Integrand size = 23, antiderivative size = 118 \[ \int x (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=-\frac {d^2 \left (d^2-e^2 x^2\right )^{1+p}}{e^2 (1+p)}+\frac {\left (d^2-e^2 x^2\right )^{2+p}}{2 e^2 (2+p)}+\frac {2}{3} d e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1666, 455, 45, 12, 372, 371} \[ \int x (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\frac {2}{3} d e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )-\frac {d^2 \left (d^2-e^2 x^2\right )^{p+1}}{e^2 (p+1)}+\frac {\left (d^2-e^2 x^2\right )^{p+2}}{2 e^2 (p+2)} \]
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Rule 12
Rule 45
Rule 371
Rule 372
Rule 455
Rule 1666
Rubi steps \begin{align*} \text {integral}& = \int 2 d e x^2 \left (d^2-e^2 x^2\right )^p \, dx+\int x \left (d^2-e^2 x^2\right )^p \left (d^2+e^2 x^2\right ) \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \left (d^2-e^2 x\right )^p \left (d^2+e^2 x\right ) \, dx,x,x^2\right )+(2 d e) \int x^2 \left (d^2-e^2 x^2\right )^p \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \left (2 d^2 \left (d^2-e^2 x\right )^p-\left (d^2-e^2 x\right )^{1+p}\right ) \, dx,x,x^2\right )+\left (2 d e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p}\right ) \int x^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{1+p}}{e^2 (1+p)}+\frac {\left (d^2-e^2 x^2\right )^{2+p}}{2 e^2 (2+p)}+\frac {2}{3} d e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right ) \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.93 \[ \int x (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {3 \left (d^2-e^2 x^2\right ) \left (d^2 (3+p)+e^2 (1+p) x^2\right )}{(1+p) (2+p)}+4 d e^3 x^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )\right )}{6 e^2} \]
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\[\int x \left (e x +d \right )^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}d x\]
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\[ \int x (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x \,d x } \]
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Time = 1.58 (sec) , antiderivative size = 440, normalized size of antiderivative = 3.73 \[ \int x (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=d^{2} \left (\begin {cases} \frac {x^{2} \left (d^{2}\right )^{p}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\begin {cases} \frac {\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (d^{2} - e^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + \frac {2 d d^{2 p} e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} + e^{2} \left (\begin {cases} \frac {x^{4} \left (d^{2}\right )^{p}}{4} & \text {for}\: e = 0 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 e^{4}} - \frac {x^{2}}{2 e^{2}} & \text {for}\: p = -1 \\- \frac {d^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac {d^{2} e^{2} p x^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} p x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text {otherwise} \end {cases}\right ) \]
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\[ \int x (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x \,d x } \]
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\[ \int x (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x \,d x } \]
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Timed out. \[ \int x (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int x\,{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]
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