Optimal. Leaf size=33 \[ (1+x) \left (5-\frac {4+5 (-2 x+\log (5))}{x}+\log \left (\frac {4 x}{x-\log (2)}\right )\right ) \]
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Rubi [A] time = 0.43, antiderivative size = 54, normalized size of antiderivative = 1.64, number of steps used = 7, number of rules used = 5, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {1593, 6742, 1620, 2486, 31} \begin {gather*} 15 x+x \log \left (\frac {4 x}{x-\log (2)}\right )+\log (x)-(1+\log (2)) \log (x-\log (2))+\log (2) \log (x-\log (2))-\frac {4+5 \log (5)}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 1593
Rule 1620
Rule 2486
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 x-15 x^3+\left (4+x+16 x^2\right ) \log (2)+(-5 x+5 \log (2)) \log (5)+\left (-x^3+x^2 \log (2)\right ) \log \left (-\frac {4 x}{-x+\log (2)}\right )}{x^2 (-x+\log (2))} \, dx\\ &=\int \left (\frac {15 x^3-16 x^2 \log (2)-\log (2) (4+5 \log (5))+x \left (4+\log \left (\frac {3125}{2}\right )\right )}{x^2 (x-\log (2))}+\log \left (\frac {4 x}{x-\log (2)}\right )\right ) \, dx\\ &=\int \frac {15 x^3-16 x^2 \log (2)-\log (2) (4+5 \log (5))+x \left (4+\log \left (\frac {3125}{2}\right )\right )}{x^2 (x-\log (2))} \, dx+\int \log \left (\frac {4 x}{x-\log (2)}\right ) \, dx\\ &=x \log \left (\frac {4 x}{x-\log (2)}\right )+\log (2) \int \frac {1}{x-\log (2)} \, dx+\int \left (15+\frac {1}{x}+\frac {-1-\log (2)}{x-\log (2)}+\frac {4+5 \log (5)}{x^2}\right ) \, dx\\ &=15 x-\frac {4+5 \log (5)}{x}+\log (x)+x \log \left (\frac {4 x}{x-\log (2)}\right )+\log (2) \log (x-\log (2))-(1+\log (2)) \log (x-\log (2))\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 41, normalized size = 1.24 \begin {gather*} -\frac {4}{x}+15 x-\frac {5 \log (5)}{x}+\log (x)+x \log \left (\frac {4 x}{x-\log (2)}\right )-\log (x-\log (2)) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 33, normalized size = 1.00 \begin {gather*} \frac {15 \, x^{2} + {\left (x^{2} + x\right )} \log \left (\frac {4 \, x}{x - \log \relax (2)}\right ) - 5 \, \log \relax (5) - 4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 45, normalized size = 1.36 \begin {gather*} x {\left (2 \, \log \relax (2) + 15\right )} - x \log \left (x - \log \relax (2)\right ) + x \log \relax (x) - \frac {5 \, \log \relax (5) + 4}{x} - \log \left (x - \log \relax (2)\right ) + \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.43, size = 45, normalized size = 1.36
method | result | size |
risch | \(x \ln \left (-\frac {4 x}{\ln \relax (2)-x}\right )+\frac {x \ln \relax (x )-\ln \left (x -\ln \relax (2)\right ) x +15 x^{2}-5 \ln \relax (5)-4}{x}\) | \(45\) |
norman | \(\frac {x^{2} \ln \left (-\frac {4 x}{\ln \relax (2)-x}\right )+x \ln \left (-\frac {4 x}{\ln \relax (2)-x}\right )+15 x^{2}-5 \ln \relax (5)-4}{x}\) | \(46\) |
derivativedivides | \(-4 \ln \relax (2) \left (-\frac {\ln \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right ) \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right ) \left (x -\ln \relax (2)\right )}{16 \ln \relax (2)}-\frac {4}{\ln \relax (2)^{2} \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}-\frac {5 \ln \relax (5)}{\ln \relax (2)^{2} \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}-\frac {\ln \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}{4 \ln \relax (2)}-\frac {15 \left (x -\ln \relax (2)\right )}{4 \ln \relax (2)}\right )\) | \(126\) |
default | \(-4 \ln \relax (2) \left (-\frac {\ln \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right ) \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right ) \left (x -\ln \relax (2)\right )}{16 \ln \relax (2)}-\frac {4}{\ln \relax (2)^{2} \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}-\frac {5 \ln \relax (5)}{\ln \relax (2)^{2} \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}-\frac {\ln \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}{4 \ln \relax (2)}-\frac {15 \left (x -\ln \relax (2)\right )}{4 \ln \relax (2)}\right )\) | \(126\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.39, size = 178, normalized size = 5.39 \begin {gather*} -5 \, {\left (\frac {\log \left (x - \log \relax (2)\right )}{\log \relax (2)^{2}} - \frac {\log \relax (x)}{\log \relax (2)^{2}} + \frac {1}{x \log \relax (2)}\right )} \log \relax (5) \log \relax (2) + 5 \, {\left (\frac {\log \left (x - \log \relax (2)\right )}{\log \relax (2)} - \frac {\log \relax (x)}{\log \relax (2)}\right )} \log \relax (5) + 2 \, x \log \relax (2) - {\left (\frac {\log \left (x - \log \relax (2)\right )}{\log \relax (2)} - \frac {\log \relax (x)}{\log \relax (2)}\right )} \log \relax (2) - 4 \, {\left (\frac {\log \left (x - \log \relax (2)\right )}{\log \relax (2)^{2}} - \frac {\log \relax (x)}{\log \relax (2)^{2}} + \frac {1}{x \log \relax (2)}\right )} \log \relax (2) - {\left (x - \log \relax (2)\right )} \log \left (x - \log \relax (2)\right ) - \log \relax (2) \log \left (x - \log \relax (2)\right ) + x \log \relax (x) + 15 \, x + \frac {4 \, \log \left (x - \log \relax (2)\right )}{\log \relax (2)} - \frac {4 \, \log \relax (x)}{\log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.78, size = 49, normalized size = 1.48 \begin {gather*} 15\,x+x\,\ln \left (\frac {x}{x-\ln \relax (2)}\right )+2\,x\,\ln \relax (2)-\frac {5\,\ln \relax (5)}{x}-\frac {4}{x}-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{\ln \relax (2)}-\mathrm {i}\right )\,2{}\mathrm {i} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.38, size = 36, normalized size = 1.09 \begin {gather*} x \log {\left (- \frac {4 x}{- x + \log {\relax (2 )}} \right )} + 15 x + \log {\relax (x )} - \log {\left (x - \log {\relax (2 )} \right )} + \frac {- 5 \log {\relax (5 )} - 4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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