Integrand size = 25, antiderivative size = 143 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^2} \, dx=-\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{2 x^2}+\frac {2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{d^3 x}+\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (1,-1+p,p,1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (1-p)} \]
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Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {866, 1821, 778, 372, 371, 272, 67} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^2} \, dx=\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{p-1} \operatorname {Hypergeometric2F1}\left (1,p-1,p,1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (1-p)}-\frac {\left (d^2-e^2 x^2\right )^{p-1}}{2 x^2}+\frac {2 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{d^3 x} \]
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Rule 67
Rule 272
Rule 371
Rule 372
Rule 778
Rule 866
Rule 1821
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d-e x)^2 \left (d^2-e^2 x^2\right )^{-2+p}}{x^3} \, dx \\ & = -\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{2 x^2}-\frac {\int \frac {\left (4 d^3 e-2 d^2 e^2 (3-p) x\right ) \left (d^2-e^2 x^2\right )^{-2+p}}{x^2} \, dx}{2 d^2} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{2 x^2}-(2 d e) \int \frac {\left (d^2-e^2 x^2\right )^{-2+p}}{x^2} \, dx+\left (e^2 (3-p)\right ) \int \frac {\left (d^2-e^2 x^2\right )^{-2+p}}{x} \, dx \\ & = -\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{2 x^2}+\frac {1}{2} \left (e^2 (3-p)\right ) \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-2+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 \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^{-2+p}}{x^2} \, dx}{d^3} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{2 x^2}+\frac {2 e \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 {1}{2};\frac {e^2 x^2}{d^2}\right )}{d^3 x}+\frac {e^2 (3-p) \left (d^2-e^2 x^2\right )^{-1+p} \, _2F_1\left (1,-1+p;p;1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (1-p)} \\ \end{align*}
Time = 0.73 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.98 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^2} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (\frac {8 d^2 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x}+\frac {2 d^3 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {d^2}{e^2 x^2}\right )}{(-1+p) x^2}+\frac {3\ 2^{1+p} e^2 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {2^p e^2 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {6 d e^2 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )}{p}\right )}{4 d^5} \]
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\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{3} \left (e x +d \right )^{2}}d x\]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{2} x^{3}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^2} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{3} \left (d + e x\right )^{2}}\, dx \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{2} x^{3}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{2} x^{3}} \,d x } \]
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Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^2} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^3\,{\left (d+e\,x\right )}^2} \,d x \]
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