Integrand size = 87, antiderivative size = 24 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=3+5 e^{-x}+2 x+\frac {3 x}{\log (-x+\log (x))} \]
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\[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=\int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (2-5 e^{-x}-\frac {3 (-1+x)}{(x-\log (x)) \log ^2(-x+\log (x))}+\frac {3}{\log (-x+\log (x))}\right ) \, dx \\ & = 2 x-3 \int \frac {-1+x}{(x-\log (x)) \log ^2(-x+\log (x))} \, dx+3 \int \frac {1}{\log (-x+\log (x))} \, dx-5 \int e^{-x} \, dx \\ & = 5 e^{-x}+2 x-3 \int \left (-\frac {1}{(x-\log (x)) \log ^2(-x+\log (x))}+\frac {x}{(x-\log (x)) \log ^2(-x+\log (x))}\right ) \, dx+3 \int \frac {1}{\log (-x+\log (x))} \, dx \\ & = 5 e^{-x}+2 x+3 \int \frac {1}{(x-\log (x)) \log ^2(-x+\log (x))} \, dx-3 \int \frac {x}{(x-\log (x)) \log ^2(-x+\log (x))} \, dx+3 \int \frac {1}{\log (-x+\log (x))} \, dx \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=5 e^{-x}+2 x+\frac {3 x}{\log (-x+\log (x))} \]
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Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\left (2 \,{\mathrm e}^{x} x +5\right ) {\mathrm e}^{-x}+\frac {3 x}{\ln \left (\ln \left (x \right )-x \right )}\) | \(26\) |
default | \(5 \,{\mathrm e}^{-x}+\frac {3 x +2 \ln \left (\ln \left (x \right )-x \right ) x}{\ln \left (\ln \left (x \right )-x \right )}\) | \(32\) |
parts | \(5 \,{\mathrm e}^{-x}+\frac {3 x +2 \ln \left (\ln \left (x \right )-x \right ) x}{\ln \left (\ln \left (x \right )-x \right )}\) | \(32\) |
parallelrisch | \(\frac {\left (4 \ln \left ({\mathrm e}^{x}\right ) {\mathrm e}^{x} \ln \left (\ln \left (x \right )-x \right )+6 \,{\mathrm e}^{x} x +10 \ln \left (\ln \left (x \right )-x \right )\right ) {\mathrm e}^{-x}}{2 \ln \left (\ln \left (x \right )-x \right )}\) | \(45\) |
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=\frac {{\left (3 \, x e^{x} + {\left (2 \, x e^{x} + 5\right )} \log \left (-x + \log \left (x\right )\right )\right )} e^{\left (-x\right )}}{\log \left (-x + \log \left (x\right )\right )} \]
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Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=2 x + \frac {3 x}{\log {\left (- x + \log {\left (x \right )} \right )}} + 5 e^{- x} \]
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=\frac {{\left (3 \, x e^{x} + {\left (2 \, x e^{x} + 5\right )} \log \left (-x + \log \left (x\right )\right )\right )} e^{\left (-x\right )}}{\log \left (-x + \log \left (x\right )\right )} \]
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Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=\frac {2 \, x \log \left (-x + \log \left (x\right )\right ) + 5 \, e^{\left (-x\right )} \log \left (-x + \log \left (x\right )\right ) + 3 \, x}{\log \left (-x + \log \left (x\right )\right )} \]
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Time = 13.61 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.58 \[ \int \frac {e^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx=5\,x+5\,{\mathrm {e}}^{-x}-3\,\ln \left (x\right )+\frac {3}{x-1}-\frac {3\,x^2}{x-1}-\frac {3\,\ln \left (x\right )}{x-1}+\frac {3\,x}{\ln \left (\ln \left (x\right )-x\right )}+\frac {3\,x\,\ln \left (x\right )}{x-1} \]
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