Optimal. Leaf size=24 \[ \frac {64 (x-\log (2)) \log ^2((-4+x) (3+x))}{x+\log (x)} \]
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Rubi [F] time = 6.40, 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 {\left (-128 x^3+256 x^4+\left (128 x^2-256 x^3\right ) \log (2)+\left (-128 x^2+256 x^3+\left (128 x-256 x^2\right ) \log (2)\right ) \log (x)\right ) \log \left (-12-x+x^2\right )+\left (768 x+64 x^2-64 x^3+\left (-768-832 x+64 x^3\right ) \log (2)+\left (-768 x-64 x^2+64 x^3\right ) \log (x)\right ) \log ^2\left (-12-x+x^2\right )}{-12 x^3-x^4+x^5+\left (-24 x^2-2 x^3+2 x^4\right ) \log (x)+\left (-12 x-x^2+x^3\right ) \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 {64 \log \left (-12-x+x^2\right ) \left (-x^2 (-1+2 x) (2 x-\log (4))-\left (-12-x+x^2\right ) (x (-1+\log (2))+\log (2)) \log \left (-12-x+x^2\right )-x \log (x) \left ((-1+2 x) (2 x-\log (4))+\left (-12-x+x^2\right ) \log \left (-12-x+x^2\right )\right )\right )}{x \left (12+x-x^2\right ) (x+\log (x))^2} \, dx\\ &=64 \int \frac {\log \left (-12-x+x^2\right ) \left (-x^2 (-1+2 x) (2 x-\log (4))-\left (-12-x+x^2\right ) (x (-1+\log (2))+\log (2)) \log \left (-12-x+x^2\right )-x \log (x) \left ((-1+2 x) (2 x-\log (4))+\left (-12-x+x^2\right ) \log \left (-12-x+x^2\right )\right )\right )}{x \left (12+x-x^2\right ) (x+\log (x))^2} \, dx\\ &=64 \int \left (\frac {(-1+2 x) (2 x-\log (4)) \log \left (-12-x+x^2\right )}{(-4+x) (3+x) (x+\log (x))}+\frac {(-x (1-\log (2))+\log (2)+x \log (x)) \log ^2\left (-12-x+x^2\right )}{x (x+\log (x))^2}\right ) \, dx\\ &=64 \int \frac {(-1+2 x) (2 x-\log (4)) \log \left (-12-x+x^2\right )}{(-4+x) (3+x) (x+\log (x))} \, dx+64 \int \frac {(-x (1-\log (2))+\log (2)+x \log (x)) \log ^2\left (-12-x+x^2\right )}{x (x+\log (x))^2} \, dx\\ &=64 \int \left (\frac {4 \log \left (-12-x+x^2\right )}{x+\log (x)}-\frac {(-8+\log (4)) \log \left (-12-x+x^2\right )}{(-4+x) (x+\log (x))}-\frac {(6+\log (4)) \log \left (-12-x+x^2\right )}{(3+x) (x+\log (x))}\right ) \, dx+64 \int \left (\frac {(-1+\log (2)) \log ^2\left (-12-x+x^2\right )}{(x+\log (x))^2}+\frac {\log (2) \log ^2\left (-12-x+x^2\right )}{x (x+\log (x))^2}+\frac {\log (x) \log ^2\left (-12-x+x^2\right )}{(x+\log (x))^2}\right ) \, dx\\ &=64 \int \frac {\log (x) \log ^2\left (-12-x+x^2\right )}{(x+\log (x))^2} \, dx+256 \int \frac {\log \left (-12-x+x^2\right )}{x+\log (x)} \, dx-(64 (1-\log (2))) \int \frac {\log ^2\left (-12-x+x^2\right )}{(x+\log (x))^2} \, dx+(64 \log (2)) \int \frac {\log ^2\left (-12-x+x^2\right )}{x (x+\log (x))^2} \, dx+(64 (8-\log (4))) \int \frac {\log \left (-12-x+x^2\right )}{(-4+x) (x+\log (x))} \, dx-(64 (6+\log (4))) \int \frac {\log \left (-12-x+x^2\right )}{(3+x) (x+\log (x))} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 0.19, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-128 x^3+256 x^4+\left (128 x^2-256 x^3\right ) \log (2)+\left (-128 x^2+256 x^3+\left (128 x-256 x^2\right ) \log (2)\right ) \log (x)\right ) \log \left (-12-x+x^2\right )+\left (768 x+64 x^2-64 x^3+\left (-768-832 x+64 x^3\right ) \log (2)+\left (-768 x-64 x^2+64 x^3\right ) \log (x)\right ) \log ^2\left (-12-x+x^2\right )}{-12 x^3-x^4+x^5+\left (-24 x^2-2 x^3+2 x^4\right ) \log (x)+\left (-12 x-x^2+x^3\right ) \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.46, size = 25, normalized size = 1.04 \begin {gather*} \frac {64 \, {\left (x - \log \relax (2)\right )} \log \left (x^{2} - x - 12\right )^{2}}{x + \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 25, normalized size = 1.04 \begin {gather*} \frac {64 \, {\left (x - \log \relax (2)\right )} \log \left (x^{2} - x - 12\right )^{2}}{x + \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 26, normalized size = 1.08
method | result | size |
risch | \(-\frac {64 \left (\ln \relax (2)-x \right ) \ln \left (x^{2}-x -12\right )^{2}}{x +\ln \relax (x )}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.49, size = 51, normalized size = 2.12 \begin {gather*} \frac {64 \, {\left ({\left (x - \log \relax (2)\right )} \log \left (x + 3\right )^{2} + 2 \, {\left (x - \log \relax (2)\right )} \log \left (x + 3\right ) \log \left (x - 4\right ) + {\left (x - \log \relax (2)\right )} \log \left (x - 4\right )^{2}\right )}}{x + \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.32, size = 67, normalized size = 2.79 \begin {gather*} -{\ln \left (x^2-x-12\right )}^2\,\left (\frac {\frac {64\,x}{x+1}+\ln \relax (x)\,\left (\frac {64\,x}{x+1}+\frac {64}{x+1}\right )+\frac {64\,\left (\ln \relax (2)-x+x\,\ln \relax (2)\right )}{x+1}}{x+\ln \relax (x)}-64\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.52, size = 22, normalized size = 0.92 \begin {gather*} \frac {\left (64 x - 64 \log {\relax (2 )}\right ) \log {\left (x^{2} - x - 12 \right )}^{2}}{x + \log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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