Optimal. Leaf size=179 \[ -\frac{e^3 (10-3 p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^2 (2-p)}-\frac{2 e^2 (8-p) \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (-\frac{1}{2},3-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^5 x}+\frac{3 e \left (d^2-e^2 x^2\right )^{p-2}}{2 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{p-2}}{3 x^3} \]
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Rubi [A] time = 0.274783, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {852, 1807, 764, 365, 364, 266, 65} \[ -\frac{e^3 (10-3 p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^2 (2-p)}-\frac{2 e^2 (8-p) \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (-\frac{1}{2},3-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^5 x}+\frac{3 e \left (d^2-e^2 x^2\right )^{p-2}}{2 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{p-2}}{3 x^3} \]
Antiderivative was successfully verified.
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Rule 852
Rule 1807
Rule 764
Rule 365
Rule 364
Rule 266
Rule 65
Rubi steps
\begin{align*} \int \frac{\left (d^2-e^2 x^2\right )^p}{x^4 (d+e x)^3} \, dx &=\int \frac{(d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p}}{x^4} \, dx\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{3 x^3}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-3+p} \left (9 d^4 e-2 d^3 e^2 (8-p) x+3 d^2 e^3 x^2\right )}{x^3} \, dx}{3 d^2}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{3 x^3}+\frac{3 e \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac{\int \frac{\left (4 d^5 e^2 (8-p)-6 d^4 e^3 (10-3 p) x\right ) \left (d^2-e^2 x^2\right )^{-3+p}}{x^2} \, dx}{6 d^4}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{3 x^3}+\frac{3 e \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}-\left (e^3 (10-3 p)\right ) \int \frac{\left (d^2-e^2 x^2\right )^{-3+p}}{x} \, dx+\frac{1}{3} \left (2 d e^2 (8-p)\right ) \int \frac{\left (d^2-e^2 x^2\right )^{-3+p}}{x^2} \, dx\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{3 x^3}+\frac{3 e \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}-\frac{1}{2} \left (e^3 (10-3 p)\right ) \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right )+\frac{\left (2 e^2 (8-p) \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p}\right ) \int \frac{\left (1-\frac{e^2 x^2}{d^2}\right )^{-3+p}}{x^2} \, dx}{3 d^5}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{3 x^3}+\frac{3 e \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}-\frac{2 e^2 (8-p) \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},3-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^5 x}-\frac{e^3 (10-3 p) \left (d^2-e^2 x^2\right )^{-2+p} \, _2F_1\left (1,-2+p;-1+p;1-\frac{e^2 x^2}{d^2}\right )}{2 d^2 (2-p)}\\ \end{align*}
Mathematica [B] time = 0.544901, size = 393, normalized size = 2.2 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (-\frac{36 d^3 e \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )}{(p-1) x^2}-\frac{120 d e^3 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )}{p}-\frac{8 d^4 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{3}{2},-p;-\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x^3}-\frac{144 d^2 e^2 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x}+\frac{15 e^3 2^{p+3} (e x-d) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^3 2^{p+3} (e x-d) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^3 2^p (e x-d) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}\right )}{24 d^7} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.711, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{{x}^{4} \left ( ex+d \right ) ^{3}}}\, 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}}{{\left (e x + d\right )}^{3} x^{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}}{e^{3} x^{7} + 3 \, d e^{2} x^{6} + 3 \, d^{2} e x^{5} + d^{3} x^{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{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{4} \left (d + e x\right )^{3}}\, 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}}{{\left (e x + d\right )}^{3} x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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