\(\int \frac {x^5 (a+b x^2)^p}{(c+d x)^2} \, dx\) [39]

Optimal result

Integrand size = 20, antiderivative size = 324 \[ \int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=-\frac {\left (a^2 d^4+2 a b c^2 d^2 p-b^2 c^4 (5+2 p)\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 d^4 \left (b c^2+a d^2\right ) (1+p)}+\frac {\left (a d^2+b c^2 (5+2 p)\right ) \left (a+b x^2\right )^{2+p}}{2 b^2 d^2 \left (b c^2+a d^2\right ) (2+p)}+\frac {c^2 x^4 \left (a+b x^2\right )^{1+p}}{\left (b c^2+a d^2\right ) \left (c^2-d^2 x^2\right )}-\frac {2 d x^7 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {7}{2},-p,2,\frac {9}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{7 c^3}-\frac {c^4 \left (5 a d^2+b c^2 (5+2 p)\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 d^4 \left (b c^2+a d^2\right )^2 (1+p)} \] Output:

-1/2*(a^2*d^4+2*a*b*c^2*d^2*p-b^2*c^4*(5+2*p))*(b*x^2+a)^(p+1)/b^2/d^4/(a* 
d^2+b*c^2)/(p+1)+1/2*(a*d^2+b*c^2*(5+2*p))*(b*x^2+a)^(2+p)/b^2/d^2/(a*d^2+ 
b*c^2)/(2+p)+c^2*x^4*(b*x^2+a)^(p+1)/(a*d^2+b*c^2)/(-d^2*x^2+c^2)-2/7*d*x^ 
7*(b*x^2+a)^p*AppellF1(7/2,2,-p,9/2,d^2*x^2/c^2,-b*x^2/a)/c^3/((1+b*x^2/a) 
^p)-1/2*c^4*(5*a*d^2+b*c^2*(5+2*p))*(b*x^2+a)^(p+1)*hypergeom([1, p+1],[2+ 
p],d^2*(b*x^2+a)/(a*d^2+b*c^2))/d^4/(a*d^2+b*c^2)^2/(p+1)
 

Mathematica [F]

\[ \int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx \] Input:

Integrate[(x^5*(a + b*x^2)^p)/(c + d*x)^2,x]
 

Output:

Integrate[(x^5*(a + b*x^2)^p)/(c + d*x)^2, x]
 

Rubi [A] (warning: unable to verify)

Time = 2.69 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.49, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {603, 25, 2185, 27, 2185, 25, 2185, 27, 719, 238, 237, 504, 334, 333, 353, 78}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx\)

\(\Big \downarrow \) 603

\(\displaystyle \frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}-\frac {\int -\frac {\left (b x^2+a\right )^p \left (\frac {a c^4}{d^3}-\frac {\left (2 b (p+1) c^2+a d^2\right ) x c^3}{d^4}+\frac {\left (b c^2+a d^2\right ) x^2 c^2}{d^3}-\left (\frac {b c^2}{d^2}+a\right ) x^3 c+\frac {\left (b c^2+a d^2\right ) x^4}{d}\right )}{c+d x}dx}{a d^2+b c^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\left (b x^2+a\right )^p \left (\frac {a c^4}{d^3}-\frac {\left (2 b (p+1) c^2+a d^2\right ) x c^3}{d^4}+\frac {\left (b c^2+a d^2\right ) x^2 c^2}{d^3}-\left (\frac {b c^2}{d^2}+a\right ) x^3 c+\frac {\left (b c^2+a d^2\right ) x^4}{d}\right )}{c+d x}dx}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {\int -\frac {2 \left (b x^2+a\right )^p \left (b c d^2 \left (b c^2+a d^2\right ) (4 p+7) x^3+d \left (b c^2+a d^2\right ) \left (2 b (p+1) c^2+a d^2\right ) x^2+c \left (b^2 \left (2 p^2+7 p+5\right ) c^4+a b d^2 (2 p+5) c^2+2 a^2 d^4\right ) x+a c^2 d \left (a d^2-b c^2 (p+1)\right )\right )}{c+d x}dx}{2 b d^4 (p+2)}+\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\int \frac {\left (b x^2+a\right )^p \left (b c d^2 \left (b c^2+a d^2\right ) (4 p+7) x^3+d \left (b c^2+a d^2\right ) \left (2 b (p+1) c^2+a d^2\right ) x^2+c \left (b^2 \left (2 p^2+7 p+5\right ) c^4+a b d^2 (2 p+5) c^2+2 a^2 d^4\right ) x+a c^2 d \left (a d^2-b c^2 (p+1)\right )\right )}{c+d x}dx}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {\int -\frac {\left (b x^2+a\right )^p \left (-b \left (b c^2+a d^2\right ) \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right ) x^2 d^4+a b c^2 (p+2) \left (b (2 p+5) c^2+2 a d^2\right ) d^4+b c \left (-b^2 \left (4 p^3+12 p^2+9 p+1\right ) c^4+2 a b d^2 \left (2 p^2+5 p+3\right ) c^2+a^2 d^4\right ) x d^3\right )}{c+d x}dx}{b d^3 (2 p+3)}+\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {\int \frac {\left (b x^2+a\right )^p \left (-b \left (b c^2+a d^2\right ) \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right ) x^2 d^4+a b c^2 (p+2) \left (b (2 p+5) c^2+2 a d^2\right ) d^4+b c \left (-b^2 \left (4 p^3+12 p^2+9 p+1\right ) c^4+2 a b d^2 \left (2 p^2+5 p+3\right ) c^2+a^2 d^4\right ) x d^3\right )}{c+d x}dx}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {\frac {\int \frac {2 b^2 c d^5 \left (p^2+3 p+2\right ) \left (a c d \left (b (2 p+5) c^2+2 a d^2\right )+\left (-b^2 \left (4 p^2+16 p+15\right ) c^4-2 a b d^2 (4 p+5) c^2+2 a^2 d^4\right ) x\right ) \left (b x^2+a\right )^p}{c+d x}dx}{2 b d^2 (p+1)}-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {b c d^3 (p+2) \int \frac {\left (a c d \left (b (2 p+5) c^2+2 a d^2\right )+\left (-b^2 \left (4 p^2+16 p+15\right ) c^4-2 a b d^2 (4 p+5) c^2+2 a^2 d^4\right ) x\right ) \left (b x^2+a\right )^p}{c+d x}dx-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {b c d^3 (p+2) \left (\frac {\left (2 a^2 d^4-2 a b c^2 d^2 (4 p+5)-b^2 c^4 \left (4 p^2+16 p+15\right )\right ) \int \left (b x^2+a\right )^pdx}{d}+\frac {b c^3 (2 p+3) \left (5 a d^2+b c^2 (2 p+5)\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}\right )-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {b c d^3 (p+2) \left (\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a^2 d^4-2 a b c^2 d^2 (4 p+5)-b^2 c^4 \left (4 p^2+16 p+15\right )\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{d}+\frac {b c^3 (2 p+3) \left (5 a d^2+b c^2 (2 p+5)\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}\right )-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {b c d^3 (p+2) \left (\frac {b c^3 (2 p+3) \left (5 a d^2+b c^2 (2 p+5)\right ) \int \frac {\left (b x^2+a\right )^p}{c+d x}dx}{d}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a^2 d^4-2 a b c^2 d^2 (4 p+5)-b^2 c^4 \left (4 p^2+16 p+15\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 504

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {b c d^3 (p+2) \left (\frac {b c^3 (2 p+3) \left (5 a d^2+b c^2 (2 p+5)\right ) \left (c \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a^2 d^4-2 a b c^2 d^2 (4 p+5)-b^2 c^4 \left (4 p^2+16 p+15\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {b c d^3 (p+2) \left (\frac {b c^3 (2 p+3) \left (5 a d^2+b c^2 (2 p+5)\right ) \left (c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{c^2-d^2 x^2}dx-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a^2 d^4-2 a b c^2 d^2 (4 p+5)-b^2 c^4 \left (4 p^2+16 p+15\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {b c d^3 (p+2) \left (\frac {b c^3 (2 p+3) \left (5 a d^2+b c^2 (2 p+5)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a^2 d^4-2 a b c^2 d^2 (4 p+5)-b^2 c^4 \left (4 p^2+16 p+15\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {b c d^3 (p+2) \left (\frac {b c^3 (2 p+3) \left (5 a d^2+b c^2 (2 p+5)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {1}{2} d \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx^2\right )}{d}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a^2 d^4-2 a b c^2 d^2 (4 p+5)-b^2 c^4 \left (4 p^2+16 p+15\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}\right )-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\frac {(c+d x)^2 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 b d^4 (p+2)}-\frac {\frac {c (4 p+7) (c+d x) \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{2 p+3}-\frac {b c d^3 (p+2) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a^2 d^4-2 a b c^2 d^2 (4 p+5)-b^2 c^4 \left (4 p^2+16 p+15\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}+\frac {b c^3 (2 p+3) \left (5 a d^2+b c^2 (2 p+5)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {d \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{2 (p+1) \left (a d^2+b c^2\right )}\right )}{d}\right )-\frac {d^3 \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1} \left (a d^2 (2 p+3)-b c^2 \left (12 p^2+38 p+29\right )\right )}{2 (p+1)}}{b d^3 (2 p+3)}}{b d^4 (p+2)}}{a d^2+b c^2}+\frac {c^5 \left (a+b x^2\right )^{p+1}}{d^4 (c+d x) \left (a d^2+b c^2\right )}\)

Input:

Int[(x^5*(a + b*x^2)^p)/(c + d*x)^2,x]
 

Output:

(c^5*(a + b*x^2)^(1 + p))/(d^4*(b*c^2 + a*d^2)*(c + d*x)) + (((b*c^2 + a*d 
^2)*(c + d*x)^2*(a + b*x^2)^(1 + p))/(2*b*d^4*(2 + p)) - ((c*(b*c^2 + a*d^ 
2)*(7 + 4*p)*(c + d*x)*(a + b*x^2)^(1 + p))/(3 + 2*p) - (-1/2*(d^3*(b*c^2 
+ a*d^2)*(a*d^2*(3 + 2*p) - b*c^2*(29 + 38*p + 12*p^2))*(a + b*x^2)^(1 + p 
))/(1 + p) + b*c*d^3*(2 + p)*(((2*a^2*d^4 - 2*a*b*c^2*d^2*(5 + 4*p) - b^2* 
c^4*(15 + 16*p + 4*p^2))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, - 
((b*x^2)/a)])/(d*(1 + (b*x^2)/a)^p) + (b*c^3*(3 + 2*p)*(5*a*d^2 + b*c^2*(5 
 + 2*p))*((x*(a + b*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((b*x^2)/a), (d^2*x^ 
2)/c^2])/(c*(1 + (b*x^2)/a)^p) - (d*(a + b*x^2)^(1 + p)*Hypergeometric2F1[ 
1, 1 + p, 2 + p, (d^2*(a + b*x^2))/(b*c^2 + a*d^2)])/(2*(b*c^2 + a*d^2)*(1 
 + p))))/d))/(b*d^3*(3 + 2*p)))/(b*d^4*(2 + p)))/(b*c^2 + a*d^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 603

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[x^m, c + d*x, x], R = PolynomialRemainde 
r[x^m, c + d*x, x]}, Simp[d*R*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 
1)*(b*c^2 + a*d^2))), x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x) 
^(n + 1)*(a + b*x^2)^p*ExpandToSum[(n + 1)*(b*c^2 + a*d^2)*Qx + b*c*R*(n + 
1) - b*d*R*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IGt 
Q[m, 1] && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [F]

\[\int \frac {x^{5} \left (b \,x^{2}+a \right )^{p}}{\left (d x +c \right )^{2}}d x\]

Input:

int(x^5*(b*x^2+a)^p/(d*x+c)^2,x)
 

Output:

int(x^5*(b*x^2+a)^p/(d*x+c)^2,x)
 

Fricas [F]

\[ \int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{5}}{{\left (d x + c\right )}^{2}} \,d x } \] Input:

integrate(x^5*(b*x^2+a)^p/(d*x+c)^2,x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^p*x^5/(d^2*x^2 + 2*c*d*x + c^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\text {Timed out} \] Input:

integrate(x**5*(b*x**2+a)**p/(d*x+c)**2,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{5}}{{\left (d x + c\right )}^{2}} \,d x } \] Input:

integrate(x^5*(b*x^2+a)^p/(d*x+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^p*x^5/(d*x + c)^2, x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^5*(b*x^2+a)^p/(d*x+c)^2,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-1,[0,5,1,0]%%%} / %%%{1,[0,0,0,5]%%%} Error: Bad Argument 
 Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\int \frac {x^5\,{\left (b\,x^2+a\right )}^p}{{\left (c+d\,x\right )}^2} \,d x \] Input:

int((x^5*(a + b*x^2)^p)/(c + d*x)^2,x)
 

Output:

int((x^5*(a + b*x^2)^p)/(c + d*x)^2, x)
 

Reduce [F]

\[ \int \frac {x^5 \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\text {too large to display} \] Input:

int(x^5*(b*x^2+a)^p/(d*x+c)^2,x)
 

Output:

(8*(a + b*x**2)**p*a**2*c*d**2*p**4 + 36*(a + b*x**2)**p*a**2*c*d**2*p**3 
+ 42*(a + b*x**2)**p*a**2*c*d**2*p**2 + 13*(a + b*x**2)**p*a**2*c*d**2*p - 
 16*(a + b*x**2)**p*a**2*d**3*p**4*x - 36*(a + b*x**2)**p*a**2*d**3*p**3*x 
 - 6*(a + b*x**2)**p*a**2*d**3*p**2*x + 13*(a + b*x**2)**p*a**2*d**3*p*x + 
 8*(a + b*x**2)**p*a*b*c**3*p**4 + 52*(a + b*x**2)**p*a*b*c**3*p**3 + 118* 
(a + b*x**2)**p*a*b*c**3*p**2 + 107*(a + b*x**2)**p*a*b*c**3*p + 30*(a + b 
*x**2)**p*a*b*c**3 - 16*(a + b*x**2)**p*a*b*c**2*d*p**4*x - 80*(a + b*x**2 
)**p*a*b*c**2*d*p**3*x - 110*(a + b*x**2)**p*a*b*c**2*d*p**2*x - 13*(a + b 
*x**2)**p*a*b*c**2*d*p*x + 30*(a + b*x**2)**p*a*b*c**2*d*x - 8*(a + b*x**2 
)**p*a*b*c*d**2*p**5*x**2 - 36*(a + b*x**2)**p*a*b*c*d**2*p**4*x**2 - 42*( 
a + b*x**2)**p*a*b*c*d**2*p**3*x**2 - 13*(a + b*x**2)**p*a*b*c*d**2*p**2*x 
**2 + 8*(a + b*x**2)**p*a*b*d**3*p**5*x**3 + 20*(a + b*x**2)**p*a*b*d**3*p 
**4*x**3 + 14*(a + b*x**2)**p*a*b*d**3*p**3*x**3 + 3*(a + b*x**2)**p*a*b*d 
**3*p**2*x**3 - 8*(a + b*x**2)**p*b**2*c**3*p**5*x**2 - 52*(a + b*x**2)**p 
*b**2*c**3*p**4*x**2 - 118*(a + b*x**2)**p*b**2*c**3*p**3*x**2 - 107*(a + 
b*x**2)**p*b**2*c**3*p**2*x**2 - 30*(a + b*x**2)**p*b**2*c**3*p*x**2 + 8*( 
a + b*x**2)**p*b**2*c**2*d*p**5*x**3 + 44*(a + b*x**2)**p*b**2*c**2*d*p**4 
*x**3 + 78*(a + b*x**2)**p*b**2*c**2*d*p**3*x**3 + 49*(a + b*x**2)**p*b**2 
*c**2*d*p**2*x**3 + 10*(a + b*x**2)**p*b**2*c**2*d*p*x**3 - 8*(a + b*x**2) 
**p*b**2*c*d**2*p**5*x**4 - 36*(a + b*x**2)**p*b**2*c*d**2*p**4*x**4 - ...