Integrand size = 52, antiderivative size = 30 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\frac {1}{5} \log \left (\left (2 x-4 e^{-x+x \log (2)} \left (-e^x+x\right )\right )^2\right ) \]
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Timed out. \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\text {\$Aborted} \]
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Rubi steps Aborted
Time = 3.52 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=-\frac {2 x}{5}+\frac {2}{5} \log \left (2^{1+x} e^x-2^{1+x} x+e^x x\right ) \]
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Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(\frac {2 x \ln \left (2\right )}{5}-\frac {2 x}{5}+\frac {2 \ln \left (x \,{\mathrm e}^{-x \ln \left (2\right )+x}+2 \,{\mathrm e}^{x}-2 x \right )}{5}\) | \(31\) |
norman | \(\left (\frac {2 \ln \left (2\right )}{5}-\frac {2}{5}\right ) x +\frac {2 \ln \left (10 \,{\mathrm e}^{x}+5 x \,{\mathrm e}^{-x \ln \left (2\right )+x}-10 x \right )}{5}\) | \(32\) |
risch | \(\frac {2 \ln \left (x \right )}{5}+\frac {2 x \ln \left (2\right )}{5}-\frac {2 x}{5}+\frac {2 \ln \left (\left (\frac {1}{2}\right )^{x} {\mathrm e}^{x}-\frac {2 \left (x -{\mathrm e}^{x}\right )}{x}\right )}{5}\) | \(35\) |
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Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\frac {2}{5} \, x \log \left (2\right ) - \frac {2}{5} \, x + \frac {2}{5} \, \log \left (x e^{\left (-x \log \left (2\right ) + x\right )} - 2 \, x + 2 \, e^{x}\right ) \]
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Time = 0.70 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=- \frac {2 x}{5} + \frac {2 x \log {\left (2 \right )}}{5} + \frac {2 \log {\left (x e^{x} e^{- x \log {\left (2 \right )}} - 2 x + 2 e^{x} \right )}}{5} \]
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Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=-\frac {2}{5} \, x + \frac {2}{5} \, \log \left (-x + e^{x}\right ) + \frac {2}{5} \, \log \left (\frac {2 \cdot 2^{x} {\left (x - e^{x}\right )} - x e^{x}}{2 \, {\left (x - e^{x}\right )}}\right ) \]
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Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\frac {2}{5} \, x \log \left (2\right ) - \frac {2}{5} \, x + \frac {2}{5} \, \log \left (x e^{\left (-x \log \left (2\right ) + x\right )} - 2 \, x + 2 \, e^{x}\right ) \]
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Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-4+2 e^{x-x \log (2)}+4 x+4 e^x \log (2)-4 x \log (2)}{10 e^x-10 x+5 e^{x-x \log (2)} x} \, dx=\frac {2\,\ln \left (2\,{\mathrm {e}}^x-2\,x+\frac {x\,{\mathrm {e}}^x}{2^x}\right )}{5}+x\,\left (\frac {\ln \left (4\right )}{5}-\frac {2}{5}\right ) \]
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