Integrand size = 23, antiderivative size = 120 \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=-\frac {d^3 \left (d^2-e^2 x^2\right )^{1+p}}{2 e^4 (1+p)}+\frac {d \left (d^2-e^2 x^2\right )^{2+p}}{2 e^4 (2+p)}+\frac {1}{5} e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {778, 272, 45, 372, 371} \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\frac {1}{5} e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )+\frac {d \left (d^2-e^2 x^2\right )^{p+2}}{2 e^4 (p+2)}-\frac {d^3 \left (d^2-e^2 x^2\right )^{p+1}}{2 e^4 (p+1)} \]
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Rule 45
Rule 272
Rule 371
Rule 372
Rule 778
Rubi steps \begin{align*} \text {integral}& = d \int x^3 \left (d^2-e^2 x^2\right )^p \, dx+e \int x^4 \left (d^2-e^2 x^2\right )^p \, dx \\ & = \frac {1}{2} d \text {Subst}\left (\int x \left (d^2-e^2 x\right )^p \, dx,x,x^2\right )+\left (e \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 )^p \, dx \\ & = \frac {1}{5} e 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},-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )+\frac {1}{2} d \text {Subst}\left (\int \left (\frac {d^2 \left (d^2-e^2 x\right )^p}{e^2}-\frac {\left (d^2-e^2 x\right )^{1+p}}{e^2}\right ) \, dx,x,x^2\right ) \\ & = -\frac {d^3 \left (d^2-e^2 x^2\right )^{1+p}}{2 e^4 (1+p)}+\frac {d \left (d^2-e^2 x^2\right )^{2+p}}{2 e^4 (2+p)}+\frac {1}{5} e 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},-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right ) \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88 \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {5 d \left (d^2-e^2 x^2\right ) \left (d^2+e^2 (1+p) x^2\right )}{(1+p) (2+p)}+2 e^5 x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )\right )}{10 e^4} \]
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\[\int x^{3} \left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{p}d x\]
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\[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{3} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (97) = 194\).
Time = 1.59 (sec) , antiderivative size = 382, normalized size of antiderivative = 3.18 \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=d \left (\begin {cases} \frac {x^{4} \left (d^{2}\right )^{p}}{4} & \text {for}\: e = 0 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 e^{4}} - \frac {x^{2}}{2 e^{2}} & \text {for}\: p = -1 \\- \frac {d^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac {d^{2} e^{2} p x^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} p x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text {otherwise} \end {cases}\right ) + \frac {d^{2 p} e x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{5} \]
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\[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{3} \,d x } \]
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\[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{3} \,d x } \]
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Timed out. \[ \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\int x^3\,{\left (d^2-e^2\,x^2\right )}^p\,\left (d+e\,x\right ) \,d x \]
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