Integrand size = 25, antiderivative size = 207 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^4} \, dx=-\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}+\frac {4 e^2 \left (16-9 p+p^2\right ) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^6 (5-2 p)}-\frac {2 e \left (d^2-e^2 x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (1,-2+p,-1+p,1-\frac {e^2 x^2}{d^2}\right )}{d (2-p)} \]
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Time = 0.19 (sec) , antiderivative size = 207, 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, 396, 252, 251} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^4} \, dx=-\frac {2 e \left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{d (2-p)}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{p-3}}{5-2 p}-\frac {4 d e \left (d^2-e^2 x^2\right )^{p-3}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{p-3}}{x}+\frac {4 e^2 \left (p^2-9 p+16\right ) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^6 (5-2 p)} \]
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Rule 67
Rule 80
Rule 251
Rule 252
Rule 396
Rule 457
Rule 866
Rule 1666
Rule 1821
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p}}{x^2} \, dx \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-4+p} \left (4 d^5 e-d^4 e^2 (13-2 p) x+4 d^3 e^3 x^2-d^2 e^4 x^3\right )}{x} \, dx}{d^2} \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-4+p} \left (4 d^5 e+4 d^3 e^3 x^2\right )}{x} \, dx}{d^2}-\frac {\int \left (d^2-e^2 x^2\right )^{-4+p} \left (-d^4 e^2 (13-2 p)-d^2 e^4 x^2\right ) \, dx}{d^2} \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-\frac {\text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-4+p} \left (4 d^5 e+4 d^3 e^3 x\right )}{x} \, dx,x,x^2\right )}{2 d^2}+\frac {\left (4 d^2 e^2 \left (16-9 p+p^2\right )\right ) \int \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{5-2 p} \\ & = -\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-(2 d e) \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right )+\frac {\left (4 e^2 \left (16-9 p+p^2\right ) \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 )^{-4+p} \, dx}{d^6 (5-2 p)} \\ & = -\frac {4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac {e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}+\frac {4 e^2 \left (16-9 p+p^2\right ) 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},4-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^6 (5-2 p)}-\frac {2 e \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 )}{d (2-p)} \\ \end{align*}
Time = 0.58 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.63 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^4} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (-16 d^2 p (1+p) \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )+2^{5+p} e p x (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )+3\ 2^{2+p} e p x (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )+2^{2+p} e p x (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )+2^p e p x (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (4-p,1+p,2+p,\frac {d-e x}{2 d}\right )-32 d e (1+p) \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} x \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )\right )}{16 d^6 p (1+p) x} \]
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\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{2} \left (e x +d \right )^{4}}d x\]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{2}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^4} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{2} \left (d + e x\right )^{4}}\, dx \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{2}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^4} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^2\,{\left (d+e\,x\right )}^4} \,d x \]
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