Optimal. Leaf size=33 \[ \log ^2\left (\frac {1}{5} \left (-1-x^2-x^2 \log (4)+\frac {2 x \log (x)}{5-x}\right )\right ) \]
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Rubi [F] time = 13.14, 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 (-20+104 x-40 x^2+4 x^3+\left (100 x-40 x^2+4 x^3\right ) \log (4)-20 \log (x)\right ) \log \left (\frac {5-x+5 x^2-x^3+\left (5 x^2-x^3\right ) \log (4)-2 x \log (x)}{-25+5 x}\right )}{25-10 x+26 x^2-10 x^3+x^4+\left (25 x^2-10 x^3+x^4\right ) \log (4)+\left (-10 x+2 x^2\right ) \log (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 {4 \left (-\left ((-5+x) \left (1-5 x (1+\log (4))+x^2 (1+\log (4))\right )\right )+5 \log (x)\right ) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx\\ &=4 \int \frac {\left (-\left ((-5+x) \left (1-5 x (1+\log (4))+x^2 (1+\log (4))\right )\right )+5 \log (x)\right ) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx\\ &=4 \int \left (\frac {5 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {10 x^2 (-1-\log (4)) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {x^3 (1+\log (4)) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {x (1+25 (1+\log (4))) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {5 \log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}\right ) \, dx\\ &=20 \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+20 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (1+\log (4))) \int \frac {x^3 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx-(40 (1+\log (4))) \int \frac {x^2 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (26+25 \log (4))) \int \frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx\\ &=20 \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+20 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (1+\log (4))) \int \frac {x^3 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx-(40 (1+\log (4))) \int \frac {x^2 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx+(4 (26+25 \log (4))) \int \frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left ((-5+x) \left (1+x^2 (1+\log (4))\right )+2 x \log (x)\right )} \, dx\\ &=20 \int \frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+20 \int \frac {\log (x) \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(-5+x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )} \, dx+(4 (1+\log (4))) \int \left (\frac {125 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {25 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}+\frac {5 x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}+\frac {x^2 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}\right ) \, dx-(40 (1+\log (4))) \int \left (\frac {25 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {5 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}+\frac {x \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}\right ) \, dx+(4 (26+25 \log (4))) \int \left (\frac {5 \log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{(5-x) \left (5-x+5 x^2 (1+\log (4))-x^3 (1+\log (4))-2 x \log (x)\right )}+\frac {\log \left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right )}{-5+x-5 x^2 (1+\log (4))+x^3 (1+\log (4))+2 x \log (x)}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.06, size = 28, normalized size = 0.85 \begin {gather*} \log ^2\left (\frac {1}{5} \left (-1-x^2 (1+\log (4))-\frac {2 x \log (x)}{-5+x}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 39, normalized size = 1.18 \begin {gather*} \log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \relax (2) + 2 \, x \log \relax (x) + x - 5}{5 \, {\left (x - 5\right )}}\right )^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, {\left (x^{3} - 10 \, x^{2} + 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} \log \relax (2) + 26 \, x - 5 \, \log \relax (x) - 5\right )} \log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \relax (2) + 2 \, x \log \relax (x) + x - 5}{5 \, {\left (x - 5\right )}}\right )}{x^{4} - 10 \, x^{3} + 26 \, x^{2} + 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} \log \relax (2) + 2 \, {\left (x^{2} - 5 \, x\right )} \log \relax (x) - 10 \, x + 25}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {\left (-20 \ln \relax (x )+2 \left (4 x^{3}-40 x^{2}+100 x \right ) \ln \relax (2)+4 x^{3}-40 x^{2}+104 x -20\right ) \ln \left (\frac {-2 x \ln \relax (x )+2 \left (-x^{3}+5 x^{2}\right ) \ln \relax (2)-x^{3}+5 x^{2}-x +5}{5 x -25}\right )}{\left (2 x^{2}-10 x \right ) \ln \relax (x )+2 \left (x^{4}-10 x^{3}+25 x^{2}\right ) \ln \relax (2)+x^{4}-10 x^{3}+26 x^{2}-10 x +25}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 4 \, \int \frac {{\left (x^{3} - 10 \, x^{2} + 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} \log \relax (2) + 26 \, x - 5 \, \log \relax (x) - 5\right )} \log \left (-\frac {x^{3} - 5 \, x^{2} + 2 \, {\left (x^{3} - 5 \, x^{2}\right )} \log \relax (2) + 2 \, x \log \relax (x) + x - 5}{5 \, {\left (x - 5\right )}}\right )}{x^{4} - 10 \, x^{3} + 26 \, x^{2} + 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} \log \relax (2) + 2 \, {\left (x^{2} - 5 \, x\right )} \log \relax (x) - 10 \, x + 25}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.54, size = 43, normalized size = 1.30 \begin {gather*} {\ln \left (-\frac {x-2\,\ln \relax (2)\,\left (5\,x^2-x^3\right )+2\,x\,\ln \relax (x)-5\,x^2+x^3-5}{5\,x-25}\right )}^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.72, size = 39, normalized size = 1.18 \begin {gather*} \log {\left (\frac {- x^{3} + 5 x^{2} - 2 x \log {\relax (x )} - x + \left (- 2 x^{3} + 10 x^{2}\right ) \log {\relax (2 )} + 5}{5 x - 25} \right )}^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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