Integrand size = 18, antiderivative size = 336 \[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\frac {d \left (a+b x^2\right )^{1+p}}{2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\left (a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right )^2 (d+e x)}-\frac {b p \left (3 a e^2+b d^2 (1+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e \left (b d^2+a e^2\right )^2}+\frac {b (1+2 p) \left (a e^2+b d^2 p\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e \left (b d^2+a e^2\right )^2}+\frac {b d p \left (3 a e^2+b d^2 (1+2 p)\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 \left (b d^2+a e^2\right )^3 (1+p)} \]
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Time = 0.26 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {849, 858, 252, 251, 771, 441, 440, 455, 70} \[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=-\frac {b p x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+1)\right ) \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e \left (a e^2+b d^2\right )^2}+\frac {b (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2+b d^2 p\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{e \left (a e^2+b d^2\right )^2}+\frac {b d p \left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 (2 p+1)\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}-\frac {\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right )}{(d+e x) \left (a e^2+b d^2\right )^2}+\frac {d \left (a+b x^2\right )^{p+1}}{2 (d+e x)^2 \left (a e^2+b d^2\right )} \]
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Rule 70
Rule 251
Rule 252
Rule 440
Rule 441
Rule 455
Rule 771
Rule 849
Rule 858
Rubi steps \begin{align*} \text {integral}& = \frac {d \left (a+b x^2\right )^{1+p}}{2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\int \frac {(-2 a e+2 b d p x) \left (a+b x^2\right )^p}{(d+e x)^2} \, dx}{2 \left (b d^2+a e^2\right )} \\ & = \frac {d \left (a+b x^2\right )^{1+p}}{2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\left (a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\int \frac {\left (2 a b d e (1-p)+2 b (1+2 p) \left (a e^2+b d^2 p\right ) x\right ) \left (a+b x^2\right )^p}{d+e x} \, dx}{2 \left (b d^2+a e^2\right )^2} \\ & = \frac {d \left (a+b x^2\right )^{1+p}}{2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\left (a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\left (b (1+2 p) \left (a e^2+b d^2 p\right )\right ) \int \left (a+b x^2\right )^p \, dx}{e \left (b d^2+a e^2\right )^2}-\frac {\left (b d p \left (3 a e^2+b d^2 (1+2 p)\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d+e x} \, dx}{e \left (b d^2+a e^2\right )^2} \\ & = \frac {d \left (a+b x^2\right )^{1+p}}{2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\left (a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right )^2 (d+e x)}-\frac {\left (b d p \left (3 a e^2+b d^2 (1+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 \left (b d^2+a e^2\right )^2}+\frac {\left (b (1+2 p) \left (a e^2+b d^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 \left (b d^2+a e^2\right )^2} \\ & = \frac {d \left (a+b x^2\right )^{1+p}}{2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\left (a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right )^2 (d+e x)}+\frac {b (1+2 p) \left (a e^2+b d^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 \left (b d^2+a e^2\right )^2}-\frac {\left (b d p \left (3 a e^2+b d^2 (1+2 p)\right )\right ) \int \frac {x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{\left (b d^2+a e^2\right )^2}-\frac {\left (b d^2 p \left (3 a e^2+b d^2 (1+2 p)\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{e \left (b d^2+a e^2\right )^2} \\ & = \frac {d \left (a+b x^2\right )^{1+p}}{2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\left (a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right )^2 (d+e x)}+\frac {b (1+2 p) \left (a e^2+b d^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 \left (b d^2+a e^2\right )^2}-\frac {\left (b d p \left (3 a e^2+b d^2 (1+2 p)\right )\right ) \text {Subst}\left (\int \frac {(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{2 \left (b d^2+a e^2\right )^2}-\frac {\left (b d^2 p \left (3 a e^2+b d^2 (1+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 \left (b d^2+a e^2\right )^2} \\ & = \frac {d \left (a+b x^2\right )^{1+p}}{2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\left (a e^2+b d^2 p\right ) \left (a+b x^2\right )^{1+p}}{\left (b d^2+a e^2\right )^2 (d+e x)}-\frac {b p \left (3 a e^2+b d^2 (1+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 \left (b d^2+a e^2\right )^2}+\frac {b (1+2 p) \left (a e^2+b d^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 \left (b d^2+a e^2\right )^2}+\frac {b d p \left (3 a e^2+b d^2 (1+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 \left (b d^2+a e^2\right )^3 (1+p)} \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.68 \[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\frac {\left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (a+b x^2\right )^p \left (2 (-1+p) (d+e x) \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )+d (1-2 p) \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )\right )}{2 e^2 (-1+p) (-1+2 p) (d+e x)^2} \]
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\[\int \frac {x \left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{3}}d x\]
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\[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x}{{\left (e x + d\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\text {Timed out} \]
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\[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x}{{\left (e x + d\right )}^{3}} \,d x } \]
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\[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} x}{{\left (e x + d\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {x \left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int \frac {x\,{\left (b\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \]
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