Optimal. Leaf size=156 \[ -\frac {d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^3 (1-p)}-\frac {x^3 \left (d^2-e^2 x^2\right )^{-1+p}}{1+2 p}-\frac {d \left (d^2-e^2 x^2\right )^p}{e^3 p}+\frac {2 (2+p) x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},2-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )}{3 d^2 (1+2 p)} \]
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Rubi [A]
time = 0.12, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {866, 1666, 470,
372, 371, 12, 272, 45} \begin {gather*} \frac {2 (p+2) x^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {3}{2},2-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )}{3 d^2 (2 p+1)}-\frac {x^3 \left (d^2-e^2 x^2\right )^{p-1}}{2 p+1}-\frac {d \left (d^2-e^2 x^2\right )^p}{e^3 p}-\frac {d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^3 (1-p)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 45
Rule 272
Rule 371
Rule 372
Rule 470
Rule 866
Rule 1666
Rubi steps
\begin {align*} \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx &=\int x^2 (d-e x)^2 \left (d^2-e^2 x^2\right )^{-2+p} \, dx\\ &=\int -2 d e x^3 \left (d^2-e^2 x^2\right )^{-2+p} \, dx+\int x^2 \left (d^2-e^2 x^2\right )^{-2+p} \left (d^2+e^2 x^2\right ) \, dx\\ &=-\frac {x^3 \left (d^2-e^2 x^2\right )^{-1+p}}{1+2 p}-(2 d e) \int x^3 \left (d^2-e^2 x^2\right )^{-2+p} \, dx+\frac {\left (2 d^2 (2+p)\right ) \int x^2 \left (d^2-e^2 x^2\right )^{-2+p} \, dx}{1+2 p}\\ &=-\frac {x^3 \left (d^2-e^2 x^2\right )^{-1+p}}{1+2 p}-(d e) \text {Subst}\left (\int x \left (d^2-e^2 x\right )^{-2+p} \, dx,x,x^2\right )+\frac {\left (2 (2+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 )^{-2+p} \, dx}{d^2 (1+2 p)}\\ &=-\frac {x^3 \left (d^2-e^2 x^2\right )^{-1+p}}{1+2 p}+\frac {2 (2+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},2-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )}{3 d^2 (1+2 p)}-(d e) \text {Subst}\left (\int \left (\frac {d^2 \left (d^2-e^2 x\right )^{-2+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}}{1+2 p}-\frac {d \left (d^2-e^2 x^2\right )^p}{e^3 p}+\frac {2 (2+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},2-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )}{3 d^2 (1+2 p)}\\ \end {align*}
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Mathematica [A]
time = 0.31, size = 177, normalized size = 1.13 \begin {gather*} \frac {2^{-2+p} \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (4 e (1+p) x \left (\frac {1}{2}+\frac {e x}{2 d}\right )^p \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )+(d-e x) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (4 \, _2F_1\left (1-p,1+p;2+p;\frac {d-e x}{2 d}\right )-\, _2F_1\left (2-p,1+p;2+p;\frac {d-e x}{2 d}\right )\right )\right )}{e^3 (1+p)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{2}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{2}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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