Integrand size = 82, antiderivative size = 27 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=-e^{-4+e^{2 x}} x \log (x \log (5))+\log \left (-e^x+\log (x)\right ) \]
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\[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=\int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+e^x x-e^{-4+e^{2 x}} x \left (e^x-\log (x)\right ) \left (1+\left (1+2 e^{2 x} x\right ) \log (x \log (5))\right )}{x \left (e^x-\log (x)\right )} \, dx \\ & = \int \left (\frac {-1+x \log (x)}{x \left (e^x-\log (x)\right )}-2 e^{-4+e^{2 x}+2 x} x \log (x \log (5))+\frac {e^4-e^{e^{2 x}}-e^{e^{2 x}} \log (x \log (5))}{e^4}\right ) \, dx \\ & = -\left (2 \int e^{-4+e^{2 x}+2 x} x \log (x \log (5)) \, dx\right )+\frac {\int \left (e^4-e^{e^{2 x}}-e^{e^{2 x}} \log (x \log (5))\right ) \, dx}{e^4}+\int \frac {-1+x \log (x)}{x \left (e^x-\log (x)\right )} \, dx \\ & = x+2 \int \frac {\int e^{-4+e^{2 x}+2 x} x \, dx}{x} \, dx-\frac {\int e^{e^{2 x}} \, dx}{e^4}-\frac {\int e^{e^{2 x}} \log (x \log (5)) \, dx}{e^4}-(2 \log (x \log (5))) \int e^{-4+e^{2 x}+2 x} x \, dx+\int \left (-\frac {1}{x \left (e^x-\log (x)\right )}+\frac {\log (x)}{e^x-\log (x)}\right ) \, dx \\ & = x-\frac {\operatorname {ExpIntegralEi}\left (e^{2 x}\right ) \log (x \log (5))}{2 e^4}+2 \int \frac {\int e^{-4+e^{2 x}+2 x} x \, dx}{x} \, dx-\frac {\text {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^{2 x}\right )}{2 e^4}+\frac {\int \frac {\operatorname {ExpIntegralEi}\left (e^{2 x}\right )}{2 x} \, dx}{e^4}-(2 \log (x \log (5))) \int e^{-4+e^{2 x}+2 x} x \, dx-\int \frac {1}{x \left (e^x-\log (x)\right )} \, dx+\int \frac {\log (x)}{e^x-\log (x)} \, dx \\ & = x-\frac {\operatorname {ExpIntegralEi}\left (e^{2 x}\right )}{2 e^4}-\frac {\operatorname {ExpIntegralEi}\left (e^{2 x}\right ) \log (x \log (5))}{2 e^4}+2 \int \frac {\int e^{-4+e^{2 x}+2 x} x \, dx}{x} \, dx+\frac {\int \frac {\operatorname {ExpIntegralEi}\left (e^{2 x}\right )}{x} \, dx}{2 e^4}-(2 \log (x \log (5))) \int e^{-4+e^{2 x}+2 x} x \, dx-\int \frac {1}{x \left (e^x-\log (x)\right )} \, dx+\int \frac {\log (x)}{e^x-\log (x)} \, dx \\ \end{align*}
Time = 2.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=-e^{-4+e^{2 x}} x \log (x \log (5))+\log \left (e^x-\log (x)\right ) \]
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Time = 8.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
parallelrisch | \(\ln \left (\ln \left (x \right )-{\mathrm e}^{x}\right )-{\mathrm e}^{{\mathrm e}^{2 x}-4} x \ln \left (x \ln \left (5\right )\right )\) | \(25\) |
risch | \(\ln \left (\ln \left (x \right )-{\mathrm e}^{x}\right )+\left (-x \ln \left (\ln \left (5\right )\right )-x \ln \left (x \right )\right ) {\mathrm e}^{{\mathrm e}^{2 x}-4}\) | \(30\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=-{\left (x \log \left (x\right ) + x \log \left (\log \left (5\right )\right )\right )} e^{\left (e^{\left (2 \, x\right )} - 4\right )} + \log \left (-e^{x} + \log \left (x\right )\right ) \]
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Time = 134.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=\left (- x \log {\left (x \right )} - x \log {\left (\log {\left (5 \right )} \right )}\right ) e^{e^{2 x} - 4} + \log {\left (e^{x} - \log {\left (x \right )} \right )} \]
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Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=-{\left (x \log \left (x\right ) + x \log \left (\log \left (5\right )\right )\right )} e^{\left (e^{\left (2 \, x\right )} - 4\right )} + \log \left (e^{x} - \log \left (x\right )\right ) \]
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\[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=\int { \frac {x e^{x} - {\left (x e^{x} + {\left (2 \, x^{2} e^{\left (3 \, x\right )} + x e^{x} - {\left (2 \, x^{2} e^{\left (2 \, x\right )} + x\right )} \log \left (x\right )\right )} \log \left (x \log \left (5\right )\right ) - x \log \left (x\right )\right )} e^{\left (e^{\left (2 \, x\right )} - 4\right )} - 1}{x e^{x} - x \log \left (x\right )} \,d x } \]
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Time = 9.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=\ln \left (\ln \left (x\right )-{\mathrm {e}}^x\right )-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}-4}\,\left (x\,\ln \left (\ln \left (5\right )\right )+x\,\ln \left (x\right )\right ) \]
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