Integrand size = 20, antiderivative size = 368 \[ \int \frac {\left (a+b x^2\right )^p}{x (d+e x)^2} \, dx=-\frac {e x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {e x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {e^3 x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,2,\frac {5}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{3 d^5}+\frac {e^2 \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 d^2 \left (b d^2+a e^2\right ) (1+p)}-\frac {\left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a d^2 (1+p)}+\frac {b e^2 \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{\left (b d^2+a e^2\right )^2 (1+p)} \]
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Time = 0.29 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {975, 272, 67, 771, 441, 440, 455, 70, 525, 524} \[ \int \frac {\left (a+b x^2\right )^p}{x (d+e x)^2} \, dx=-\frac {e^3 x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,2,\frac {5}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{3 d^5}-\frac {e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}+\frac {e^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^2 (p+1) \left (a e^2+b d^2\right )}+\frac {b e^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (2,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{(p+1) \left (a e^2+b d^2\right )^2}-\frac {\left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{2 a d^2 (p+1)} \]
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Rule 67
Rule 70
Rule 272
Rule 440
Rule 441
Rule 455
Rule 524
Rule 525
Rule 771
Rule 975
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (a+b x^2\right )^p}{d^2 x}-\frac {e \left (a+b x^2\right )^p}{d (d+e x)^2}-\frac {e \left (a+b x^2\right )^p}{d^2 (d+e x)}\right ) \, dx \\ & = \frac {\int \frac {\left (a+b x^2\right )^p}{x} \, dx}{d^2}-\frac {e \int \frac {\left (a+b x^2\right )^p}{d+e x} \, dx}{d^2}-\frac {e \int \frac {\left (a+b x^2\right )^p}{(d+e x)^2} \, dx}{d} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,x^2\right )}{2 d^2}-\frac {e \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}{d^2}-\frac {e \int \left (\frac {d^2 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^2}-\frac {2 d e x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^2}+\frac {e^2 x^2 \left (a+b x^2\right )^p}{\left (-d^2+e^2 x^2\right )^2}\right ) \, dx}{d} \\ & = -\frac {\left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac {b x^2}{a}\right )}{2 a d^2 (1+p)}-\frac {e \int \frac {\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{d}-(d e) \int \frac {\left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^2} \, dx+\left (2 e^2\right ) \int \frac {x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^2} \, dx-\frac {e^2 \int \frac {x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{d^2}-\frac {e^3 \int \frac {x^2 \left (a+b x^2\right )^p}{\left (-d^2+e^2 x^2\right )^2} \, dx}{d} \\ & = -\frac {\left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac {b x^2}{a}\right )}{2 a d^2 (1+p)}+e^2 \text {Subst}\left (\int \frac {(a+b x)^p}{\left (d^2-e^2 x\right )^2} \, dx,x,x^2\right )-\frac {e^2 \text {Subst}\left (\int \frac {(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{2 d^2}-\frac {\left (e \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}{d}-\left (d e \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}{\left (d^2-e^2 x^2\right )^2} \, dx-\frac {\left (e^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {x^2 \left (1+\frac {b x^2}{a}\right )^p}{\left (-d^2+e^2 x^2\right )^2} \, dx}{d} \\ & = -\frac {e 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 )}{d^3}-\frac {e x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,2;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {e^3 x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {3}{2};-p,2;\frac {5}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{3 d^5}+\frac {e^2 \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 d^2 \left (b d^2+a e^2\right ) (1+p)}-\frac {\left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac {b x^2}{a}\right )}{2 a d^2 (1+p)}+\frac {b e^2 \left (a+b x^2\right )^{1+p} \, _2F_1\left (2,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{\left (b d^2+a e^2\right )^2 (1+p)} \\ \end{align*}
Time = 0.41 (sec) , antiderivative size = 303, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b x^2\right )^p}{x (d+e x)^2} \, dx=\frac {\left (a+b x^2\right )^p \left (-\frac {2 d \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(-1+2 p) (d+e x)}+\frac {-\left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )+\left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {a}{b x^2}\right )}{p}\right )}{2 d^2} \]
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\[\int \frac {\left (b \,x^{2}+a \right )^{p}}{x \left (e x +d \right )^{2}}d x\]
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\[ \int \frac {\left (a+b x^2\right )^p}{x (d+e x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{2} x} \,d x } \]
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\[ \int \frac {\left (a+b x^2\right )^p}{x (d+e x)^2} \, dx=\int \frac {\left (a + b x^{2}\right )^{p}}{x \left (d + e x\right )^{2}}\, dx \]
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\[ \int \frac {\left (a+b x^2\right )^p}{x (d+e x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{2} x} \,d x } \]
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\[ \int \frac {\left (a+b x^2\right )^p}{x (d+e x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{2} x} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{x (d+e x)^2} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{x\,{\left (d+e\,x\right )}^2} \,d x \]
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