Integrand size = 25, antiderivative size = 173 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {2 e^3 (8-3 p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^6}+\frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (1,-2+p,-1+p,1-\frac {e^2 x^2}{d^2}\right )}{2 d (2-p)} \]
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Time = 0.18 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {866, 1821, 778, 272, 67, 252, 251} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=\frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{2 d (2-p)}+\frac {3 e \left (d^2-e^2 x^2\right )^{p-2}}{x}-\frac {d \left (d^2-e^2 x^2\right )^{p-2}}{2 x^2}-\frac {2 e^3 (8-3 p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^6} \]
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Rule 67
Rule 251
Rule 252
Rule 272
Rule 778
Rule 866
Rule 1821
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p}}{x^3} \, dx \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-3+p} \left (6 d^4 e-2 d^3 e^2 (6-p) x+2 d^2 e^3 x^2\right )}{x^2} \, dx}{2 d^2} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}+\frac {\int \frac {\left (2 d^5 e^2 (6-p)-4 d^4 e^3 (8-3 p) x\right ) \left (d^2-e^2 x^2\right )^{-3+p}}{x} \, dx}{2 d^4} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\left (2 e^3 (8-3 p)\right ) \int \left (d^2-e^2 x^2\right )^{-3+p} \, dx+\left (d e^2 (6-p)\right ) \int \frac {\left (d^2-e^2 x^2\right )^{-3+p}}{x} \, dx \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}+\frac {1}{2} \left (d e^2 (6-p)\right ) \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right )-\frac {\left (2 e^3 (8-3 p) \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 )^{-3+p} \, dx}{d^6} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac {3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {2 e^3 (8-3 p) 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},3-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^6}+\frac {e^2 (6-p) \left (d^2-e^2 x^2\right )^{-2+p} \, _2F_1\left (1,-2+p;-1+p;1-\frac {e^2 x^2}{d^2}\right )}{2 d (2-p)} \\ \end{align*}
Time = 0.67 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.97 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (\frac {24 d^2 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x}+\frac {4 d^3 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {d^2}{e^2 x^2}\right )}{(-1+p) x^2}+\frac {3\ 2^{3+p} e^2 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {3\ 2^{1+p} e^2 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {2^p e^2 (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {24 d e^2 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )}{p}\right )}{8 d^6} \]
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\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{3} \left (e x +d \right )^{3}}d x\]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{3}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{3} \left (d + e x\right )^{3}}\, dx \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{3}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{3}} \,d x } \]
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Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^3} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^3\,{\left (d+e\,x\right )}^3} \,d x \]
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