Optimal. Leaf size=29 \[ \frac {2}{5} \left (\log \left (-e^x+\frac {1}{x}\right )-\frac {1}{60} \log (2-x-\log (x))\right ) \]
________________________________________________________________________________________
Rubi [F] time = 1.36, 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 {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{150 x \left (1-e^x x\right ) (2-x-\log (x))} \, dx\\ &=\frac {1}{150} \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{x \left (1-e^x x\right ) (2-x-\log (x))} \, dx\\ &=\frac {1}{150} \int \left (\frac {60 (1+x)}{x \left (-1+e^x x\right )}+\frac {-1-121 x+60 x^2+60 x \log (x)}{x (-2+x+\log (x))}\right ) \, dx\\ &=\frac {1}{150} \int \frac {-1-121 x+60 x^2+60 x \log (x)}{x (-2+x+\log (x))} \, dx+\frac {2}{5} \int \frac {1+x}{x \left (-1+e^x x\right )} \, dx\\ &=\frac {1}{150} \int \left (60+\frac {-1-x}{x (-2+x+\log (x))}\right ) \, dx+\frac {2}{5} \int \left (\frac {1}{-1+e^x x}+\frac {1}{x \left (-1+e^x x\right )}\right ) \, dx\\ &=\frac {2 x}{5}+\frac {1}{150} \int \frac {-1-x}{x (-2+x+\log (x))} \, dx+\frac {2}{5} \int \frac {1}{-1+e^x x} \, dx+\frac {2}{5} \int \frac {1}{x \left (-1+e^x x\right )} \, dx\\ &=\frac {2 x}{5}-\frac {1}{150} \log (2-x-\log (x))+\frac {2}{5} \int \frac {1}{-1+e^x x} \, dx+\frac {2}{5} \int \frac {1}{x \left (-1+e^x x\right )} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.37, size = 32, normalized size = 1.10 \begin {gather*} \frac {1}{150} \left (-60 \log (x)+60 \log \left (1-e^x x\right )-\log (2-x-\log (x))\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.67, size = 22, normalized size = 0.76 \begin {gather*} -\frac {1}{150} \, \log \left (x + \log \relax (x) - 2\right ) + \frac {2}{5} \, \log \left (\frac {x e^{x} - 1}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.16, size = 22, normalized size = 0.76 \begin {gather*} \frac {2}{5} \, \log \left (x e^{x} - 1\right ) - \frac {1}{150} \, \log \left (x + \log \relax (x) - 2\right ) - \frac {2}{5} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.06, size = 21, normalized size = 0.72
method | result | size |
risch | \(\frac {2 \ln \left ({\mathrm e}^{x}-\frac {1}{x}\right )}{5}-\frac {\ln \left (x +\ln \relax (x )-2\right )}{150}\) | \(21\) |
norman | \(-\frac {2 \ln \relax (x )}{5}-\frac {\ln \left (x +\ln \relax (x )-2\right )}{150}+\frac {2 \ln \left ({\mathrm e}^{x} x -1\right )}{5}\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.43, size = 22, normalized size = 0.76 \begin {gather*} -\frac {1}{150} \, \log \left (x + \log \relax (x) - 2\right ) + \frac {2}{5} \, \log \left (\frac {x e^{x} - 1}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 2.29, size = 22, normalized size = 0.76 \begin {gather*} \frac {2\,\ln \left (\frac {x\,{\mathrm {e}}^x-1}{x}\right )}{5}-\frac {\ln \left (x+\ln \relax (x)-2\right )}{150} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.39, size = 20, normalized size = 0.69 \begin {gather*} \frac {2 \log {\left (e^{x} - \frac {1}{x} \right )}}{5} - \frac {\log {\left (x + \log {\relax (x )} - 2 \right )}}{150} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________