Optimal. Leaf size=155 \[ \frac {2 d^2 (p+4) x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )}{3 (2 p+5)}-\frac {x^3 \left (d^2-e^2 x^2\right )^{p+1}}{2 p+5}+\frac {d \left (d^2-e^2 x^2\right )^{p+2}}{e^3 (p+2)}-\frac {d^3 \left (d^2-e^2 x^2\right )^{p+1}}{e^3 (p+1)} \]
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Rubi [A] time = 0.14, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1652, 459, 365, 364, 12, 266, 43} \[ \frac {2 d^2 (p+4) x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )}{3 (2 p+5)}-\frac {x^3 \left (d^2-e^2 x^2\right )^{p+1}}{2 p+5}-\frac {d^3 \left (d^2-e^2 x^2\right )^{p+1}}{e^3 (p+1)}+\frac {d \left (d^2-e^2 x^2\right )^{p+2}}{e^3 (p+2)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 43
Rule 266
Rule 364
Rule 365
Rule 459
Rule 1652
Rubi steps
\begin {align*} \int x^2 (d+e x)^2 \left (d^2-e^2 x^2\right )^p \, dx &=\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) \operatorname {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) \operatorname {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*}
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Mathematica [A] time = 0.11, size = 168, normalized size = 1.08 \[ \frac {\left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (-15 d \left (d^2-e^2 x^2\right ) \left (d^2+e^2 (p+1) x^2\right ) \left (1-\frac {e^2 x^2}{d^2}\right )^p+3 e^5 \left (p^2+3 p+2\right ) x^5 \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )+5 d^2 e^3 \left (p^2+3 p+2\right ) x^3 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )\right )}{15 e^3 (p+1) (p+2)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{2} x^{4} + 2 \, d e x^{3} + d^{2} x^{2}\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{2} x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 6.43, size = 425, normalized size = 2.74 \[ \frac {d^{2} d^{2 p} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} + 2 d e \left (\begin {cases} \frac {x^{4} \left (d^{2}\right )^{p}}{4} & \text {for}\: e = 0 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 e^{4}} - \frac {x^{2}}{2 e^{2}} & \text {for}\: p = -1 \\- \frac {d^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac {d^{2} e^{2} p x^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} p x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text {otherwise} \end {cases}\right ) + \frac {d^{2 p} e^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{5} \]
Verification of antiderivative is not currently implemented for this CAS.
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