Optimal. Leaf size=27 \[ \frac {e^2 x^2}{-3-x+\frac {4+x}{x}+\frac {\log (x)}{2}} \]
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Rubi [F] time = 0.83, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^2 \left (48 x^2-18 x^3-4 x^4\right )+4 e^2 x^3 \log (x)}{64-64 x-16 x^2+16 x^3+4 x^4+\left (16 x-8 x^2-4 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^2 x^2 \left (24-9 x-2 x^2+2 x \log (x)\right )}{\left (8-4 x-2 x^2+x \log (x)\right )^2} \, dx\\ &=\left (2 e^2\right ) \int \frac {x^2 \left (24-9 x-2 x^2+2 x \log (x)\right )}{\left (8-4 x-2 x^2+x \log (x)\right )^2} \, dx\\ &=\left (2 e^2\right ) \int \left (\frac {x^2 \left (8-x+2 x^2\right )}{\left (-8+4 x+2 x^2-x \log (x)\right )^2}-\frac {2 x^2}{-8+4 x+2 x^2-x \log (x)}\right ) \, dx\\ &=\left (2 e^2\right ) \int \frac {x^2 \left (8-x+2 x^2\right )}{\left (-8+4 x+2 x^2-x \log (x)\right )^2} \, dx-\left (4 e^2\right ) \int \frac {x^2}{-8+4 x+2 x^2-x \log (x)} \, dx\\ &=\left (2 e^2\right ) \int \left (\frac {8 x^2}{\left (-8+4 x+2 x^2-x \log (x)\right )^2}-\frac {x^3}{\left (-8+4 x+2 x^2-x \log (x)\right )^2}+\frac {2 x^4}{\left (-8+4 x+2 x^2-x \log (x)\right )^2}\right ) \, dx-\left (4 e^2\right ) \int \frac {x^2}{-8+4 x+2 x^2-x \log (x)} \, dx\\ &=-\left (\left (2 e^2\right ) \int \frac {x^3}{\left (-8+4 x+2 x^2-x \log (x)\right )^2} \, dx\right )+\left (4 e^2\right ) \int \frac {x^4}{\left (-8+4 x+2 x^2-x \log (x)\right )^2} \, dx-\left (4 e^2\right ) \int \frac {x^2}{-8+4 x+2 x^2-x \log (x)} \, dx+\left (16 e^2\right ) \int \frac {x^2}{\left (-8+4 x+2 x^2-x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.35, size = 24, normalized size = 0.89 \begin {gather*} \frac {2 e^2 x^3}{8-4 x-2 x^2+x \log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 24, normalized size = 0.89 \begin {gather*} -\frac {2 \, x^{3} e^{2}}{2 \, x^{2} - x \log \relax (x) + 4 \, x - 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 24, normalized size = 0.89 \begin {gather*} -\frac {2 \, x^{3} e^{2}}{2 \, x^{2} - x \log \relax (x) + 4 \, x - 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 25, normalized size = 0.93
method | result | size |
norman | \(-\frac {2 x^{3} {\mathrm e}^{2}}{2 x^{2}-x \ln \relax (x )+4 x -8}\) | \(25\) |
risch | \(-\frac {2 x^{3} {\mathrm e}^{2}}{2 x^{2}-x \ln \relax (x )+4 x -8}\) | \(25\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 24, normalized size = 0.89 \begin {gather*} -\frac {2 \, x^{3} e^{2}}{2 \, x^{2} - x \log \relax (x) + 4 \, x - 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.98, size = 24, normalized size = 0.89 \begin {gather*} -\frac {2\,x^3\,{\mathrm {e}}^2}{4\,x-x\,\ln \relax (x)+2\,x^2-8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.15, size = 22, normalized size = 0.81 \begin {gather*} \frac {2 x^{3} e^{2}}{- 2 x^{2} + x \log {\relax (x )} - 4 x + 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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