Optimal. Leaf size=174 \[ \frac{e^4 (10-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (2,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{4 d^3 (2-p)}+\frac{2 e^3 (4-p) \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (-\frac{1}{2},3-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{d^6 x}+\frac{e \left (d^2-e^2 x^2\right )^{p-2}}{x^3}-\frac{d \left (d^2-e^2 x^2\right )^{p-2}}{4 x^4} \]
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Rubi [A] time = 0.27686, antiderivative size = 174, 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, 266, 65, 365, 364} \[ \frac{e^4 (10-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (2,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{4 d^3 (2-p)}+\frac{2 e^3 (4-p) \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (-\frac{1}{2},3-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{d^6 x}+\frac{e \left (d^2-e^2 x^2\right )^{p-2}}{x^3}-\frac{d \left (d^2-e^2 x^2\right )^{p-2}}{4 x^4} \]
Antiderivative was successfully verified.
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Rule 852
Rule 1807
Rule 764
Rule 266
Rule 65
Rule 365
Rule 364
Rubi steps
\begin{align*} \int \frac{\left (d^2-e^2 x^2\right )^p}{x^5 (d+e x)^3} \, dx &=\int \frac{(d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p}}{x^5} \, dx\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{4 x^4}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-3+p} \left (12 d^4 e-2 d^3 e^2 (10-p) x+4 d^2 e^3 x^2\right )}{x^4} \, dx}{4 d^2}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{4 x^4}+\frac{e \left (d^2-e^2 x^2\right )^{-2+p}}{x^3}+\frac{\int \frac{\left (6 d^5 e^2 (10-p)-24 d^4 e^3 (4-p) x\right ) \left (d^2-e^2 x^2\right )^{-3+p}}{x^3} \, dx}{12 d^4}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{4 x^4}+\frac{e \left (d^2-e^2 x^2\right )^{-2+p}}{x^3}-\left (2 e^3 (4-p)\right ) \int \frac{\left (d^2-e^2 x^2\right )^{-3+p}}{x^2} \, dx+\frac{1}{2} \left (d e^2 (10-p)\right ) \int \frac{\left (d^2-e^2 x^2\right )^{-3+p}}{x^3} \, dx\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{4 x^4}+\frac{e \left (d^2-e^2 x^2\right )^{-2+p}}{x^3}+\frac{1}{4} \left (d e^2 (10-p)\right ) \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-3+p}}{x^2} \, dx,x,x^2\right )-\frac{\left (2 e^3 (4-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}{d^6}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{4 x^4}+\frac{e \left (d^2-e^2 x^2\right )^{-2+p}}{x^3}+\frac{2 e^3 (4-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 )}{d^6 x}+\frac{e^4 (10-p) \left (d^2-e^2 x^2\right )^{-2+p} \, _2F_1\left (2,-2+p;-1+p;1-\frac{e^2 x^2}{d^2}\right )}{4 d^3 (2-p)}\\ \end{align*}
Mathematica [B] time = 0.624467, size = 446, normalized size = 2.56 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (\frac{24 d^3 e^2 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )}{(p-1) x^2}+\frac{4 d^5 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (2-p,-p;3-p;\frac{d^2}{e^2 x^2}\right )}{(p-2) x^4}+\frac{60 d e^4 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )}{p}+\frac{8 d^4 e \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{3}{2},-p;-\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x^3}+\frac{80 d^2 e^3 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x}+\frac{15 e^4 2^{p+2} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{5 e^4 2^{p+1} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{e^4 2^p (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}\right )}{8 d^8} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.693, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{{x}^{5} \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^{5}}\,{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^{8} + 3 \, d e^{2} x^{7} + 3 \, d^{2} e x^{6} + d^{3} x^{5}}, 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^{5} \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^{5}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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