Integrand size = 113, antiderivative size = 25 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{x \left (20+\log \left (\frac {1}{x}-\log \left (\frac {e^{-x} \log (2)}{x}\right )\right )\right )} \]
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Time = 1.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6838} \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{20 x} \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )^x \]
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Rule 6838
Rubi steps \begin{align*} \text {integral}& = e^{20 x} \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )^x \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{20 x} \left (\frac {1}{x}-\log \left (\frac {e^{-x} \log (2)}{x}\right )\right )^x \]
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Time = 2.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \({\mathrm e}^{x \left (\ln \left (-\frac {x \ln \left (\frac {\ln \left (2\right ) {\mathrm e}^{-x}}{x}\right )-1}{x}\right )+20\right )}\) | \(27\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{\left (x \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1}{x}\right ) + 20 \, x\right )} \]
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Timed out. \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=\text {Timed out} \]
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Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=e^{\left (x \log \left (x^{2} + x \log \left (x\right ) - x \log \left (\log \left (2\right )\right ) + 1\right ) - x \log \left (x\right ) + 20 \, x\right )} \]
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\[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx=\int { -\frac {{\left (x^{2} - 20 \, x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - {\left (x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1\right )} \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1}{x}\right ) + x + 19\right )} e^{\left (x \log \left (-\frac {x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1}{x}\right ) + 20 \, x\right )}}{x \log \left (\frac {e^{\left (-x\right )} \log \left (2\right )}{x}\right ) - 1} \,d x } \]
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Time = 11.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{20 x+x \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )} \left (-19-x-x^2+20 x \log \left (\frac {e^{-x} \log (2)}{x}\right )+\left (-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )\right ) \log \left (\frac {1-x \log \left (\frac {e^{-x} \log (2)}{x}\right )}{x}\right )\right )}{-1+x \log \left (\frac {e^{-x} \log (2)}{x}\right )} \, dx={\mathrm {e}}^{20\,x}\,{\left (x-\ln \left (\frac {\ln \left (2\right )}{x}\right )+\frac {1}{x}\right )}^x \]
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