Integrand size = 20, antiderivative size = 449 \[ \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\frac {\left (a+b x^2\right )^{1+p}}{2 b e^3 (1+p)}-\frac {d^4 \left (a+b x^2\right )^{1+p}}{2 e^3 \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d^3 \left (4 a e^2+b d^2 (3+p)\right ) \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^4 \left (b d^2+a e^2\right )^2}-\frac {d \left (3 a^2 e^4+2 a b d^2 e^2 (5+4 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e^4 \left (b d^2+a e^2\right )^2}-\frac {d^2 \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 e^3 \left (b d^2+a e^2\right )^3 (1+p)} \]
1/2*(b*x^2+a)^(p+1)/b/e^3/(p+1)-1/2*d^4*(b*x^2+a)^(p+1)/e^3/(a*e^2+b*d^2)/ (e*x+d)^2+d^3*(4*a*e^2+b*d^2*(3+p))*(b*x^2+a)^(p+1)/e^3/(a*e^2+b*d^2)^2/(e *x+d)+d*(6*a^2*e^4+3*a*b*d^2*e^2*(4+3*p)+b^2*d^4*(2*p^2+7*p+6))*x*(b*x^2+a )^p*AppellF1(1/2,1,-p,3/2,e^2*x^2/d^2,-b*x^2/a)/e^4/(a*e^2+b*d^2)^2/((1+b* x^2/a)^p)-d*(3*a^2*e^4+2*a*b*d^2*e^2*(5+4*p)+b^2*d^4*(2*p^2+7*p+6))*x*(b*x ^2+a)^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/e^4/(a*e^2+b*d^2)^2/((1+b*x^2/ a)^p)-1/2*d^2*(6*a^2*e^4+3*a*b*d^2*e^2*(4+3*p)+b^2*d^4*(2*p^2+7*p+6))*(b*x ^2+a)^(p+1)*hypergeom([1, p+1],[2+p],e^2*(b*x^2+a)/(a*e^2+b*d^2))/e^3/(a*e ^2+b*d^2)^3/(p+1)
Time = 0.89 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.03 \[ \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\frac {\left (a+b x^2\right )^p \left (\frac {a e^2}{b+b p}+\frac {e^2 x^2}{1+p}-\frac {8 d^3 \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(-1+2 p) (d+e x)}+\frac {d^4 \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(-1+p) (d+e x)^2}+\frac {6 d^2 \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}-6 d e x \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{2 e^5} \]
((a + b*x^2)^p*((a*e^2)/(b + b*p) + (e^2*x^2)/(1 + p) - (8*d^3*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + Sqrt[-(a/b)]* e)/(d + e*x)])/((-1 + 2*p)*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt [-(a/b)] + x))/(d + e*x))^p*(d + e*x)) + (d^4*AppellF1[2 - 2*p, -p, -p, 3 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/(d + e*x)])/(( -1 + p)*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p*(d + e*x)^2) + (6*d^2*AppellF1[-2*p, -p, -p, 1 - 2*p, (d - Sqrt[- (a/b)]*e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/(d + e*x)])/(p*((e*(-Sqrt[-(a/b) ] + x))/(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p) - (6*d*e*x*Hype rgeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/(2*e^5)
Time = 1.09 (sec) , antiderivative size = 433, normalized size of antiderivative = 0.96, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {603, 27, 2182, 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.
| \(\displaystyle \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 603 |
| \(\displaystyle -\frac {\int \frac {2 \left (b x^2+a\right )^p \left (\frac {a d^3}{e^2}-\frac {\left (b (p+1) d^2+a e^2\right ) x d^2}{e^3}+\left (\frac {b d^2}{e^2}+a\right ) x^2 d-\left (\frac {b d^2}{e}+a e\right ) x^3\right )}{(d+e x)^2}dx}{2 \left (a e^2+b d^2\right )}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\left (b x^2+a\right )^p \left (\frac {a d^3}{e^2}-\frac {\left (b (p+1) d^2+a e^2\right ) x d^2}{e^3}+\left (\frac {b d^2}{e^2}+a\right ) x^2 d-\frac {\left (b d^2+a e^2\right ) x^3}{e}\right )}{(d+e x)^2}dx}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 2182 |
| \(\displaystyle -\frac {-\frac {\int \frac {\left (b x^2+a\right )^p \left (a \left (\frac {b (p+2) d^2}{e^2}+3 a\right ) d^2-\frac {\left (b^2 \left (2 p^2+7 p+5\right ) d^4+8 a b e^2 (p+1) d^2+2 a^2 e^4\right ) x d}{e^3}+\frac {\left (b d^2+a e^2\right )^2 x^2}{e^2}\right )}{d+e x}dx}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 2185 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {2 b d (p+1) \left (a d e \left (b (p+2) d^2+3 a e^2\right )-\left (b^2 \left (2 p^2+7 p+6\right ) d^4+2 a b e^2 (4 p+5) d^2+3 a^2 e^4\right ) x\right ) \left (b x^2+a\right )^p}{e (d+e x)}dx}{2 b e^2 (p+1)}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\frac {d \int \frac {\left (a d e \left (b (p+2) d^2+3 a e^2\right )-\left (b^2 \left (2 p^2+7 p+6\right ) d^4+2 a b e^2 (4 p+5) d^2+3 a^2 e^4\right ) x\right ) \left (b x^2+a\right )^p}{d+e x}dx}{e^3}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 719 |
| \(\displaystyle -\frac {-\frac {\frac {d \left (\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \int \frac {\left (b x^2+a\right )^p}{d+e x}dx}{e}-\frac {\left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \int \left (b x^2+a\right )^pdx}{e}\right )}{e^3}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 238 |
| \(\displaystyle -\frac {-\frac {\frac {d \left (\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \int \frac {\left (b x^2+a\right )^p}{d+e x}dx}{e}-\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{e}\right )}{e^3}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 237 |
| \(\displaystyle -\frac {-\frac {\frac {d \left (\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \int \frac {\left (b x^2+a\right )^p}{d+e x}dx}{e}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e}\right )}{e^3}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 504 |
| \(\displaystyle -\frac {-\frac {\frac {d \left (\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \left (d \int \frac {\left (b x^2+a\right )^p}{d^2-e^2 x^2}dx-e \int \frac {x \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx\right )}{e}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e}\right )}{e^3}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 334 |
| \(\displaystyle -\frac {-\frac {\frac {d \left (\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \left (d \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{d^2-e^2 x^2}dx-e \int \frac {x \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx\right )}{e}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e}\right )}{e^3}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 333 |
| \(\displaystyle -\frac {-\frac {\frac {d \left (\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-e \int \frac {x \left (b x^2+a\right )^p}{d^2-e^2 x^2}dx\right )}{e}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e}\right )}{e^3}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle -\frac {-\frac {\frac {d \left (\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-\frac {1}{2} e \int \frac {\left (b x^2+a\right )^p}{d^2-e^2 x^2}dx^2\right )}{e}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e}\right )}{e^3}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle -\frac {-\frac {\frac {d \left (\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-\frac {e \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )}\right )}{e}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e}\right )}{e^3}+\frac {\left (a e^2+b d^2\right )^2 \left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)}}{a e^2+b d^2}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )}}{a e^2+b d^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}\) |
-1/2*(d^4*(a + b*x^2)^(1 + p))/(e^3*(b*d^2 + a*e^2)*(d + e*x)^2) - (-((d^3 *(4*a*e^2 + b*d^2*(3 + p))*(a + b*x^2)^(1 + p))/(e^3*(b*d^2 + a*e^2)*(d + e*x))) - (((b*d^2 + a*e^2)^2*(a + b*x^2)^(1 + p))/(2*b*e^3*(1 + p)) + (d*( -(((3*a^2*e^4 + 2*a*b*d^2*e^2*(5 + 4*p) + b^2*d^4*(6 + 7*p + 2*p^2))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(e*(1 + (b*x^2)/ a)^p)) + (d*(6*a^2*e^4 + 3*a*b*d^2*e^2*(4 + 3*p) + b^2*d^4*(6 + 7*p + 2*p^ 2))*((x*(a + b*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((b*x^2)/a), (e^2*x^2)/d^ 2])/(d*(1 + (b*x^2)/a)^p) - (e*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(2*(b*d^2 + a*e^2)*(1 + p) )))/e))/e^3)/(b*d^2 + a*e^2))/(b*d^2 + a*e^2)
3.5.23.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
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])
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])
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]
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]
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]
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]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[e*R*(d + e*x)^(m + 1)*((a + b*x^2)^(p + 1)/((m + 1)*(b* d^2 + a*e^2))), x] + Simp[1/((m + 1)*(b*d^2 + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[(m + 1)*(b*d^2 + a*e^2)*Qx + b*d*R*(m + 1) - b *e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] && LtQ[m, -1]
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]))
\[\int \frac {x^{4} \left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{3}}d x\]
\[ \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{4}}{{\left (e x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\text {Timed out} \]
\[ \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{4}}{{\left (e x + d\right )}^{3}} \,d x } \]
\[ \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{4}}{{\left (e x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int \frac {x^4\,{\left (b\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \]