Integrand size = 25, antiderivative size = 184 \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=-\frac {d^5 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {x^5 \left (d^2-e^2 x^2\right )^{-1+p}}{3+2 p}-\frac {2 d^3 \left (d^2-e^2 x^2\right )^p}{e^5 p}+\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{e^5 (1+p)}+\frac {2 (4+p) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},2-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^2 (3+2 p)} \]
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Time = 0.13 (sec) , antiderivative size = 184, 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, 470, 372, 371, 12, 272, 45} \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {2 (p+4) x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {5}{2},2-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^2 (2 p+3)}-\frac {x^5 \left (d^2-e^2 x^2\right )^{p-1}}{2 p+3}+\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{e^5 (p+1)}-\frac {d^5 \left (d^2-e^2 x^2\right )^{p-1}}{e^5 (1-p)}-\frac {2 d^3 \left (d^2-e^2 x^2\right )^p}{e^5 p} \]
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Rule 12
Rule 45
Rule 272
Rule 371
Rule 372
Rule 470
Rule 866
Rule 1666
Rubi steps \begin{align*} \text {integral}& = \int x^4 (d-e x)^2 \left (d^2-e^2 x^2\right )^{-2+p} \, dx \\ & = \int -2 d e x^5 \left (d^2-e^2 x^2\right )^{-2+p} \, dx+\int x^4 \left (d^2-e^2 x^2\right )^{-2+p} \left (d^2+e^2 x^2\right ) \, dx \\ & = -\frac {x^5 \left (d^2-e^2 x^2\right )^{-1+p}}{3+2 p}-(2 d e) \int x^5 \left (d^2-e^2 x^2\right )^{-2+p} \, dx+\frac {\left (2 d^2 (4+p)\right ) \int x^4 \left (d^2-e^2 x^2\right )^{-2+p} \, dx}{3+2 p} \\ & = -\frac {x^5 \left (d^2-e^2 x^2\right )^{-1+p}}{3+2 p}-(d e) \text {Subst}\left (\int x^2 \left (d^2-e^2 x\right )^{-2+p} \, dx,x,x^2\right )+\frac {\left (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 x^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{-2+p} \, dx}{d^2 (3+2 p)} \\ & = -\frac {x^5 \left (d^2-e^2 x^2\right )^{-1+p}}{3+2 p}+\frac {2 (4+p) 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},2-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^2 (3+2 p)}-(d e) \text {Subst}\left (\int \left (\frac {d^4 \left (d^2-e^2 x\right )^{-2+p}}{e^4}-\frac {2 d^2 \left (d^2-e^2 x\right )^{-1+p}}{e^4}+\frac {\left (d^2-e^2 x\right )^p}{e^4}\right ) \, dx,x,x^2\right ) \\ & = -\frac {d^5 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac {x^5 \left (d^2-e^2 x^2\right )^{-1+p}}{3+2 p}-\frac {2 d^3 \left (d^2-e^2 x^2\right )^p}{e^5 p}+\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{e^5 (1+p)}+\frac {2 (4+p) 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},2-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )}{5 d^2 (3+2 p)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.36 \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {x^5 (d-e x)^p (d+e x)^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {AppellF1}\left (5,-p,2-p,6,\frac {e x}{d},-\frac {e x}{d}\right )}{5 d^2} \]
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\[\int \frac {x^{4} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{2}}d x\]
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\[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{2}} \,d x } \]
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\[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\int \frac {x^{4} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{2}}\, dx \]
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\[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{2}} \,d x } \]
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\[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4}}{{\left (e x + d\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx=\int \frac {x^4\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^2} \,d x \]
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