Integrand size = 25, antiderivative size = 155 \[ \int x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=-\frac {d^3 \left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1+p)}-\frac {x^3 \left (d^2-e^2 x^2\right )^{1+p}}{5+2 p}+\frac {d \left (d^2-e^2 x^2\right )^{2+p}}{e^3 (2+p)}+\frac {2 d^2 (4+p) x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )}{3 (5+2 p)} \]
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Time = 0.09 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1666, 470, 372, 371, 12, 272, 45} \[ \int x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\frac {2 d^2 (p+4) x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )}{3 (2 p+5)}-\frac {x^3 \left (d^2-e^2 x^2\right )^{p+1}}{2 p+5}+\frac {d \left (d^2-e^2 x^2\right )^{p+2}}{e^3 (p+2)}-\frac {d^3 \left (d^2-e^2 x^2\right )^{p+1}}{e^3 (p+1)} \]
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Rule 12
Rule 45
Rule 272
Rule 371
Rule 372
Rule 470
Rule 1666
Rubi steps \begin{align*} \text {integral}& = \int 2 d e x^3 \left (d^2-e^2 x^2\right )^p \, dx+\int x^2 \left (d^2-e^2 x^2\right )^p \left (d^2+e^2 x^2\right ) \, dx \\ & = -\frac {x^3 \left (d^2-e^2 x^2\right )^{1+p}}{5+2 p}+(2 d e) \int x^3 \left (d^2-e^2 x^2\right )^p \, dx+\frac {\left (2 d^2 (4+p)\right ) \int x^2 \left (d^2-e^2 x^2\right )^p \, dx}{5+2 p} \\ & = -\frac {x^3 \left (d^2-e^2 x^2\right )^{1+p}}{5+2 p}+(d e) \text {Subst}\left (\int x \left (d^2-e^2 x\right )^p \, dx,x,x^2\right )+\frac {\left (2 d^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^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx}{5+2 p} \\ & = -\frac {x^3 \left (d^2-e^2 x^2\right )^{1+p}}{5+2 p}+\frac {2 d^2 (4+p) 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 )}{3 (5+2 p)}+(d e) \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}}{e^3 (1+p)}-\frac {x^3 \left (d^2-e^2 x^2\right )^{1+p}}{5+2 p}+\frac {d \left (d^2-e^2 x^2\right )^{2+p}}{e^3 (2+p)}+\frac {2 d^2 (4+p) 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 )}{3 (5+2 p)} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08 \[ \int x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (-15 d \left (d^2-e^2 x^2\right ) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (d^2+e^2 (1+p) x^2\right )+5 d^2 e^3 \left (2+3 p+p^2\right ) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )+3 e^5 \left (2+3 p+p^2\right ) x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )\right )}{15 e^3 (1+p) (2+p)} \]
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\[\int x^{2} \left (e x +d \right )^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}d x\]
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\[ \int x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (128) = 256\).
Time = 1.95 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.74 \[ \int x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\frac {d^{2} d^{2 p} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} + 2 d e \left (\begin {cases} \frac {x^{4} \left (d^{2}\right )^{p}}{4} & \text {for}\: e = 0 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 e^{4}} - \frac {x^{2}}{2 e^{2}} & \text {for}\: p = -1 \\- \frac {d^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac {d^{2} e^{2} p x^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} p x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text {otherwise} \end {cases}\right ) + \frac {d^{2 p} e^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{5} \]
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\[ \int x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2} \,d x } \]
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\[ \int x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2} \,d x } \]
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Timed out. \[ \int x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx=\int x^2\,{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]
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