Optimal. Leaf size=26 \[ x+x^2-\frac {x \log (x)}{-1+\log ^2\left (\frac {x}{5-x}\right )} \]
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Rubi [F] time = 0.92, 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 {-10-8 x+2 x^2+\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )+\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )+\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{-5+x+(10-2 x) \log ^2\left (-\frac {x}{-5+x}\right )+(-5+x) \log ^4\left (-\frac {x}{-5+x}\right )} \, 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+8 x-2 x^2-\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )-\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )-\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{(5-x) \left (1-\log ^2\left (-\frac {x}{-5+x}\right )\right )^2} \, dx\\ &=\int \left (1+2 x-\frac {5 \log (x)}{2 (-5+x) \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )^2}+\frac {-1-\log (x)}{2 \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )}+\frac {5 \log (x)}{2 (-5+x) \left (1+\log \left (-\frac {x}{-5+x}\right )\right )^2}+\frac {1+\log (x)}{2 \left (1+\log \left (-\frac {x}{-5+x}\right )\right )}\right ) \, dx\\ &=x+x^2+\frac {1}{2} \int \frac {-1-\log (x)}{-1+\log \left (-\frac {x}{-5+x}\right )} \, dx+\frac {1}{2} \int \frac {1+\log (x)}{1+\log \left (-\frac {x}{-5+x}\right )} \, dx-\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx+\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx\\ &=x+x^2+\frac {1}{2} \int \left (\frac {1}{1-\log \left (-\frac {x}{-5+x}\right )}-\frac {\log (x)}{-1+\log \left (-\frac {x}{-5+x}\right )}\right ) \, dx+\frac {1}{2} \int \left (\frac {1}{1+\log \left (-\frac {x}{-5+x}\right )}+\frac {\log (x)}{1+\log \left (-\frac {x}{-5+x}\right )}\right ) \, dx-\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx+\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx\\ &=x+x^2+\frac {1}{2} \int \frac {1}{1-\log \left (-\frac {x}{-5+x}\right )} \, dx-\frac {1}{2} \int \frac {\log (x)}{-1+\log \left (-\frac {x}{-5+x}\right )} \, dx+\frac {1}{2} \int \frac {1}{1+\log \left (-\frac {x}{-5+x}\right )} \, dx+\frac {1}{2} \int \frac {\log (x)}{1+\log \left (-\frac {x}{-5+x}\right )} \, dx-\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (-1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx+\frac {5}{2} \int \frac {\log (x)}{(-5+x) \left (1+\log \left (-\frac {x}{-5+x}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 25, normalized size = 0.96 \begin {gather*} x+x^2-\frac {x \log (x)}{-1+\log ^2\left (-\frac {x}{-5+x}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 47, normalized size = 1.81 \begin {gather*} \frac {{\left (x^{2} + x\right )} \log \left (-\frac {x}{x - 5}\right )^{2} - x^{2} - x \log \relax (x) - x}{\log \left (-\frac {x}{x - 5}\right )^{2} - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [C] time = 1.24, size = 50, normalized size = 1.92 \begin {gather*} x^{2} + x + \frac {x \log \relax (x)}{\pi ^{2} + 2 i \, \pi \log \left (x - 5\right ) - \log \left (x - 5\right )^{2} - 2 i \, \pi \log \relax (x) + 2 \, \log \left (x - 5\right ) \log \relax (x) - \log \relax (x)^{2} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 3.43, size = 740, normalized size = 28.46
method | result | size |
risch | \(x^{2}+x -\frac {4 x \ln \relax (x )}{-4+4 \ln \left (x -5\right )^{2}+4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )-\pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right )^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+2 \pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}+2 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -5}\right )^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-8 i \ln \relax (x ) \pi \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+8 i \ln \left (x -5\right ) \pi \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-4 \pi ^{2}+4 \ln \relax (x )^{2}-8 \ln \relax (x ) \ln \left (x -5\right )+4 i \ln \relax (x ) \pi \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-4 i \ln \left (x -5\right ) \pi \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}+4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}+4 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}+8 i \pi \ln \relax (x )-8 i \pi \ln \left (x -5\right )-4 i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )+4 i \ln \left (x -5\right ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )-4 \pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}+4 \pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{5}-4 \pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-\pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{6}+8 \pi ^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+4 i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+4 i \ln \relax (x ) \pi \,\mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-4 i \ln \left (x -5\right ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-4 i \ln \left (x -5\right ) \pi \,\mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-2 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{5}-4 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-4 \pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-\pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}-2 \pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{5}-\pi ^{2} \mathrm {csgn}\left (\frac {i}{x -5}\right )^{2} \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{4}}\) | \(740\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 80, normalized size = 3.08 \begin {gather*} \frac {{\left (x^{2} + x\right )} \log \relax (x)^{2} - 2 \, {\left (x^{2} + x\right )} \log \relax (x) \log \left (-x + 5\right ) + {\left (x^{2} + x\right )} \log \left (-x + 5\right )^{2} - x^{2} - x \log \relax (x) - x}{\log \relax (x)^{2} - 2 \, \log \relax (x) \log \left (-x + 5\right ) + \log \left (-x + 5\right )^{2} - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.45, size = 37, normalized size = 1.42 \begin {gather*} x+x^2+\frac {\ln \relax (x)\,\left (5\,x-x^2\right )}{\left ({\ln \left (-\frac {x}{x-5}\right )}^2-1\right )\,\left (x-5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.29, size = 20, normalized size = 0.77 \begin {gather*} x^{2} + x - \frac {x \log {\relax (x )}}{\log {\left (- \frac {x}{x - 5} \right )}^{2} - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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