Optimal. Leaf size=25 \[ \frac {x \log (x)}{5 \left (4+\frac {4 (2+3 x-\log (6))}{x}\right )} \]
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Rubi [A] time = 0.33, antiderivative size = 32, normalized size of antiderivative = 1.28, number of steps used = 12, number of rules used = 9, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.155, Rules used = {6, 6688, 12, 6742, 43, 2357, 2295, 2314, 31} \begin {gather*} \frac {1}{80} x \log (x)-\frac {x (2-\log (6)) \log (x)}{80 (4 x+2-\log (6))} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 31
Rule 43
Rule 2295
Rule 2314
Rule 2357
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x^2+x (2-\log (6))+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx\\ &=\int \frac {x \left (4 x+2 \left (1-\frac {\log (6)}{2}\right )+(4+4 x-2 \log (6)) \log (x)\right )}{20 (2+4 x-\log (6))^2} \, dx\\ &=\frac {1}{20} \int \frac {x \left (4 x+2 \left (1-\frac {\log (6)}{2}\right )+(4+4 x-2 \log (6)) \log (x)\right )}{(2+4 x-\log (6))^2} \, dx\\ &=\frac {1}{20} \int \left (\frac {x}{2+4 x-\log (6)}+\frac {2 x (2+2 x-\log (6)) \log (x)}{(2+4 x-\log (6))^2}\right ) \, dx\\ &=\frac {1}{20} \int \frac {x}{2+4 x-\log (6)} \, dx+\frac {1}{10} \int \frac {x (2+2 x-\log (6)) \log (x)}{(2+4 x-\log (6))^2} \, dx\\ &=\frac {1}{20} \int \left (\frac {1}{4}+\frac {-2+\log (6)}{4 (2+4 x-\log (6))}\right ) \, dx+\frac {1}{10} \int \left (\frac {\log (x)}{8}-\frac {(-2+\log (6))^2 \log (x)}{8 (-2-4 x+\log (6))^2}\right ) \, dx\\ &=\frac {x}{80}-\frac {1}{320} (2-\log (6)) \log (2+4 x-\log (6))+\frac {1}{80} \int \log (x) \, dx-\frac {1}{80} (2-\log (6))^2 \int \frac {\log (x)}{(-2-4 x+\log (6))^2} \, dx\\ &=\frac {1}{80} x \log (x)-\frac {x (2-\log (6)) \log (x)}{80 (2+4 x-\log (6))}-\frac {1}{320} (2-\log (6)) \log (2+4 x-\log (6))+\frac {1}{80} (-2+\log (6)) \int \frac {1}{-2-4 x+\log (6)} \, dx\\ &=\frac {1}{80} x \log (x)-\frac {x (2-\log (6)) \log (x)}{80 (2+4 x-\log (6))}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 20, normalized size = 0.80 \begin {gather*} \frac {x^2 \log (x)}{20 (2+4 x-\log (6))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 18, normalized size = 0.72 \begin {gather*} \frac {x^{2} \log \relax (x)}{20 \, {\left (4 \, x - \log \relax (6) + 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.28, size = 39, normalized size = 1.56 \begin {gather*} \frac {1}{320} \, {\left (4 \, x + \frac {\log \relax (6)^{2} - 4 \, \log \relax (6) + 4}{4 \, x - \log \relax (6) + 2}\right )} \log \relax (x) + \frac {1}{320} \, {\left (\log \relax (6) - 2\right )} \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 17, normalized size = 0.68
method | result | size |
norman | \(-\frac {x^{2} \ln \relax (x )}{20 \left (\ln \relax (6)-4 x -2\right )}\) | \(17\) |
risch | \(-\frac {\left (\ln \relax (2)^{2}+2 \ln \relax (2) \ln \relax (3)-4 x \ln \relax (2)+\ln \relax (3)^{2}-4 x \ln \relax (3)+16 x^{2}-4 \ln \relax (2)-4 \ln \relax (3)+8 x +4\right ) \ln \relax (x )}{320 \left (\ln \relax (2)+\ln \relax (3)-4 x -2\right )}-\frac {\ln \relax (x )}{160}+\frac {\ln \relax (2) \ln \relax (x )}{320}+\frac {\ln \relax (3) \ln \relax (x )}{320}\) | \(75\) |
default | \(\frac {\ln \left (-\ln \relax (6)+4 x +2\right ) \ln \relax (6)}{320}-\frac {\ln \left (-\ln \relax (6)+4 x +2\right )}{160}+\frac {x \ln \relax (x )}{80}-\frac {\ln \relax (6)^{2} \ln \left (\ln \relax (6)-4 x -2\right )}{320 \left (\ln \relax (6)-2\right )}-\frac {\ln \relax (6)^{2} \ln \relax (x ) x}{80 \left (\ln \relax (6)-2\right ) \left (\ln \relax (6)-4 x -2\right )}+\frac {\ln \relax (6) \ln \left (\ln \relax (6)-4 x -2\right )}{80 \ln \relax (6)-160}+\frac {\ln \relax (6) \ln \relax (x ) x}{20 \left (\ln \relax (6)-2\right ) \left (\ln \relax (6)-4 x -2\right )}-\frac {\ln \left (\ln \relax (6)-4 x -2\right )}{80 \left (\ln \relax (6)-2\right )}-\frac {\ln \relax (x ) x}{20 \left (\ln \relax (6)-2\right ) \left (\ln \relax (6)-4 x -2\right )}\) | \(153\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.81, size = 209, normalized size = 8.36 \begin {gather*} \frac {1}{320} \, {\left (\frac {\log \relax (6) - 2}{4 \, x - \log \relax (6) + 2} - \log \left (4 \, x - \log \relax (6) + 2\right )\right )} \log \relax (6) + \frac {1}{160} \, {\left (\log \relax (6) - 2\right )} \log \left (4 \, x - \log \relax (6) + 2\right ) - \frac {1}{320} \, {\left (\log \relax (3) + \log \relax (2) - 2\right )} \log \left (4 \, x - \log \relax (3) - \log \relax (2) + 2\right ) + \frac {1}{320} \, {\left (\log \relax (3) + \log \relax (2) - 2\right )} \log \relax (x) + \frac {1}{80} \, x - \frac {\log \relax (6)^{2} - 4 \, \log \relax (6) + 4}{320 \, {\left (4 \, x - \log \relax (6) + 2\right )}} - \frac {16 \, x^{2} - 4 \, x {\left (\log \relax (3) + \log \relax (2) - 2\right )} - {\left (16 \, x^{2} - 4 \, x {\left (\log \relax (3) + \log \relax (2) - 2\right )} + \log \relax (3)^{2} + 2 \, {\left (\log \relax (3) - 2\right )} \log \relax (2) + \log \relax (2)^{2} - 4 \, \log \relax (3) + 4\right )} \log \relax (x)}{320 \, {\left (4 \, x - \log \relax (3) - \log \relax (2) + 2\right )}} - \frac {\log \relax (6) - 2}{160 \, {\left (4 \, x - \log \relax (6) + 2\right )}} + \frac {1}{160} \, \log \left (4 \, x - \log \relax (6) + 2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.26, size = 135, normalized size = 5.40 \begin {gather*} \frac {\ln \left (\ln \relax (6)-4\,x-2\right )}{160}-\frac {\frac {{\ln \relax (6)}^2}{4}-\ln \relax (6)+1}{320\,x-80\,\ln \relax (6)+160}+\ln \left (\ln \relax (6)-4\,x-2\right )\,\left (\frac {\ln \relax (6)}{160}-\frac {1}{80}\right )-\frac {\ln \left (\ln \relax (6)-4\,x-2\right )\,\ln \relax (6)}{320}-\frac {\frac {\ln \relax (6)}{16}-\frac {1}{8}}{40\,x-10\,\ln \relax (6)+20}-\ln \left (4\,x-\ln \relax (6)+2\right )\,\left (\frac {\ln \relax (6)}{320}-\frac {1}{160}\right )+\frac {x^2\,\ln \relax (x)}{20\,\left (4\,x-\ln \relax (6)+2\right )}+\frac {\ln \relax (6)\,\left (\ln \relax (6)-2\right )}{16\,\left (80\,x-20\,\ln \relax (6)+40\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.28, size = 51, normalized size = 2.04 \begin {gather*} \left (- \frac {1}{160} + \frac {\log {\relax (6 )}}{320}\right ) \log {\relax (x )} + \frac {\left (16 x^{2} - 4 x \log {\relax (6 )} + 8 x - 4 \log {\relax (6 )} + \log {\relax (6 )}^{2} + 4\right ) \log {\relax (x )}}{1280 x - 320 \log {\relax (6 )} + 640} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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