Integrand size = 20, antiderivative size = 321 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\frac {\left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}+\frac {d^3 \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right ) (d+e x)}+\frac {d \left (3 a e^2+b d^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^3 \left (b d^2+a e^2\right )}-\frac {d \left (2 a e^2+b d^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e^3 \left (b d^2+a e^2\right )}-\frac {d^2 \left (3 a e^2+b d^2 (3+2 p)\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 e^2 \left (b d^2+a e^2\right )^2 (1+p)} \]
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Time = 0.35 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1665, 1668, 858, 252, 251, 771, 441, 440, 455, 70} \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+3)\right ) \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^3 \left (a e^2+b d^2\right )}-\frac {d^2 \left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^2 (p+1) \left (a e^2+b d^2\right )^2}-\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a e^2+b d^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e^3 \left (a e^2+b d^2\right )}+\frac {d^3 \left (a+b x^2\right )^{p+1}}{e^2 (d+e x) \left (a e^2+b d^2\right )}+\frac {\left (a+b x^2\right )^{p+1}}{2 b e^2 (p+1)} \]
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Rule 70
Rule 251
Rule 252
Rule 440
Rule 441
Rule 455
Rule 771
Rule 858
Rule 1665
Rule 1668
Rubi steps \begin{align*} \text {integral}& = \frac {d^3 \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right ) (d+e x)}-\frac {\int \frac {\left (a+b x^2\right )^p \left (-\frac {a d^2}{e}+d \left (a+\frac {2 b d^2 (1+p)}{e^2}\right ) x-\frac {\left (b d^2+a e^2\right ) x^2}{e}\right )}{d+e x} \, dx}{b d^2+a e^2} \\ & = \frac {\left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}+\frac {d^3 \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right ) (d+e x)}-\frac {\int \frac {\left (-2 a b d^2 e (1+p)+2 b d (1+p) \left (2 a e^2+b d^2 (3+2 p)\right ) x\right ) \left (a+b x^2\right )^p}{d+e x} \, dx}{2 b e^2 \left (b d^2+a e^2\right ) (1+p)} \\ & = \frac {\left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}+\frac {d^3 \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right ) (d+e x)}-\frac {\left (d \left (2 a e^2+b d^2 (3+2 p)\right )\right ) \int \left (a+b x^2\right )^p \, dx}{e^3 \left (b d^2+a e^2\right )}+\frac {\left (d^2 \left (3 a e^2+b d^2 (3+2 p)\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d+e x} \, dx}{e^3 \left (b d^2+a e^2\right )} \\ & = \frac {\left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}+\frac {d^3 \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right ) (d+e x)}+\frac {\left (d^2 \left (3 a e^2+b d^2 (3+2 p)\right )\right ) \int \left (\frac {d \left (a+b x^2\right )^p}{d^2-e^2 x^2}+\frac {e x \left (a+b x^2\right )^p}{-d^2+e^2 x^2}\right ) \, dx}{e^3 \left (b d^2+a e^2\right )}-\frac {\left (d \left (2 a e^2+b d^2 (3+2 p)\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^2}{a}\right )^p \, dx}{e^3 \left (b d^2+a e^2\right )} \\ & = \frac {\left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}+\frac {d^3 \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right ) (d+e x)}-\frac {d \left (2 a e^2+b d^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^3 \left (b d^2+a e^2\right )}+\frac {\left (d^3 \left (3 a e^2+b d^2 (3+2 p)\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{e^3 \left (b d^2+a e^2\right )}+\frac {\left (d^2 \left (3 a e^2+b d^2 (3+2 p)\right )\right ) \int \frac {x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{e^2 \left (b d^2+a e^2\right )} \\ & = \frac {\left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}+\frac {d^3 \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right ) (d+e x)}-\frac {d \left (2 a e^2+b d^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^3 \left (b d^2+a e^2\right )}+\frac {\left (d^2 \left (3 a e^2+b d^2 (3+2 p)\right )\right ) \text {Subst}\left (\int \frac {(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{2 e^2 \left (b d^2+a e^2\right )}+\frac {\left (d^3 \left (3 a e^2+b d^2 (3+2 p)\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {\left (1+\frac {b x^2}{a}\right )^p}{d^2-e^2 x^2} \, dx}{e^3 \left (b d^2+a e^2\right )} \\ & = \frac {\left (a+b x^2\right )^{1+p}}{2 b e^2 (1+p)}+\frac {d^3 \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right ) (d+e x)}+\frac {d \left (3 a e^2+b d^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^3 \left (b d^2+a e^2\right )}-\frac {d \left (2 a e^2+b d^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^3 \left (b d^2+a e^2\right )}-\frac {d^2 \left (3 a e^2+b d^2 (3+2 p)\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 e^2 \left (b d^2+a e^2\right )^2 (1+p)} \\ \end{align*}
Time = 0.59 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.07 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\frac {\left (a+b x^2\right )^p \left (-\frac {2 d^3 \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(-1+2 p) (d+e x)}+\frac {3 d^2 \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}+e \left (\frac {e \left (a+b x^2-a \left (1+\frac {b x^2}{a}\right )^{-p}\right )}{b+b p}-4 d x \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )\right )}{2 e^4} \]
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\[\int \frac {x^{3} \left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{2}}d x\]
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\[ \int \frac {x^3 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{3}}{{\left (e x + d\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\text {Timed out} \]
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\[ \int \frac {x^3 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{3}}{{\left (e x + d\right )}^{2}} \,d x } \]
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\[ \int \frac {x^3 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x^{3}}{{\left (e x + d\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx=\int \frac {x^3\,{\left (b\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^2} \,d x \]
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