Integrand size = 25, antiderivative size = 145 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^2} \, dx=-\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{3 x^3}-\frac {2 e^2 (4-p) \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^4 x}-\frac {e^3 \left (d^2-e^2 x^2\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (2,-1+p,p,1-\frac {e^2 x^2}{d^2}\right )}{d^3 (1-p)} \]
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Time = 0.12 (sec) , antiderivative size = 145, 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, 272, 67, 372, 371} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^2} \, dx=-\frac {\left (d^2-e^2 x^2\right )^{p-1}}{3 x^3}-\frac {2 e^2 (4-p) \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^4 x}-\frac {e^3 \left (d^2-e^2 x^2\right )^{p-1} \operatorname {Hypergeometric2F1}\left (2,p-1,p,1-\frac {e^2 x^2}{d^2}\right )}{d^3 (1-p)} \]
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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^4} \, dx \\ & = -\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{3 x^3}-\frac {\int \frac {\left (6 d^3 e-2 d^2 e^2 (4-p) x\right ) \left (d^2-e^2 x^2\right )^{-2+p}}{x^3} \, dx}{3 d^2} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{3 x^3}-(2 d e) \int \frac {\left (d^2-e^2 x^2\right )^{-2+p}}{x^3} \, dx+\frac {1}{3} \left (2 e^2 (4-p)\right ) \int \frac {\left (d^2-e^2 x^2\right )^{-2+p}}{x^2} \, dx \\ & = -\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{3 x^3}-(d e) \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-2+p}}{x^2} \, dx,x,x^2\right )+\frac {\left (2 e^2 (4-p) \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}{3 d^4} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{-1+p}}{3 x^3}-\frac {2 e^2 (4-p) \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 )}{3 d^4 x}-\frac {e^3 \left (d^2-e^2 x^2\right )^{-1+p} \, _2F_1\left (2,-1+p;p;1-\frac {e^2 x^2}{d^2}\right )}{d^3 (1-p)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(334\) vs. \(2(145)=290\).
Time = 0.81 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.30 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^2} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {4 d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-p,-\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x^3}-\frac {36 d^2 e^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x}-\frac {12 d^3 e \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {d^2}{e^2 x^2}\right )}{(-1+p) x^2}+\frac {3\ 2^{3+p} e^3 (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {3\ 2^p e^3 (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}-\frac {24 d e^3 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )}{p}\right )}{12 d^6} \]
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\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{4} \left (e x +d \right )^{2}}d x\]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{2} x^{4}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^2} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{4} \left (d + e x\right )^{2}}\, dx \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{2} x^{4}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{2} x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^2} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^4\,{\left (d+e\,x\right )}^2} \,d x \]
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