Integrand size = 25, antiderivative size = 204 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^4} \, dx=\frac {4 d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac {4 d e x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-\frac {\left (d^2-e^2 x^2\right )^{-2+p}}{2 (2-p)}-\frac {8 e (2-p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^5 (5-2 p)}+\frac {\left (d^2-e^2 x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (1,-2+p,-1+p,1-\frac {e^2 x^2}{d^2}\right )}{2 (2-p)} \]
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Time = 0.14 (sec) , antiderivative size = 204, 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, 1666, 1265, 965, 80, 67, 396, 252, 251} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^4} \, dx=\frac {\left (d^2-e^2 x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (1,p-2,p-1,1-\frac {e^2 x^2}{d^2}\right )}{2 (2-p)}-\frac {4 d e x \left (d^2-e^2 x^2\right )^{p-3}}{5-2 p}+\frac {4 d^2 \left (d^2-e^2 x^2\right )^{p-3}}{3-p}-\frac {\left (d^2-e^2 x^2\right )^{p-2}}{2 (2-p)}-\frac {8 e (2-p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{d^5 (5-2 p)} \]
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Rule 67
Rule 80
Rule 251
Rule 252
Rule 396
Rule 866
Rule 965
Rule 1265
Rule 1666
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p}}{x} \, dx \\ & = \int \left (d^2-e^2 x^2\right )^{-4+p} \left (-4 d^3 e-4 d e^3 x^2\right ) \, dx+\int \frac {\left (d^2-e^2 x^2\right )^{-4+p} \left (d^4+6 d^2 e^2 x^2+e^4 x^4\right )}{x} \, dx \\ & = -\frac {4 d e x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}+\frac {1}{2} \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-4+p} \left (d^4+6 d^2 e^2 x+e^4 x^2\right )}{x} \, dx,x,x^2\right )-\frac {\left (8 d^3 e (2-p)\right ) \int \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{5-2 p} \\ & = -\frac {4 d e x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-\frac {\left (d^2-e^2 x^2\right )^{-2+p}}{2 (2-p)}-\frac {\text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-4+p} \left (-d^4 e^4 (2-p)-7 d^2 e^6 (2-p) x\right )}{x} \, dx,x,x^2\right )}{2 e^4 (2-p)}-\frac {\left (8 e (2-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 )^{-4+p} \, dx}{d^5 (5-2 p)} \\ & = \frac {4 d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac {4 d e x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-\frac {\left (d^2-e^2 x^2\right )^{-2+p}}{2 (2-p)}-\frac {8 e (2-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},4-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^5 (5-2 p)}+\frac {1}{2} d^2 \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right ) \\ & = \frac {4 d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac {4 d e x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-\frac {\left (d^2-e^2 x^2\right )^{-2+p}}{2 (2-p)}-\frac {8 e (2-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},4-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^5 (5-2 p)}+\frac {\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 (2-p)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(417\) vs. \(2(204)=408\).
Time = 0.42 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.04 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^4} \, dx=\frac {2^{-4+p} \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (8 p \left (1-\frac {d^2}{e^2 x^2}\right )^p (d-e x) \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )+4 p \left (1-\frac {d^2}{e^2 x^2}\right )^p (d-e x) \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )+2 d p \left (1-\frac {d^2}{e^2 x^2}\right )^p \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )-2 e p \left (1-\frac {d^2}{e^2 x^2}\right )^p x \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )+d p \left (1-\frac {d^2}{e^2 x^2}\right )^p \operatorname {Hypergeometric2F1}\left (4-p,1+p,2+p,\frac {d-e x}{2 d}\right )-e p \left (1-\frac {d^2}{e^2 x^2}\right )^p x \operatorname {Hypergeometric2F1}\left (4-p,1+p,2+p,\frac {d-e x}{2 d}\right )+8 d \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )+8 d p \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )\right )}{d^5 p (1+p)} \]
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\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x \left (e x +d \right )^{4}}d x\]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^4} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x \left (d + e x\right )^{4}}\, dx \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x} \,d x } \]
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Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x (d+e x)^4} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x\,{\left (d+e\,x\right )}^4} \,d x \]
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