Integrand size = 25, antiderivative size = 210 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^4} \, dx=-\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}+\frac {2 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x^2}-\frac {e^2 (27-2 p) \left (d^2-e^2 x^2\right )^{-3+p}}{3 x}+\frac {4 e^4 \left (48-17 p+p^2\right ) x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^8}-\frac {2 e^3 (5-p) \left (d^2-e^2 x^2\right )^{-3+p} \operatorname {Hypergeometric2F1}\left (1,-3+p,-2+p,1-\frac {e^2 x^2}{d^2}\right )}{d (3-p)} \]
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Time = 0.27 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {866, 1821, 778, 272, 67, 252, 251} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^4} \, dx=-\frac {e^2 (27-2 p) \left (d^2-e^2 x^2\right )^{p-3}}{3 x}+\frac {2 d e \left (d^2-e^2 x^2\right )^{p-3}}{x^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{p-3}}{3 x^3}-\frac {2 e^3 (5-p) \left (d^2-e^2 x^2\right )^{p-3} \operatorname {Hypergeometric2F1}\left (1,p-3,p-2,1-\frac {e^2 x^2}{d^2}\right )}{d (3-p)}+\frac {4 e^4 \left (p^2-17 p+48\right ) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )}{3 d^8} \]
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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)^4 \left (d^2-e^2 x^2\right )^{-4+p}}{x^4} \, dx \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-4+p} \left (12 d^5 e-d^4 e^2 (27-2 p) x+12 d^3 e^3 x^2-3 d^2 e^4 x^3\right )}{x^3} \, dx}{3 d^2} \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}+\frac {2 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x^2}+\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-4+p} \left (2 d^6 e^2 (27-2 p)-24 d^5 e^3 (5-p) x+6 d^4 e^4 x^2\right )}{x^2} \, dx}{6 d^4} \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}+\frac {2 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x^2}-\frac {e^2 (27-2 p) \left (d^2-e^2 x^2\right )^{-3+p}}{3 x}-\frac {\int \frac {\left (24 d^7 e^3 (5-p)-8 d^6 e^4 \left (48-17 p+p^2\right ) x\right ) \left (d^2-e^2 x^2\right )^{-4+p}}{x} \, dx}{6 d^6} \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}+\frac {2 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x^2}-\frac {e^2 (27-2 p) \left (d^2-e^2 x^2\right )^{-3+p}}{3 x}-\left (4 d e^3 (5-p)\right ) \int \frac {\left (d^2-e^2 x^2\right )^{-4+p}}{x} \, dx+\frac {1}{3} \left (4 e^4 \left (48-17 p+p^2\right )\right ) \int \left (d^2-e^2 x^2\right )^{-4+p} \, dx \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}+\frac {2 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x^2}-\frac {e^2 (27-2 p) \left (d^2-e^2 x^2\right )^{-3+p}}{3 x}-\left (2 d e^3 (5-p)\right ) \text {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-4+p}}{x} \, dx,x,x^2\right )+\frac {\left (4 e^4 \left (48-17 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}{3 d^8} \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{3 x^3}+\frac {2 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x^2}-\frac {e^2 (27-2 p) \left (d^2-e^2 x^2\right )^{-3+p}}{3 x}+\frac {4 e^4 \left (48-17 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 )}{3 d^8}-\frac {2 e^3 (5-p) \left (d^2-e^2 x^2\right )^{-3+p} \, _2F_1\left (1,-3+p;-2+p;1-\frac {e^2 x^2}{d^2}\right )}{d (3-p)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(452\) vs. \(2(210)=420\).
Time = 0.78 (sec) , antiderivative size = 452, normalized size of antiderivative = 2.15 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^4} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {16 d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-p,-\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x^3}-\frac {480 d^2 e^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x}-\frac {96 d^3 e \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {d^2}{e^2 x^2}\right )}{(-1+p) x^2}+\frac {15\ 2^{5+p} e^3 (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {15\ 2^{3+p} e^3 (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {3\ 2^{3+p} e^3 (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {3\ 2^p e^3 (-d+e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (4-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}-\frac {480 d e^3 \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )}{p}\right )}{48 d^8} \]
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\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{4} \left (e x +d \right )^{4}}d x\]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{4}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^4} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{4} \left (d + e x\right )^{4}}\, dx \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{4}} \,d x } \]
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\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^4} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^4\,{\left (d+e\,x\right )}^4} \,d x \]
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