Integrand size = 78, antiderivative size = 29 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {2}{5} \left (\log \left (-e^x+\frac {1}{x}\right )-\frac {1}{60} \log (2-x-\log (x))\right ) \]
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\[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx \]
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Rubi steps \begin{align*} \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{align*}
Time = 0.46 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {1}{150} \left (-60 \log (x)+60 \log \left (1-e^x x\right )-\log (2-x-\log (x))\right ) \]
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Time = 0.68 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {2 \ln \left ({\mathrm e}^{x}-\frac {1}{x}\right )}{5}-\frac {\ln \left (x +\ln \left (x \right )-2\right )}{150}\) | \(21\) |
norman | \(-\frac {2 \ln \left (x \right )}{5}-\frac {\ln \left (x +\ln \left (x \right )-2\right )}{150}+\frac {2 \ln \left ({\mathrm e}^{x} x -1\right )}{5}\) | \(23\) |
parallelrisch | \(-\frac {2 \ln \left (x \right )}{5}-\frac {\ln \left (x +\ln \left (x \right )-2\right )}{150}+\frac {2 \ln \left ({\mathrm e}^{x} x -1\right )}{5}\) | \(23\) |
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=-\frac {1}{150} \, \log \left (x + \log \left (x\right ) - 2\right ) + \frac {2}{5} \, \log \left (\frac {x e^{x} - 1}{x}\right ) \]
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Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {2 \log {\left (e^{x} - \frac {1}{x} \right )}}{5} - \frac {\log {\left (x + \log {\left (x \right )} - 2 \right )}}{150} \]
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Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=-\frac {1}{150} \, \log \left (x + \log \left (x\right ) - 2\right ) + \frac {2}{5} \, \log \left (\frac {x e^{x} - 1}{x}\right ) \]
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {2}{5} \, \log \left (x e^{x} - 1\right ) - \frac {1}{150} \, \log \left (x + \log \left (x\right ) - 2\right ) - \frac {2}{5} \, \log \left (x\right ) \]
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Time = 9.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx=\frac {2\,\ln \left (\frac {x\,{\mathrm {e}}^x-1}{x}\right )}{5}-\frac {\ln \left (x+\ln \left (x\right )-2\right )}{150} \]
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