{a(m+1)))\,\cdot \\&\qquad \int x^{m+n}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c(p+1)+a\,d\,q)+d((A\,b-a\,B)(m+1)+A\,b\,n(p+q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {(A\,b-a\,B)x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1)){b\,n(b\,c-a\,d)(p+1)))\,-\,{\frac {1}{b\,n(b\,c-a\,d)(p+1)))\,\cdot \\&\qquad \int x^{m-n}\left(c(A\,b-a\,B)(m-n+1)+(d(A\,b-a\,B)(m+n\,q+1)-b\,n(B\,c-A\,d)(p+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx\end{aligned))$

Integrands of the form (d + e x)^m (a + b x + c x²)^p when b² − 4 a c = 0

The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
These reduction formulas can be used for integrands having integer and/or fractional exponents.
Special cases of these reductions formulas can be used for integrands of the form $\left(a+b\,x+c\,x^{2}\right)^{p}$ when $b^{2}-4\,a\,c=0$ by setting m to 0.

$\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p)){e(m+1)))\,-\,{\frac {p(d+e\,x)^{m+2}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1)){e^{2}(m+1)(m+2p+1)))\,+\,{\frac {p(2p-1)(2c\,d-b\,e)}{e^{2}(m+1)(m+2p+1)))\int (d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx$
$\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p)){e(m+1)))\,-\,{\frac {p(d+e\,x)^{m+2}(b+2\,c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1)){e^{2}(m+1)(m+2)))\,+\,{\frac {2\,c\,p\,(2\,p-1)}{e^{2}(m+1)(m+2)))\int (d+e\,x)^{m+2}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx$
$\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {e(m+2p+2)(d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1)){(p+1)(2p+1)(2c\,d-b\,e)))\,+\,{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p)){(2p+1)(2c\,d-b\,e)))\,+\,{\frac {e^{2}m(m+2p+2)}{(p+1)(2p+1)(2c\,d-b\,e)))\int (d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx$
$\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {e\,m(d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p+1)){2c(p+1)(2p+1)))\,+\,{\frac {(d+e\,x)^{m}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p)){2c(2p+1)))\,+\,{\frac {e^{2}m(m-1)}{2c(p+1)(2p+1)))\int (d+e\,x)^{m-2}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx$
$\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p)){e(m+2p+1)))\,-\,{\frac {p(2c\,d-b\,e)(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p-1)){2c\,e^{2}(m+2p)(m+2p+1)))\,+\,{\frac {p(2p-1)(2c\,d-b\,e)^{2)){2c\,e^{2}(m+2p)(m+2p+1)))\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p-1}dx$
$\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {2c\,e(m+2p+2)(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p+1)){(p+1)(2p+1)(2c\,d-b\,e)^{2))}\,+\,{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p)){(2p+1)(2c\,d-b\,e)))\,+\,{\frac {2c\,e^{2}(m+2p+2)(m+2p+3)}{(p+1)(2p+1)(2c\,d-b\,e)^{2))}\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1}dx$
$\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p)){2c(m+2p+1)))\,+\,{\frac {m(2c\,d-b\,e)}{2c(m+2p+1)))\int (d+e\,x)^{m-1}\left(a+b\,x+c\,x^{2}\right)^{p}dx$
$\int (d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {(d+e\,x)^{m+1}(b+2c\,x)\left(a+b\,x+c\,x^{2}\right)^{p)){(m+1)(2c\,d-b\,e)))\,+\,{\frac {2c(m+2p+2)}{(m+1)(2c\,d-b\,e)))\int (d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p}dx$

Integrands of the form (d + e x)^m (A + B x) (a + b x + c x²)^p

The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
These reduction formulas can be used for integrands having integer and/or fractional exponents.
Special cases of these reductions formulas can be used for integrands of the form $\left(a+b\,x+c\,x^{2}\right)^{p}$ and $(d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p}$ by setting m and/or B to 0.

${\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}(A\,e(m+2p+2)-B\,d(2p+1)+e\,B(m+1)x)\left(a+b\,x+c\,x^{2}\right)^{p)){e^{2}(m+1)(m+2p+2)))\,+\,{\frac {1}{e^{2}(m+1)(m+2p+2)))p\,\cdot \\&\qquad \int (d+e\,x)^{m+1}(B(b\,d+2a\,e+2a\,e\,m+2b\,d\,p)-A\,b\,e(m+2p+2)+(B(2c\,d+b\,e+b\,em+4c\,d\,p)-2A\,c\,e(m+2p+2))x)\left(a+b\,x+c\,x^{2}\right)^{p-1}dx\end{aligned))$
${\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m}(A\,b-2a\,B-(b\,B-2A\,c)x)\left(a+b\,x+c\,x^{2}\right)^{p+1)){(p+1)\left(b^{2}-4a\,c\right)))\,+\,{\frac {1}{(p+1)\left(b^{2}-4a\,c\right)))\,\cdot \\&\qquad \int (d+e\,x)^{m-1}(B(2a\,e\,m+b\,d(2p+3))-A(b\,e\,m+2c\,d(2p+3))+e(b\,B-2A\,c)(m+2p+3)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}dx\end{aligned))$
${\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}(A\,c\,e(m+2p+2)-B(c\,d+2c\,d\,p-b\,e\,p)+B\,c\,e(m+2p+1)x)\left(a+b\,x+c\,x^{2}\right)^{p)){c\,e^{2}(m+2p+1)(m+2p+2)))\,-\,{\frac {p}{c\,e^{2}(m+2p+1)(m+2p+2)))\,\cdot \\&\qquad \int (d+e\,x)^{m}(A\,c\,e(b\,d-2a\,e)(m+2p+2)+B(a\,e(b\,e-2c\,d\,m+b\,e\,m)+b\,d(b\,e\,p-c\,d-2c\,d\,p))+\\&\qquad \qquad \left(A\,c\,e(2c\,d-b\,e)(m+2p+2)-B\left(-b^{2}e^{2}(m+p+1)+2c^{2}d^{2}(1+2p)+c\,e(b\,d(m-2p)+2a\,e(m+2p+1))\right)\right)x)\left(a+b\,x+c\,x^{2}\right)^{p-1}dx\end{aligned))$
${\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {(d+e\,x)^{m+1}\left(A\left(b\,c\,d-b^{2}e+2a\,c\,e\right)-a\,B(2c\,d-b\,e)+c(A(2c\,d-b\,e)-B(b\,d-2a\,e))x\right)\left(a+b\,x+c\,x^{2}\right)^{p+1)){(p+1)\left(b^{2}-4a\,c\right)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)))\,+\\&\qquad {\frac {1}{(p+1)\left(b^{2}-4a\,c\right)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)))\,\cdot \\&\qquad \qquad \int (d+e\,x)^{m}(A\left(b\,c\,d\,e(2p-m+2)+b^{2}e^{2}(m+p+2)-2c^{2}d^{2}(3+2p)-2a\,c\,e^{2}(m+2p+3)\right)-\\&\qquad \qquad \qquad B(a\,e(b\,e-2c\,dm+b\,e\,m)+b\,d(-3c\,d+b\,e-2c\,d\,p+b\,e\,p))+c\,e(B(b\,d-2a\,e)-A(2c\,d-b\,e))(m+2p+4)x)\left(a+b\,x+c\,x^{2}\right)^{p+1}dx\end{aligned))$
${\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx={\frac {B(d+e\,x)^{m}\left(a+b\,x+c\,x^{2}\right)^{p+1)){c(m+2p+2)))\,+\,{\frac {1}{c(m+2p+2)))\,\cdot \\&\qquad \int (d+e\,x)^{m-1}(m(A\,c\,d-a\,B\,e)-d(b\,B-2A\,c)(p+1)+((B\,c\,d-b\,B\,e+A\,c\,e)m-e(b\,B-2A\,c)(p+1))x)\left(a+b\,x+c\,x^{2}\right)^{p}dx\end{aligned))$
${\begin{aligned}&\int (d+e\,x)^{m}(A+B\,x)\left(a+b\,x+c\,x^{2}\right)^{p}dx=-{\frac {(B\,d-A\,e)(d+e\,x)^{m+1}\left(a+b\,x+c\,x^{2}\right)^{p+1)){(m+1)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)))\,+\,{\frac {1}{(m+1)\left(c\,d^{2}-b\,d\,e+a\,e^{2}\right)))\,\cdot \\&\qquad \int (d+e\,x)^{m+1}((A\,c\,d-A\,b\,e+a\,B\,e)(m+1)+b(B\,d-A\,e)(p+1)+c(B\,d-A\,e)(m+2p+3)x)\left(a+b\,x+c\,x^{2}\right)^{p}dx\end{aligned))$

Integrands of the form x^m (a + b xⁿ + c x²ⁿ)^p when b² − 4 a c = 0

The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
These reduction formulas can be used for integrands having integer and/or fractional exponents.
Special cases of these reductions formulas can be used for integrands of the form $\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}$ when $b^{2}-4\,a\,c=0$ by setting m to 0.

$\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){m+2n\,p+1))\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1)){(m+1)(m+2n\,p+1)))\,-\,{\frac {b\,n^{2}p(2p-1)}{(m+1)(m+2n\,p+1)))\int x^{m+n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx$
$\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {(m+n(2p-1)+1)x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){(m+1)(m+n+1)))\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1)){(m+1)(m+n+1)))\,+\,{\frac {2c\,p\,n^{2}(2p-1)}{(m+1)(m+n+1)))\int x^{m+2n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx$
$\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {(m+n(2p+1)+1)x^{m-n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1)){b\,n^{2}(p+1)(2p+1)))\,-\,{\frac {x^{m+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){b\,n(2p+1)))\,-\,{\frac {(m-n+1)(m+n(2p+1)+1)}{b\,n^{2}(p+1)(2p+1)))\int x^{m-n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx$
$\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {(m-3n-2n\,p+1)x^{m-2n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1)){2c\,n^{2}(p+1)(2p+1)))\,-\,{\frac {x^{m-2n+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){2c\,n(2p+1)))\,+\,{\frac {(m-n+1)(m-2n+1)}{2c\,n^{2}(p+1)(2p+1)))\int x^{m-2n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx$
$\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){m+2n\,p+1))\,+\,{\frac {n\,p\,x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1)){(m+2n\,p+1)(m+n(2p-1)+1)))\,+\,{\frac {2a\,n^{2}p(2p-1)}{(m+2n\,p+1)(m+n(2p-1)+1)))\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx$
$\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {(m+n+2n\,p+1)x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1)){2a\,n^{2}(p+1)(2p+1)))\,-\,{\frac {x^{m+1}\left(2a+b\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){2a\,n(2p+1)))\,+\,{\frac {(m+n(2p+1)+1)(m+2n(p+1)+1)}{2a\,n^{2}(p+1)(2p+1)))\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx$
$\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m-n+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){2c(m+2n\,p+1)))\,-\,{\frac {b(m-n+1)}{2c(m+2n\,p+1)))\int x^{m-n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx$
$\int x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(b+2c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){b(m+1)))\,-\,{\frac {2c(m+n(2p+1)+1)}{b(m+1)))\int x^{m+n}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx$

Integrands of the form x^m (A + B xⁿ) (a + b xⁿ + c x²ⁿ)^p

The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
These reduction formulas can be used for integrands having integer and/or fractional exponents.
Special cases of these reductions formulas can be used for integrands of the form $\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}$ and $x^{m}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}$ by setting m and/or B to 0.

${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(A(m+n(2p+1)+1)+B(m+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){(m+1)(m+n(2p+1)+1)))\,+\,{\frac {n\,p}{(m+1)(m+n(2p+1)+1)))\,\cdot \\&\qquad \int x^{m+n}\left(2a\,B(m+1)-A\,b(m+n(2p+1)+1)+(b\,B(m+1)-2\,A\,c(m+n(2p+1)+1))x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m-n+1}\left(A\,b-2a\,B-(b\,B-2A\,c)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1)){n(p+1)\left(b^{2}-4a\,c\right)))\,+\,{\frac {1}{n(p+1)\left(b^{2}-4a\,c\right)))\,\cdot \\&\qquad \int x^{m-n}\left((m-n+1)(2a\,B-A\,b)+(m+2n(p+1)+1)(b\,B-2A\,c)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {x^{m+1}\left(b\,B\,n\,p+A\,c(m+n(2p+1)+1)+B\,c(m+2n\,p+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p)){c(m+2n\,p+1)(m+n(2p+1)+1)))\,+\,{\frac {n\,p}{c(m+2n\,p+1)(m+n(2p+1)+1)))\,\cdot \\&\qquad \int x^{m}\left(2a\,A\,c(m+n(2p+1)+1)-a\,b\,B(m+1)+\left(2a\,B\,c(m+2n\,p+1)+A\,b\,c(m+n(2p+1)+1)-b^{2}B(m+n\,p+1)\right)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p-1}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx=-{\frac {x^{m+1}\left(A\,b^{2}-a\,b\,B-2a\,A\,c+(A\,b-2a\,B)c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1)){a\,n(p+1)\left(b^{2}-4a\,c\right)))\,+\,{\frac {1}{a\,n(p+1)\left(b^{2}-4a\,c\right)))\,\cdot \\&\qquad \int x^{m}\left((m+n(p+1)+1)A\,b^{2}-a\,b\,B(m+1)-2(m+2n(p+1)+1)a\,A\,c+(m+n(2p+3)+1)(A\,b-2a\,B)c\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {B\,x^{m-n+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1)){c(m+n(2p+1)+1)))\,-\,{\frac {1}{c(m+n(2p+1)+1)))\,\cdot \\&\qquad \int x^{m-n}\left(a\,B(m-n+1)+(b\,B(m+n\,p+1)-A\,c(m+n(2p+1)+1))x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx\end{aligned))$
${\begin{aligned}&\int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}+c\,x^{2n}\right)^{p+1)){a(m+1)))\,+\,{\frac {1}{a(m+1)))\,\cdot \\&\qquad \int x^{m+n}\left(a\,B(m+1)-A\,b(m+n(p+1)+1)-A\,c(m+2n(p+1)+1)x^{n}\right)\left(a+b\,x^{n}+c\,x^{2n}\right)^{p}dx\end{aligned))$

References

^ "Reader Survey: log|x| + C", Tom Leinster, The n-category Café, March 19, 2012

v t e Lists of integrals
Rational functions Irrational functions Trigonometric functions Inverse trigonometric functions Hyperbolic functions Inverse hyperbolic functions Exponential functions Logarithmic functions Gaussian functions Definite integrals

Category