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:

${\displaystyle {\frac {a}{(x-b)^{n))))$, and ${\displaystyle {\frac {ax+b}{\left((x-c)^{2}+d^{2}\right)^{n))}.}$

which can then be integrated term by term.

For other types of functions, see lists of integrals.

Miscellaneous integrands

${\displaystyle \int {\frac {f'(x)}{f(x)))\,dx=\ln \left|f(x)\right|+C}$

${\displaystyle \int {\frac {1}{x^{2}+a^{2))}\,dx={\frac {1}{a))\arctan {\frac {x}{a))\,\!+C}$

${\displaystyle \int {\frac {1}{x^{2}-a^{2))}\,dx={\frac {1}{2a))\ln \left|{\frac {x-a}{x+a))\right|+C={\begin{cases}\displaystyle -{\frac {1}{a))\,\operatorname {artanh} {\frac {x}{a))+C={\frac {1}{2a))\ln {\frac {a-x}{a+x))+C&{\text{(for ))|x|<|a|{\mbox{)))\\[12pt]\displaystyle -{\frac {1}{a))\,\operatorname {arcoth} {\frac {x}{a))+C={\frac {1}{2a))\ln {\frac {x-a}{x+a))+C&{\text{(for ))|x|>|a|{\mbox{)))\end{cases))}$

${\displaystyle \int {\frac {1}{a^{2}-x^{2))}\,dx={\frac {1}{2a))\ln \left|{\frac {a+x}{a-x))\right|+C={\begin{cases}\displaystyle {\frac {1}{a))\,\operatorname {artanh} {\frac {x}{a))+C={\frac {1}{2a))\ln {\frac {a+x}{a-x))+C&{\text{(for ))|x|<|a|{\mbox{)))\\[12pt]\displaystyle {\frac {1}{a))\,\operatorname {arcoth} {\frac {x}{a))+C={\frac {1}{2a))\ln {\frac {x+a}{x-a))+C&{\text{(for ))|x|>|a|{\mbox{)))\end{cases))}$

${\displaystyle \int {\frac {dx}{x^{2^{n))+1))={\frac {1}{2^{n-1))}\sum _{k=1}^{2^{n-1))\sin \left({\frac {2k-1}{2^{n))}\pi \right)\arctan \left[\left(x-\cos \left({\frac {2k-1}{2^{n))}\pi \right)\right)\csc \left({\frac {2k-1}{2^{n))}\pi \right)\right]-{\frac {1}{2))\cos \left({\frac {2k-1}{2^{n))}\pi \right)\ln \left|x^{2}-2x\cos \left({\frac {2k-1}{2^{n))}\pi \right)+1\right|+C}$

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,

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

is usually abbreviated as

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

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

${\displaystyle \int (ax+b)^{n}\,dx={\frac {(ax+b)^{n+1)){a(n+1)))+C\qquad {\text{(for ))n\neq -1{\mbox{)))}$ (Cavalieri's quadrature formula)

${\displaystyle \int {\frac {x}{ax+b))\,dx={\frac {x}{a))-{\frac {b}{a^{2))}\ln \left|ax+b\right|+C}$

${\displaystyle \int {\frac {mx+n}{ax+b))\,dx={\frac {m}{a))x+{\frac {an-bm}{a^{2))}\ln \left|ax+b\right|+C}$

${\displaystyle \int {\frac {x}{(ax+b)^{2))}\,dx={\frac {b}{a^{2}(ax+b)))+{\frac {1}{a^{2))}\ln \left|ax+b\right|+C}$

${\displaystyle \int {\frac {x}{(ax+b)^{n))}\,dx={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1))}+C\qquad {\text{(for ))n\not \in \{1,2\}{\mbox{)))}$

${\displaystyle \int x(ax+b)^{n}\,dx={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)))(ax+b)^{n+1}+C\qquad {\text{(for ))n\not \in \{-1,-2\}{\mbox{)))}$

${\displaystyle \int {\frac {x^{2)){ax+b))\,dx={\frac {b^{2}\ln(\left|ax+b\right|)}{a^{3))}+{\frac {ax^{2}-2bx}{2a^{2))}+C}$

${\displaystyle \int {\frac {x^{2)){(ax+b)^{2))}\,dx={\frac {1}{a^{3))}\left(ax-2b\ln \left|ax+b\right|-{\frac {b^{2)){ax+b))\right)+C}$

${\displaystyle \int {\frac {x^{2)){(ax+b)^{3))}\,dx={\frac {1}{a^{3))}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b))-{\frac {b^{2)){2(ax+b)^{2))}\right)+C}$

${\displaystyle \int {\frac {x^{2)){(ax+b)^{n))}\,dx={\frac {1}{a^{3))}\left(-{\frac {(ax+b)^{3-n)){(n-3)))+{\frac {2b(ax+b)^{2-n)){(n-2)))-{\frac {b^{2}(ax+b)^{1-n)){(n-1)))\right)+C\qquad {\text{(for ))n\not \in \{1,2,3\}{\mbox{)))}$

${\displaystyle \int {\frac {1}{x(ax+b)))\,dx=-{\frac {1}{b))\ln \left|{\frac {ax+b}{x))\right|+C}$

${\displaystyle \int {\frac {1}{x^{2}(ax+b)))\,dx=-{\frac {1}{bx))+{\frac {a}{b^{2))}\ln \left|{\frac {ax+b}{x))\right|+C}$

${\displaystyle \int {\frac {1}{x^{2}(ax+b)^{2))}\,dx=-a\left({\frac {1}{b^{2}(ax+b)))+{\frac {1}{ab^{2}x))-{\frac {2}{b^{3))}\ln \left|{\frac {ax+b}{x))\right|\right)+C}$

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

For ${\displaystyle a\neq 0:}$

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

${\displaystyle \int {\frac {x}{ax^{2}+bx+c))\,dx={\frac {1}{2a))\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a))\int {\frac {dx}{ax^{2}+bx+c))+C}$

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

${\displaystyle \int {\frac {1}{(ax^{2}+bx+c)^{n))}\,dx={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1))}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})))\int {\frac {1}{(ax^{2}+bx+c)^{n-1))}\,dx+C}$

${\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n))}\,dx=-{\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1))}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})))\int {\frac {1}{(ax^{2}+bx+c)^{n-1))}\,dx+C}$

${\displaystyle \int {\frac {1}{x(ax^{2}+bx+c)))\,dx={\frac {1}{2c))\ln \left|{\frac {x^{2)){ax^{2}+bx+c))\right|-{\frac {b}{2c))\int {\frac {1}{ax^{2}+bx+c))\,dx+C}$

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.

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

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

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

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

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

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

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 ${\displaystyle (a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p))$ by setting B to 0.

${\displaystyle \int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx=-{\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p+1)){b(m+1)(a\,f-b\,e)))\,+\,{\frac {1}{b(m+1)(a\,f-b\,e)))\,\cdot }$

${\displaystyle \int (b\,c(m+1)(A\,f-B\,e)+(A\,b-a\,B)(n\,d\,e+c\,f(p+1))+d(b(m+1)(A\,f-B\,e)+f(n+p+1)(A\,b-a\,B))x)(a+b\,x)^{m+1}(c+d\,x)^{n-1}(e+f\,x)^{p}dx}$

${\displaystyle \int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {B(a+b\,x)^{m}(c+d\,x)^{n+1}(e+f\,x)^{p+1)){d\,f(m+n+p+2)))\,+\,{\frac {1}{d\,f(m+n+p+2)))\,\cdot }$

${\displaystyle \int (A\,a\,d\,f(m+n+p+2)-B(b\,c\,e\,m+a(d\,e(n+1)+c\,f(p+1)))+(A\,b\,d\,f(m+n+p+2)+B(a\,d\,f\,m-b(d\,e(m+n+1)+c\,f(m+p+1))))x)(a+b\,x)^{m-1}(c+d\,x)^{n}(e+f\,x)^{p}dx}$

${\displaystyle \int (A+B\,x)(a+b\,x)^{m}(c+d\,x)^{n}(e+f\,x)^{p}dx={\frac {(A\,b-a\,B)(a+b\,x)^{m+1}(c+d\,x)^{n+1}(e+f\,x)^{p+1)){(m+1)(a\,d-b\,c)(a\,f-b\,e)))\,+\,{\frac {1}{(m+1)(a\,d-b\,c)(a\,f-b\,e)))\,\cdot }$

${\displaystyle \int ((m+1)(A(a\,d\,f-b(c\,f+d\,e))+B\,b\,c\,e)-(A\,b-a\,B)(d\,e(n+1)+c\,f(p+1))-d\,f(m+n+p+3)(A\,b-a\,B)x)(a+b\,x)^{m+1}(c+d\,x)^{n}(e+f\,x)^{p}dx}$

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 ${\displaystyle \left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q))$ and ${\displaystyle x^{m}\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q))$ by setting m and/or B to 0.

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

${\displaystyle \int x^{m}\left(c(A\,b\,n(p+1)+(A\,b-a\,B)(m+1))+d(A\,b\,n(p+1)+(A\,b-a\,B)(m+n\,q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q-1}dx}$

${\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q)){b(m+n(p+q+1)+1)))\,+\,{\frac {1}{b(m+n(p+q+1)+1)))\,\cdot }$

${\displaystyle \int x^{m}\left(c((A\,b-a\,B)(1+m)+A\,b\,n(1+p+q))+(d(A\,b-a\,B)(1+m)+B\,n\,q(b\,c-a\,d)+A\,b\,d\,n(1+p+q))\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx}$

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

${\displaystyle \int x^{m}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c-a\,d)(p+1)+d(A\,b-a\,B)(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx}$

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

${\displaystyle \int x^{m-n}\left(a\,B\,c(m-n+1)+(a\,B\,d(m+n\,q+1)-b(-B\,c(m+n\,p+1)+A\,d(m+n(p+q+1)+1)))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx}$

${\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1)){a\,c(m+1)))\,+\,{\frac {1}{a\,c(m+1)))\,\cdot }$

${\displaystyle \int x^{m+n}\left(a\,B\,c(m+1)-A(b\,c+a\,d)(m+n+1)-A\,n(b\,c\,p+a\,d\,q)-A\,b\,d(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx}$

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

${\displaystyle \int x^{m+n}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c(p+1)+a\,d\,q)+d((A\,b-a\,B)(m+1)+A\,b\,n(p+q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx}$

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

${\displaystyle \int x^{m-n}\left(c(A\,b-a\,B)(m-n+1)+(d(A\,b-a\,B)(m+n\,q+1)-b\,n(B\,c-A\,d)(p+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx}$

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 ${\displaystyle \left(a+b\,x+c\,x^{2}\right)^{p))$ when ${\displaystyle b^{2}-4\,a\,c=0}$ by setting m to 0.

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

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

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

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

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

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