Integrand size = 25, antiderivative size = 128 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^2} \, dx=\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{1-p}-\frac {2 e x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},2-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {\left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (1,p,1+p,1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 p} \]
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Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {866, 1666, 457, 80, 67, 12, 252, 251} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^2} \, dx=-\frac {\left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (1,p,p+1,1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 p}+\frac {\left (d^2-e^2 x^2\right )^{p-1}}{1-p}-\frac {2 e x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},2-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^3} \]
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Rule 12
Rule 67
Rule 80
Rule 251
Rule 252
Rule 457
Rule 866
Rule 1666
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{-2+p}}{x} \, dx \\ & = \int -2 d e \left (d^2-e^2 x^2\right )^{-2+p} \, dx+\int \frac {\left (d^2-e^2 x^2\right )^{-2+p} \left (d^2+e^2 x^2\right )}{x} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-2+p} \left (d^2+e^2 x\right )}{x} \, dx,x,x^2\right )-(2 d e) \int \left (d^2-e^2 x^2\right )^{-2+p} \, dx \\ & = \frac {\left (d^2-e^2 x^2\right )^{-1+p}}{1-p}+\frac {1}{2} \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-1+p}}{x} \, dx,x,x^2\right )-\frac {\left (2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p}\right ) \int \left (1-\frac {e^2 x^2}{d^2}\right )^{-2+p} \, dx}{d^3} \\ & = \frac {\left (d^2-e^2 x^2\right )^{-1+p}}{1-p}-\frac {2 e x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {1}{2},2-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {\left (d^2-e^2 x^2\right )^p \, _2F_1\left (1,p;1+p;1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 p} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.57 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^2} \, dx=\frac {2^{-2+p} \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (2 p \left (1-\frac {d^2}{e^2 x^2}\right )^p (d-e x) \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )+p \left (1-\frac {d^2}{e^2 x^2}\right )^p (d-e x) \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )+2 d (1+p) \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )\right )}{d^3 p (1+p)} \]
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\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x \left (e x +d \right )^{2}}d x\]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{2} x} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^2} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x \left (d + e x\right )^{2}}\, dx \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{2} x} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{2} x} \,d x } \]
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Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^2} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x\,{\left (d+e\,x\right )}^2} \,d x \]
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