Optimal. Leaf size=265 \[ -\frac{4 \left (p^2+15 p+16\right ) x^5 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{5}{2},4-p;\frac{7}{2};\frac{e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )}+\frac{d^2 (12 p+13) x^5 \left (d^2-e^2 x^2\right )^{p-3}}{1-4 p^2}-\frac{e^2 x^7 \left (d^2-e^2 x^2\right )^{p-3}}{2 p+1}-\frac{4 d^7 \left (d^2-e^2 x^2\right )^{p-3}}{e^5 (3-p)}+\frac{10 d^5 \left (d^2-e^2 x^2\right )^{p-2}}{e^5 (2-p)}-\frac{8 d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^5 (1-p)}-\frac{2 d \left (d^2-e^2 x^2\right )^p}{e^5 p} \]
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Rubi [A] time = 0.305566, antiderivative size = 265, 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, 1267, 459, 365, 364, 446, 77} \[ -\frac{4 \left (p^2+15 p+16\right ) x^5 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{5}{2},4-p;\frac{7}{2};\frac{e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )}+\frac{d^2 (12 p+13) x^5 \left (d^2-e^2 x^2\right )^{p-3}}{1-4 p^2}-\frac{e^2 x^7 \left (d^2-e^2 x^2\right )^{p-3}}{2 p+1}-\frac{4 d^7 \left (d^2-e^2 x^2\right )^{p-3}}{e^5 (3-p)}+\frac{10 d^5 \left (d^2-e^2 x^2\right )^{p-2}}{e^5 (2-p)}-\frac{8 d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^5 (1-p)}-\frac{2 d \left (d^2-e^2 x^2\right )^p}{e^5 p} \]
Antiderivative was successfully verified.
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Rule 852
Rule 1652
Rule 1267
Rule 459
Rule 365
Rule 364
Rule 446
Rule 77
Rubi steps
\begin{align*} \int \frac{x^4 \left (d^2-e^2 x^2\right )^p}{(d+e x)^4} \, dx &=\int x^4 (d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p} \, dx\\ &=\int x^5 \left (d^2-e^2 x^2\right )^{-4+p} \left (-4 d^3 e-4 d e^3 x^2\right ) \, dx+\int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \left (d^4+6 d^2 e^2 x^2+e^4 x^4\right ) \, dx\\ &=-\frac{e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac{1}{2} \operatorname{Subst}\left (\int x^2 \left (d^2-e^2 x\right )^{-4+p} \left (-4 d^3 e-4 d e^3 x\right ) \, dx,x,x^2\right )-\frac{\int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \left (-d^4 e^2 (1+2 p)-d^2 e^4 (13+12 p) x^2\right ) \, dx}{e^2 (1+2 p)}\\ &=\frac{d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac{e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{8 d^7 \left (d^2-e^2 x\right )^{-4+p}}{e^3}+\frac{20 d^5 \left (d^2-e^2 x\right )^{-3+p}}{e^3}-\frac{16 d^3 \left (d^2-e^2 x\right )^{-2+p}}{e^3}+\frac{4 d \left (d^2-e^2 x\right )^{-1+p}}{e^3}\right ) \, dx,x,x^2\right )-\frac{\left (4 d^4 \left (16+15 p+p^2\right )\right ) \int x^4 \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{1-4 p^2}\\ &=-\frac{4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac{d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac{e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac{10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac{8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac{2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac{\left (4 \left (16+15 p+p^2\right ) \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p}\right ) \int x^4 \left (1-\frac{e^2 x^2}{d^2}\right )^{-4+p} \, dx}{d^4 \left (1-4 p^2\right )}\\ &=-\frac{4 d^7 \left (d^2-e^2 x^2\right )^{-3+p}}{e^5 (3-p)}+\frac{d^2 (13+12 p) x^5 \left (d^2-e^2 x^2\right )^{-3+p}}{1-4 p^2}-\frac{e^2 x^7 \left (d^2-e^2 x^2\right )^{-3+p}}{1+2 p}+\frac{10 d^5 \left (d^2-e^2 x^2\right )^{-2+p}}{e^5 (2-p)}-\frac{8 d^3 \left (d^2-e^2 x^2\right )^{-1+p}}{e^5 (1-p)}-\frac{2 d \left (d^2-e^2 x^2\right )^p}{e^5 p}-\frac{4 \left (16+15 p+p^2\right ) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{5}{2},4-p;\frac{7}{2};\frac{e^2 x^2}{d^2}\right )}{5 d^4 \left (1-4 p^2\right )}\\ \end{align*}
Mathematica [C] time = 0.15664, size = 66, normalized size = 0.25 \[ \frac{x^5 (d-e x)^p (d+e x)^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} F_1\left (5;-p,4-p;6;\frac{e x}{d},-\frac{e x}{d}\right )}{5 d^4} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.722, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{4} \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^{4}}{{\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^{4}}{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^{4} \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^{4}}{{\left (e x + d\right )}^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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