Optimal. Leaf size=156 \[ \frac{2 (p+2) x^3 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{3}{2},2-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^2 (2 p+1)}-\frac{x^3 \left (d^2-e^2 x^2\right )^{p-1}}{2 p+1}-\frac{d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^3 (1-p)}-\frac{d \left (d^2-e^2 x^2\right )^p}{e^3 p} \]
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Rubi [A] time = 0.185173, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {852, 1652, 459, 365, 364, 12, 266, 43} \[ \frac{2 (p+2) x^3 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{3}{2},2-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^2 (2 p+1)}-\frac{x^3 \left (d^2-e^2 x^2\right )^{p-1}}{2 p+1}-\frac{d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^3 (1-p)}-\frac{d \left (d^2-e^2 x^2\right )^p}{e^3 p} \]
Antiderivative was successfully verified.
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Rule 852
Rule 1652
Rule 459
Rule 365
Rule 364
Rule 12
Rule 266
Rule 43
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) \operatorname{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) \operatorname{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*}
Mathematica [A] time = 0.175926, size = 177, normalized size = 1.13 \[ \frac{2^{p-2} \left (\frac{e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (4 e (p+1) x \left (\frac{e x}{2 d}+\frac{1}{2}\right )^p \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )+(d-e x) \left (1-\frac{e^2 x^2}{d^2}\right )^p \left (4 \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )-\, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )\right )\right )}{e^3 (p+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.662, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{ \left ( ex+d \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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