Integrand size = 23, antiderivative size = 116 \[ \int x (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=-\frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^{1+p}}{e^2 (5+2 p)}-\frac {3\ 2^{3+p} d^3 \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{e^2 (1+p) (5+2 p)} \]
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Time = 0.05 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {809, 692, 71} \[ \int x (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=-\frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^{p+1}}{e^2 (2 p+5)}-\frac {3 d^3 2^{p+3} \left (d^2-e^2 x^2\right )^{p+1} \left (\frac {e x}{d}+1\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (-p-3,p+1,p+2,\frac {d-e x}{2 d}\right )}{e^2 (p+1) (2 p+5)} \]
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Rule 71
Rule 692
Rule 809
Rubi steps \begin{align*} \text {integral}& = -\frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^{1+p}}{e^2 (5+2 p)}+\frac {(3 d) \int (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx}{e (5+2 p)} \\ & = -\frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^{1+p}}{e^2 (5+2 p)}+\frac {\left (3 d^3 (d-e x)^{-1-p} \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p}\right ) \int (d-e x)^p \left (1+\frac {e x}{d}\right )^{3+p} \, dx}{e (5+2 p)} \\ & = -\frac {(d+e x)^3 \left (d^2-e^2 x^2\right )^{1+p}}{e^2 (5+2 p)}-\frac {3\ 2^{3+p} d^3 \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \, _2F_1\left (-3-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{e^2 (1+p) (5+2 p)} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.37 \[ \int x (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {5 d \left (d^2-e^2 x^2\right ) \left (d^2 (5+p)+3 e^2 (1+p) x^2\right )}{(1+p) (2+p)}+10 d^2 e^3 x^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )+2 e^5 x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )\right )}{10 e^2} \]
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\[\int x \left (e x +d \right )^{3} \left (-e^{2} x^{2}+d^{2}\right )^{p}d x\]
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\[ \int x (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x \,d x } \]
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Time = 2.09 (sec) , antiderivative size = 479, normalized size of antiderivative = 4.13 \[ \int x (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=d^{3} \left (\begin {cases} \frac {x^{2} \left (d^{2}\right )^{p}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\begin {cases} \frac {\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (d^{2} - e^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + d^{2} d^{2 p} e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + 3 d e^{2} \left (\begin {cases} \frac {x^{4} \left (d^{2}\right )^{p}}{4} & \text {for}\: e = 0 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 e^{4}} - \frac {x^{2}}{2 e^{2}} & \text {for}\: p = -1 \\- \frac {d^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac {d^{2} e^{2} p x^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} p x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text {otherwise} \end {cases}\right ) + \frac {d^{2 p} e^{3} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{5} \]
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\[ \int x (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x \,d x } \]
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\[ \int x (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x \,d x } \]
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Timed out. \[ \int x (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int x\,{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \]
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