Integrand size = 25, antiderivative size = 166 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^3} \, dx=-\frac {2 e \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}+\frac {2 e^2 (4-p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (1,-1+p,p,1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (1-p)} \]
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Time = 0.15 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {866, 1821, 1666, 457, 80, 67, 12, 252, 251} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^3} \, dx=-\frac {3 e \left (d^2-e^2 x^2\right )^{p-1} \operatorname {Hypergeometric2F1}\left (1,p-1,p,1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (1-p)}-\frac {2 e \left (d^2-e^2 x^2\right )^{p-2}}{2-p}-\frac {d \left (d^2-e^2 x^2\right )^{p-2}}{x}+\frac {2 e^2 (4-p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^5} \]
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Rule 12
Rule 67
Rule 80
Rule 251
Rule 252
Rule 457
Rule 866
Rule 1666
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^2} \, dx \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-3+p} \left (3 d^4 e-2 d^3 e^2 (4-p) x+d^2 e^3 x^2\right )}{x} \, dx}{d^2} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {\int -2 d^3 e^2 (4-p) \left (d^2-e^2 x^2\right )^{-3+p} \, dx}{d^2}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-3+p} \left (3 d^4 e+d^2 e^3 x^2\right )}{x} \, dx}{d^2} \\ & = -\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {\text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-3+p} \left (3 d^4 e+d^2 e^3 x\right )}{x} \, dx,x,x^2\right )}{2 d^2}+\left (2 d e^2 (4-p)\right ) \int \left (d^2-e^2 x^2\right )^{-3+p} \, dx \\ & = -\frac {2 e \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {1}{2} (3 e) \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-2+p}}{x} \, 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 \left (1-\frac {e^2 x^2}{d^2}\right )^{-3+p} \, dx}{d^5} \\ & = -\frac {2 e \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}+\frac {2 e^2 (4-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^5}-\frac {3 e \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 = 0.51 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.69 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^3} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {8 d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x}+\frac {3\ 2^{2+p} e (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {2^{2+p} e (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {2^p e (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}-\frac {12 d e \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )}{p}\right )}{8 d^5} \]
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\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{2} \left (e x +d \right )^{3}}d x\]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{2}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^3} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{2} \left (d + e x\right )^{3}}\, dx \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{2}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^3} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^3} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^2\,{\left (d+e\,x\right )}^3} \,d x \]
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