Integrand size = 23, antiderivative size = 57 \[ \int (d+e x)^3 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=-\frac {2^{3+p} d^4 \left (\frac {d-e x}{d}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{e (1+p)} \]
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Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {690, 71} \[ \int (d+e x)^3 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=-\frac {d^4 2^{p+3} \left (\frac {d-e x}{d}\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p-3,p+1,p+2,\frac {d-e x}{2 d}\right )}{e (p+1)} \]
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Rule 71
Rule 690
Rubi steps \begin{align*} \text {integral}& = \left (d^2 \left (\frac {d-e x}{d}\right )^{1+p} \left (\frac {1}{d}-\frac {e x}{d^2}\right )^{-1-p}\right ) \int \left (\frac {1}{d}-\frac {e x}{d^2}\right )^p \left (1+\frac {e x}{d}\right )^{3+p} \, dx \\ & = -\frac {2^{3+p} d^4 \left (\frac {d-e x}{d}\right )^{1+p} \, _2F_1\left (-3-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{e (1+p)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(116\) vs. \(2(57)=114\).
Time = 0.46 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.04 \[ \int (d+e x)^3 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=-\frac {2 d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{1+p}}{e (1+p)}+\frac {d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{2+p}}{2 e (2+p)}+d^3 x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )+d e^2 x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right ) \]
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Time = 4.41 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.82
method | result | size |
meijerg | \(\frac {e^{3} x^{4} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (2,-p ;3;\frac {e^{2} x^{2}}{d^{2}}\right )}{4}+d \,e^{2} x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{2},-p ;\frac {5}{2};\frac {e^{2} x^{2}}{d^{2}}\right )+\frac {3 e \,d^{2} x^{2} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (1,-p ;2;\frac {e^{2} x^{2}}{d^{2}}\right )}{2}+d^{3} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},-p ;\frac {3}{2};\frac {e^{2} x^{2}}{d^{2}}\right )\) | \(104\) |
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\[ \int (d+e x)^3 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (42) = 84\).
Time = 2.19 (sec) , antiderivative size = 479, normalized size of antiderivative = 8.40 \[ \int (d+e x)^3 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=d^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + 3 d^{2} e \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: e^{2} = 0 \\- \frac {d^{2} \left (\begin {cases} \frac {\left (1 - \frac {e^{2} x^{2}}{d^{2}}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (1 - \frac {e^{2} x^{2}}{d^{2}} \right )} & \text {otherwise} \end {cases}\right )}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + d e^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + e^{3} \left (\begin {cases} \frac {x^{4}}{4} & \text {for}\: e = 0 \\- \frac {d^{6} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{6} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{6}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {d^{4} e^{2} x^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {d^{4} e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {d^{4} \log {\left (- \frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{4} \log {\left (\frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{2} x^{2}}{2 e^{2}} & \text {for}\: p = -1 \\- \frac {d^{4} \left (1 - \frac {e^{2} x^{2}}{d^{2}}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac {d^{2} e^{2} p x^{2} \left (1 - \frac {e^{2} x^{2}}{d^{2}}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} p x^{4} \left (1 - \frac {e^{2} x^{2}}{d^{2}}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} x^{4} \left (1 - \frac {e^{2} x^{2}}{d^{2}}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text {otherwise} \end {cases}\right ) \]
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\[ \int (d+e x)^3 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p} \,d x } \]
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\[ \int (d+e x)^3 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p} \,d x } \]
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Timed out. \[ \int (d+e x)^3 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=\int {\left (1-\frac {e^2\,x^2}{d^2}\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \]
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