Optimal. Leaf size=163 \[ -\frac{2^{p-3} (p+7) \left (d^2-e^2 x^2\right )^{p+1} \left (\frac{e x}{d}+1\right )^{-p-1} \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{d^3 e^3 (1-2 p) (3-p) (p+1)}+\frac{\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}-\frac{d \left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 (3-p) (d+e x)^4} \]
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Rubi [A] time = 0.1906, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1639, 793, 678, 69} \[ -\frac{2^{p-3} (p+7) \left (d^2-e^2 x^2\right )^{p+1} \left (\frac{e x}{d}+1\right )^{-p-1} \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{d^3 e^3 (1-2 p) (3-p) (p+1)}+\frac{\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}-\frac{d \left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 (3-p) (d+e x)^4} \]
Antiderivative was successfully verified.
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Rule 1639
Rule 793
Rule 678
Rule 69
Rubi steps
\begin{align*} \int \frac{x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx &=\frac{\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}+\frac{\int \frac{\left (3 d^2 e^2+2 d e^3 (1+p) x\right ) \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx}{e^4 (1-2 p)}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (3-p) (d+e x)^4}+\frac{\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}+\frac{(d (7+p)) \int \frac{\left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx}{e^2 (1-2 p) (3-p)}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (3-p) (d+e x)^4}+\frac{\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}+\frac{\left ((7+p) (d-e x)^{-1-p} \left (1+\frac{e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p}\right ) \int (d-e x)^p \left (1+\frac{e x}{d}\right )^{-3+p} \, dx}{d^3 e^2 (1-2 p) (3-p)}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (3-p) (d+e x)^4}+\frac{\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}-\frac{2^{-3+p} (7+p) \left (1+\frac{e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \, _2F_1\left (3-p,1+p;2+p;\frac{d-e x}{2 d}\right )}{d^3 e^3 (1-2 p) (3-p) (1+p)}\\ \end{align*}
Mathematica [A] time = 0.135175, size = 130, normalized size = 0.8 \[ -\frac{2^{p-4} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (4 \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )-4 \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )+\, _2F_1\left (4-p,p+1;p+2;\frac{d-e x}{2 d}\right )\right )}{d^2 e^3 (p+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.742, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{ \left ( ex+d \right ) ^{4}}}\, 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 )}^{4}}\,{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^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, 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 )^{4}}\, 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 )}^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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