Integrand size = 25, antiderivative size = 265 \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=-\frac {4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {4 \left (16+15 p+p^2\right ) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},4-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )} \]
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Time = 0.21 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {866, 1666, 1281, 470, 372, 371, 457, 78} \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\frac {d^2 (12 p+13) x^5 \left (d^2-e^2 x^2\right )^{p-3}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{p-3}}{2 p+1}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {4 d^7 \left (d^2-e^2 x^2\right )^{p-3}}{e^5 (3-p)}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{p-2}}{e^5 (2-p)}-\frac {4 \left (p^2+15 p+16\right ) x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {5}{2},4-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^5 (1-p)} \]
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Rule 78
Rule 371
Rule 372
Rule 457
Rule 470
Rule 866
Rule 1281
Rule 1666
Rubi steps \begin{align*} \text {integral}& = \int x^4 (d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p} \, dx \\ & = \int x^5 \left (d^2-e^2 x^2\right )^{-4+p} \left (-4 d^3 e-4 d e^3 x^2\right ) \, dx+\int x^4 \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 ) \, dx \\ & = -\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {1}{2} \text {Subst}\left (\int x^2 \left (d^2-e^2 x\right )^{-4+p} \left (-4 d^3 e-4 d e^3 x\right ) \, dx,x,x^2\right )-\frac {\int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \left (-d^4 e^2 (1+2 p)-d^2 e^4 (13+12 p) x^2\right ) \, dx}{e^2 (1+2 p)} \\ & = \frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {1}{2} \text {Subst}\left (\int \left (-\frac {8 d^7 \left (d^2-e^2 x\right )^{-4+p}}{e^3}+\frac {20 d^5 \left (d^2-e^2 x\right )^{-3+p}}{e^3}-\frac {16 d^3 \left (d^2-e^2 x\right )^{-2+p}}{e^3}+\frac {4 d \left (d^2-e^2 x\right )^{-1+p}}{e^3}\right ) \, dx,x,x^2\right )-\frac {\left (4 d^4 \left (16+15 p+p^2\right )\right ) \int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{1-4 p^2} \\ & = -\frac {4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {\left (4 \left (16+15 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 x^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{-4+p} \, dx}{d^4 \left (1-4 p^2\right )} \\ & = -\frac {4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac {d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac {e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac {10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac {8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac {4 \left (16+15 p+p^2\right ) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {5}{2},4-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )} \\ \end{align*}
Time = 0.53 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.87 \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\frac {2^{-4+p} \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (16 e (1+p) x \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )+(d-e x) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (32 \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )-24 \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )+8 \operatorname {Hypergeometric2F1}\left (3-p,1+p,2+p,\frac {d-e x}{2 d}\right )-\operatorname {Hypergeometric2F1}\left (4-p,1+p,2+p,\frac {d-e x}{2 d}\right )\right )\right )}{e^5 (1+p)} \]
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\[\int \frac {x^{4} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{4}}d x\]
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\[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{4}} \,d x } \]
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\[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int \frac {x^{4} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{4}}\, dx \]
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\[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{4}} \,d x } \]
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\[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{4}} \,d x } \]
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Timed out. \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx=\int \frac {x^4\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^4} \,d x \]
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