Integrand size = 74, antiderivative size = 20 \[ \int \frac {6 x-2 e^x x+\left (e (1-x)+e^x (1-x)-3 x+3 x^2\right ) \log \left (e+e^x-3 x\right )}{\left (e^{2 x}+e^x (e-3 x)\right ) \log ^3\left (e+e^x-3 x\right )} \, dx=1+\frac {e^{-x} x}{\log ^2\left (e+e^x-3 x\right )} \]
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\[ \int \frac {6 x-2 e^x x+\left (e (1-x)+e^x (1-x)-3 x+3 x^2\right ) \log \left (e+e^x-3 x\right )}{\left (e^{2 x}+e^x (e-3 x)\right ) \log ^3\left (e+e^x-3 x\right )} \, dx=\int \frac {6 x-2 e^x x+\left (e (1-x)+e^x (1-x)-3 x+3 x^2\right ) \log \left (e+e^x-3 x\right )}{\left (e^{2 x}+e^x (e-3 x)\right ) \log ^3\left (e+e^x-3 x\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (-\frac {2 \left (-3+e^x\right ) x}{e+e^x-3 x}-(-1+x) \log \left (e+e^x-3 x\right )\right )}{\log ^3\left (e+e^x-3 x\right )} \, dx \\ & = \int \left (\frac {2 e^{-x} (3+e-3 x) x}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )}+\frac {e^{-x} \left (-2 x+\log \left (e+e^x-3 x\right )-x \log \left (e+e^x-3 x\right )\right )}{\log ^3\left (e+e^x-3 x\right )}\right ) \, dx \\ & = 2 \int \frac {e^{-x} (3+e-3 x) x}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )} \, dx+\int \frac {e^{-x} \left (-2 x+\log \left (e+e^x-3 x\right )-x \log \left (e+e^x-3 x\right )\right )}{\log ^3\left (e+e^x-3 x\right )} \, dx \\ & = 2 \int \left (\frac {e^{-x} (3+e) x}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )}-\frac {3 e^{-x} x^2}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )}\right ) \, dx+\int \frac {e^{-x} \left (-2 x-(-1+x) \log \left (e+e^x-3 x\right )\right )}{\log ^3\left (e+e^x-3 x\right )} \, dx \\ & = -\left (6 \int \frac {e^{-x} x^2}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )} \, dx\right )+(2 (3+e)) \int \frac {e^{-x} x}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )} \, dx+\int \left (-\frac {2 e^{-x} x}{\log ^3\left (e+e^x-3 x\right )}+\frac {e^{-x} (1-x)}{\log ^2\left (e+e^x-3 x\right )}\right ) \, dx \\ & = -\left (2 \int \frac {e^{-x} x}{\log ^3\left (e+e^x-3 x\right )} \, dx\right )-6 \int \frac {e^{-x} x^2}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )} \, dx+(2 (3+e)) \int \frac {e^{-x} x}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )} \, dx+\int \frac {e^{-x} (1-x)}{\log ^2\left (e+e^x-3 x\right )} \, dx \\ & = -\left (2 \int \frac {e^{-x} x}{\log ^3\left (e+e^x-3 x\right )} \, dx\right )-6 \int \frac {e^{-x} x^2}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )} \, dx+(2 (3+e)) \int \frac {e^{-x} x}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )} \, dx+\int \left (\frac {e^{-x}}{\log ^2\left (e+e^x-3 x\right )}-\frac {e^{-x} x}{\log ^2\left (e+e^x-3 x\right )}\right ) \, dx \\ & = -\left (2 \int \frac {e^{-x} x}{\log ^3\left (e+e^x-3 x\right )} \, dx\right )-6 \int \frac {e^{-x} x^2}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )} \, dx+(2 (3+e)) \int \frac {e^{-x} x}{\left (e+e^x-3 x\right ) \log ^3\left (e+e^x-3 x\right )} \, dx+\int \frac {e^{-x}}{\log ^2\left (e+e^x-3 x\right )} \, dx-\int \frac {e^{-x} x}{\log ^2\left (e+e^x-3 x\right )} \, dx \\ \end{align*}
Time = 2.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {6 x-2 e^x x+\left (e (1-x)+e^x (1-x)-3 x+3 x^2\right ) \log \left (e+e^x-3 x\right )}{\left (e^{2 x}+e^x (e-3 x)\right ) \log ^3\left (e+e^x-3 x\right )} \, dx=\frac {e^{-x} x}{\log ^2\left (e+e^x-3 x\right )} \]
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Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\frac {x \,{\mathrm e}^{-x}}{\ln \left ({\mathrm e}^{x}+{\mathrm e}-3 x \right )^{2}}\) | \(18\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{-x}}{\ln \left ({\mathrm e}^{x}+{\mathrm e}-3 x \right )^{2}}\) | \(18\) |
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Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {6 x-2 e^x x+\left (e (1-x)+e^x (1-x)-3 x+3 x^2\right ) \log \left (e+e^x-3 x\right )}{\left (e^{2 x}+e^x (e-3 x)\right ) \log ^3\left (e+e^x-3 x\right )} \, dx=\frac {x e^{\left (-x\right )}}{\log \left (-3 \, x + e + e^{x}\right )^{2}} \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {6 x-2 e^x x+\left (e (1-x)+e^x (1-x)-3 x+3 x^2\right ) \log \left (e+e^x-3 x\right )}{\left (e^{2 x}+e^x (e-3 x)\right ) \log ^3\left (e+e^x-3 x\right )} \, dx=\frac {x e^{- x}}{\log {\left (- 3 x + e^{x} + e \right )}^{2}} \]
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Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {6 x-2 e^x x+\left (e (1-x)+e^x (1-x)-3 x+3 x^2\right ) \log \left (e+e^x-3 x\right )}{\left (e^{2 x}+e^x (e-3 x)\right ) \log ^3\left (e+e^x-3 x\right )} \, dx=\frac {x e^{\left (-x\right )}}{\log \left (-3 \, x + e + e^{x}\right )^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {6 x-2 e^x x+\left (e (1-x)+e^x (1-x)-3 x+3 x^2\right ) \log \left (e+e^x-3 x\right )}{\left (e^{2 x}+e^x (e-3 x)\right ) \log ^3\left (e+e^x-3 x\right )} \, dx=\frac {x e^{\left (-x\right )}}{\log \left (-3 \, x + e + e^{x}\right )^{2}} \]
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Time = 9.12 (sec) , antiderivative size = 377, normalized size of antiderivative = 18.85 \[ \int \frac {6 x-2 e^x x+\left (e (1-x)+e^x (1-x)-3 x+3 x^2\right ) \log \left (e+e^x-3 x\right )}{\left (e^{2 x}+e^x (e-3 x)\right ) \log ^3\left (e+e^x-3 x\right )} \, dx=\frac {\frac {\mathrm {e}}{6}+\frac {{\mathrm {e}}^2}{9}-\frac {5\,x\,\mathrm {e}}{6}-\frac {x\,{\mathrm {e}}^2}{18}+\frac {x^2\,\mathrm {e}}{3}+\frac {3\,x^2}{2}-\frac {x^3}{2}-1}{{\mathrm {e}}^x-3}+\frac {x\,{\mathrm {e}}^{-x}+\frac {{\mathrm {e}}^{-x}\,\ln \left (\mathrm {e}-3\,x+{\mathrm {e}}^x\right )\,\left (x-1\right )\,\left (\mathrm {e}-3\,x+{\mathrm {e}}^x\right )}{2\,\left ({\mathrm {e}}^x-3\right )}}{{\ln \left (\mathrm {e}-3\,x+{\mathrm {e}}^x\right )}^2}-\frac {\frac {{\mathrm {e}}^{-x}\,\left (x-1\right )\,\left (\mathrm {e}-3\,x+{\mathrm {e}}^x\right )}{2\,\left ({\mathrm {e}}^x-3\right )}+\frac {{\mathrm {e}}^{-x}\,\ln \left (\mathrm {e}-3\,x+{\mathrm {e}}^x\right )\,\left (\mathrm {e}-3\,x+{\mathrm {e}}^x\right )\,\left (6\,\mathrm {e}-2\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{x+1}-27\,x+x\,{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^{x+1}-6\,x^2\,{\mathrm {e}}^x-3\,x\,\mathrm {e}+12\,x\,{\mathrm {e}}^x+9\,x^2+9\right )}{2\,{\left ({\mathrm {e}}^x-3\right )}^3}}{\ln \left (\mathrm {e}-3\,x+{\mathrm {e}}^x\right )}-{\mathrm {e}}^{-x}\,\left (-\frac {x^3}{2}+\left (\frac {\mathrm {e}}{3}+\frac {3}{2}\right )\,x^2+\left (-\frac {5\,\mathrm {e}}{6}-\frac {{\mathrm {e}}^2}{18}-\frac {1}{2}\right )\,x+\frac {\mathrm {e}}{6}+\frac {{\mathrm {e}}^2}{9}\right )+\frac {3\,\mathrm {e}-\frac {27\,x}{2}+\frac {{\mathrm {e}}^2}{2}-6\,x\,\mathrm {e}-\frac {x\,{\mathrm {e}}^2}{2}+3\,x^2\,\mathrm {e}+\frac {27\,x^2}{2}-\frac {9\,x^3}{2}+\frac {9}{2}}{9\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{3\,x}-27\,{\mathrm {e}}^x+27}-\frac {3\,\mathrm {e}-\frac {27\,x}{2}+\frac {{\mathrm {e}}^2}{3}-4\,x\,\mathrm {e}-\frac {x\,{\mathrm {e}}^2}{6}+x^2\,\mathrm {e}+9\,x^2-\frac {3\,x^3}{2}+6}{{\mathrm {e}}^{2\,x}-6\,{\mathrm {e}}^x+9} \]
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