Optimal. Leaf size=211 \[ \frac{e^2 (10-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^2 (2-p)}-\frac{8 e^3 (4-p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^7}+\frac{e^2 (11-p) \left (d^2-e^2 x^2\right )^{p-3}}{2 (3-p)}+\frac{4 d e \left (d^2-e^2 x^2\right )^{p-3}}{x}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{2 x^2} \]
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Rubi [A] time = 0.367507, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {852, 1807, 1652, 446, 79, 65, 12, 246, 245} \[ \frac{e^2 (10-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^2 (2-p)}-\frac{8 e^3 (4-p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^7}+\frac{e^2 (11-p) \left (d^2-e^2 x^2\right )^{p-3}}{2 (3-p)}+\frac{4 d e \left (d^2-e^2 x^2\right )^{p-3}}{x}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{2 x^2} \]
Antiderivative was successfully verified.
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Rule 852
Rule 1807
Rule 1652
Rule 446
Rule 79
Rule 65
Rule 12
Rule 246
Rule 245
Rubi steps
\begin{align*} \int \frac{\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^4} \, dx &=\int \frac{(d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p}}{x^3} \, dx\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-4+p} \left (8 d^5 e-2 d^4 e^2 (10-p) x+8 d^3 e^3 x^2-2 d^2 e^4 x^3\right )}{x^2} \, dx}{2 d^2}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-4+p} \left (2 d^6 e^2 (10-p)-16 d^5 e^3 (4-p) x+2 d^4 e^4 x^2\right )}{x} \, dx}{2 d^4}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac{\int -16 d^5 e^3 (4-p) \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{2 d^4}+\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-4+p} \left (2 d^6 e^2 (10-p)+2 d^4 e^4 x^2\right )}{x} \, dx}{2 d^4}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac{\operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-4+p} \left (2 d^6 e^2 (10-p)+2 d^4 e^4 x\right )}{x} \, dx,x,x^2\right )}{4 d^4}-\left (8 d e^3 (4-p)\right ) \int \left (d^2-e^2 x^2\right )^{-4+p} \, dx\\ &=\frac{e^2 (11-p) \left (d^2-e^2 x^2\right )^{-3+p}}{2 (3-p)}-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac{1}{2} \left (e^2 (10-p)\right ) \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right )-\frac{\left (8 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 \left (1-\frac{e^2 x^2}{d^2}\right )^{-4+p} \, dx}{d^7}\\ &=\frac{e^2 (11-p) \left (d^2-e^2 x^2\right )^{-3+p}}{2 (3-p)}-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{2 x^2}+\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{x}-\frac{8 e^3 (4-p) x \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},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^7}+\frac{e^2 (10-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 [A] time = 0.856493, size = 399, normalized size = 1.89 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (\frac{8 d^3 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )}{(p-1) x^2}+\frac{80 d e^2 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )}{p}+\frac{64 d^2 e \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x}+\frac{5 e^2 2^{p+4} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^2 2^{p+3} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^2 2^{p+1} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{e^2 2^p (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (4-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}\right )}{16 d^7} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.704, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{{x}^{3} \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}}{{\left (e x + d\right )}^{4} x^{3}}\,{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^{4} x^{7} + 4 \, d e^{3} x^{6} + 6 \, d^{2} e^{2} x^{5} + 4 \, d^{3} e x^{4} + d^{4} x^{3}}, 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^{3} \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}}{{\left (e x + d\right )}^{4} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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