Integrand size = 74, antiderivative size = 27 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=\log \left (-\frac {2}{4+x+\frac {2-e^{-1+x}-e^x}{\log (x)}}\right ) \]
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\[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=\int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e \left (2-e^{-1+x} (1+e)+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)\right )}{x \log (x) \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )} \, dx \\ & = e \int \frac {2-e^{-1+x} (1+e)+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{x \log (x) \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )} \, dx \\ & = e \int \left (\frac {1-x \log (x)}{e x \log (x)}+\frac {-4+x+3 x \log (x)+x^2 \log (x)}{x \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )}\right ) \, dx \\ & = e \int \frac {-4+x+3 x \log (x)+x^2 \log (x)}{x \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )} \, dx+\int \frac {1-x \log (x)}{x \log (x)} \, dx \\ & = e \int \left (\frac {4}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )}+\frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)}+\frac {3 \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)}+\frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)}\right ) \, dx+\int \left (-1+\frac {1}{x \log (x)}\right ) \, dx \\ & = -x+e \int \frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+e \int \frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(3 e) \int \frac {\log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(4 e) \int \frac {1}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = -x+e \int \frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+e \int \frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(3 e) \int \frac {\log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(4 e) \int \frac {1}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )} \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -x+\log (\log (x))+e \int \frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+e \int \frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(3 e) \int \frac {\log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(4 e) \int \frac {1}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )} \, dx \\ \end{align*}
Time = 3.82 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=\log (\log (x))-\log \left (-2 e+e^x+e^{1+x}-4 e \log (x)-e x \log (x)\right ) \]
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Time = 1.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\ln \left (\ln \left (x \right )\right )-\ln \left (x \ln \left (x \right )+4 \ln \left (x \right )-{\mathrm e}^{x}-{\mathrm e}^{-1+x}+2\right )\) | \(28\) |
risch | \(-\ln \left (4+x \right )+\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (x \right )-\frac {{\mathrm e}^{-1+x}+{\mathrm e}^{x}-2}{4+x}\right )\) | \(32\) |
norman | \(-\ln \left (x \,{\mathrm e} \ln \left (x \right )-{\mathrm e} \,{\mathrm e}^{x}+4 \,{\mathrm e} \ln \left (x \right )+2 \,{\mathrm e}-{\mathrm e}^{x}\right )+\ln \left (\ln \left (x \right )\right )\) | \(35\) |
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Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\log \left (x + 4\right ) - \log \left (\frac {{\left (x + 4\right )} e \log \left (x\right ) - {\left (e + 1\right )} e^{x} + 2 \, e}{x + 4}\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=- \log {\left (\frac {- e x \log {\left (x \right )} - 4 e \log {\left (x \right )} - 2 e}{1 + e} + e^{x} \right )} + \log {\left (\log {\left (x \right )} \right )} \]
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\log \left (\frac {{\left (e + 1\right )} e^{x} - {\left (x e + 4 \, e\right )} \log \left (x\right ) - 2 \, e}{e + 1}\right ) + \log \left (\log \left (x\right )\right ) \]
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Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\log \left (x e \log \left (x\right ) + 4 \, e \log \left (x\right ) + 2 \, e - e^{\left (x + 1\right )} - e^{x}\right ) + \log \left (\log \left (x\right )\right ) \]
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Timed out. \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\int -\frac {x\,{\ln \left (x\right )}^2+\left (-x\,{\mathrm {e}}^{x-1}-x\,{\mathrm {e}}^x\right )\,\ln \left (x\right )+{\mathrm {e}}^{x-1}+{\mathrm {e}}^x-2}{\ln \left (x\right )\,\left (x\,{\mathrm {e}}^{x-1}-2\,x+x\,{\mathrm {e}}^x\right )-{\ln \left (x\right )}^2\,\left (x^2+4\,x\right )} \,d x \]
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