Integrand size = 20, antiderivative size = 178 \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)} \, dx=-\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{2},-p,1,\frac {1}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d x}-\frac {e^3 \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 d^2 \left (b d^2+a e^2\right ) (1+p)}+\frac {e \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a d^2 (1+p)} \]
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Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {973, 525, 524, 457, 88, 67, 70} \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)} \, dx=-\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{2},-p,1,\frac {1}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d x}-\frac {e^3 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^2 (p+1) \left (a e^2+b d^2\right )}+\frac {e \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{2 a d^2 (p+1)} \]
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Rule 67
Rule 70
Rule 88
Rule 457
Rule 524
Rule 525
Rule 973
Rubi steps \begin{align*} \text {integral}& = d \int \frac {\left (a+b x^2\right )^p}{x^2 \left (d^2-e^2 x^2\right )} \, dx-e \int \frac {\left (a+b x^2\right )^p}{x \left (d^2-e^2 x^2\right )} \, dx \\ & = -\left (\frac {1}{2} e \text {Subst}\left (\int \frac {(a+b x)^p}{x \left (d^2-e^2 x\right )} \, dx,x,x^2\right )\right )+\left (d \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}{x^2 \left (d^2-e^2 x^2\right )} \, dx \\ & = -\frac {\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 {1}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d x}-\frac {e \text {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,x^2\right )}{2 d^2}-\frac {e^3 \text {Subst}\left (\int \frac {(a+b x)^p}{d^2-e^2 x} \, dx,x,x^2\right )}{2 d^2} \\ & = -\frac {\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 {1}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d x}-\frac {e^3 \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 {e \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)} \\ \end{align*}
Time = 0.40 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)} \, dx=\frac {\left (a+b x^2\right )^p \left (\frac {e \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}-\frac {2 d \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}-\frac {e \left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {a}{b x^2}\right )}{p}\right )}{2 d^2} \]
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\[\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2} \left (e x +d \right )}d x\]
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\[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )} x^{2}} \,d x } \]
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\[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{x^2\,\left (d+e\,x\right )} \,d x \]
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