Optimal. Leaf size=22 \[ \frac {-4-x+\log \left (\frac {x}{2-x}\right )}{1+\log (x)} \]
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Rubi [F] time = 0.96, 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 {-10+4 x+\left (-2+2 x-x^2\right ) \log (x)+(2-x) \log \left (-\frac {x}{-2+x}\right )}{-2 x+x^2+\left (-4 x+2 x^2\right ) \log (x)+\left (-2 x+x^2\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 {10-4 x+\left (2-2 x+x^2\right ) \log (x)+(-2+x) \log \left (-\frac {x}{-2+x}\right )}{(2-x) x (1+\log (x))^2} \, dx\\ &=\int \left (\frac {4}{(-2+x) (1+\log (x))^2}-\frac {10}{(-2+x) x (1+\log (x))^2}-\frac {\left (2-2 x+x^2\right ) \log (x)}{(-2+x) x (1+\log (x))^2}-\frac {\log \left (-\frac {x}{-2+x}\right )}{x (1+\log (x))^2}\right ) \, dx\\ &=4 \int \frac {1}{(-2+x) (1+\log (x))^2} \, dx-10 \int \frac {1}{(-2+x) x (1+\log (x))^2} \, dx-\int \frac {\left (2-2 x+x^2\right ) \log (x)}{(-2+x) x (1+\log (x))^2} \, dx-\int \frac {\log \left (-\frac {x}{-2+x}\right )}{x (1+\log (x))^2} \, dx\\ &=4 \int \frac {1}{(-2+x) (1+\log (x))^2} \, dx-10 \int \frac {1}{(-2+x) x (1+\log (x))^2} \, dx-\int \left (\frac {-2+2 x-x^2}{(-2+x) x (1+\log (x))^2}+\frac {2-2 x+x^2}{(-2+x) x (1+\log (x))}\right ) \, dx-\int \frac {\log \left (-\frac {x}{-2+x}\right )}{x (1+\log (x))^2} \, dx\\ &=4 \int \frac {1}{(-2+x) (1+\log (x))^2} \, dx-10 \int \frac {1}{(-2+x) x (1+\log (x))^2} \, dx-\int \frac {-2+2 x-x^2}{(-2+x) x (1+\log (x))^2} \, dx-\int \frac {2-2 x+x^2}{(-2+x) x (1+\log (x))} \, dx-\int \frac {\log \left (-\frac {x}{-2+x}\right )}{x (1+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 22, normalized size = 1.00 \begin {gather*} -\frac {4+x-\log \left (-\frac {x}{-2+x}\right )}{1+\log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 22, normalized size = 1.00 \begin {gather*} -\frac {x - \log \left (-\frac {x}{x - 2}\right ) + 4}{\log \relax (x) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [C] time = 0.40, size = 27, normalized size = 1.23 \begin {gather*} -\frac {-i \, \pi + x + 5}{\log \relax (x) + 1} - \frac {\log \left (x - 2\right )}{\log \relax (x) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.11, size = 137, normalized size = 6.23
method | result | size |
risch | \(-\frac {\ln \left (x -2\right )}{\ln \relax (x )+1}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -2}\right ) \mathrm {csgn}\left (\frac {i x}{x -2}\right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -2}\right )^{2}+2 i \pi \mathrm {csgn}\left (\frac {i x}{x -2}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i}{x -2}\right ) \mathrm {csgn}\left (\frac {i x}{x -2}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i x}{x -2}\right )^{3}-2 i \pi +2 x +10}{2 \left (\ln \relax (x )+1\right )}\) | \(137\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.51, size = 17, normalized size = 0.77 \begin {gather*} -\frac {x + \log \left (-x + 2\right ) + 5}{\log \relax (x) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.05, size = 22, normalized size = 1.00 \begin {gather*} -\frac {x-\ln \left (-\frac {x}{x-2}\right )+4}{\ln \relax (x)+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.37, size = 22, normalized size = 1.00 \begin {gather*} \frac {- x - 4}{\log {\relax (x )} + 1} + \frac {\log {\left (- \frac {x}{x - 2} \right )}}{\log {\relax (x )} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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