Integrand size = 73, antiderivative size = 19 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2-x}}{46+x+\log \left (3+e^x\right )} \]
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\[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2-x} \left (-3 (47+x)-e^x (48+x)-\left (3+e^x\right ) \log \left (3+e^x\right )\right )}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2} \, dx \\ & = \int \left (\frac {3 e^{2-x}}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2}+\frac {e^{2-x} \left (-48-x-\log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right )^2}\right ) \, dx \\ & = 3 \int \frac {e^{2-x}}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2} \, dx+\int \frac {e^{2-x} \left (-48-x-\log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right )^2} \, dx \\ & = 3 \int \frac {e^{2-x}}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2} \, dx+\int \left (\frac {e^{2-x}}{-46-x-\log \left (3+e^x\right )}-\frac {2 e^{2-x}}{\left (46+x+\log \left (3+e^x\right )\right )^2}\right ) \, dx \\ & = -\left (2 \int \frac {e^{2-x}}{\left (46+x+\log \left (3+e^x\right )\right )^2} \, dx\right )+3 \int \frac {e^{2-x}}{\left (3+e^x\right ) \left (46+x+\log \left (3+e^x\right )\right )^2} \, dx+\int \frac {e^{2-x}}{-46-x-\log \left (3+e^x\right )} \, dx \\ \end{align*}
Time = 1.53 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2-x}}{46+x+\log \left (3+e^x\right )} \]
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Time = 0.92 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {{\mathrm e}^{2-x}}{\ln \left (3+{\mathrm e}^{x}\right )+x +46}\) | \(18\) |
parallelrisch | \({\mathrm e}^{-\ln \left (\ln \left (3+{\mathrm e}^{x}\right )+x +46\right )+2-x}\) | \(18\) |
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2}}{{\left (x + 46\right )} e^{x} + e^{x} \log \left (e^{x} + 3\right )} \]
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Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2}}{x e^{x} + e^{x} \log {\left (e^{x} + 3 \right )} + 46 e^{x}} \]
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2}}{{\left (x + 46\right )} e^{x} + e^{x} \log \left (e^{x} + 3\right )} \]
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {e^{2}}{x e^{x} + e^{x} \log \left (e^{x} + 3\right ) + 46 \, e^{x}} \]
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Time = 9.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2-x} \left (-141+e^x (-48-x)-3 x+\left (-3-e^x\right ) \log \left (3+e^x\right )\right )}{\left (46+x+\log \left (3+e^x\right )\right ) \left (138+3 x+e^x (46+x)+\left (3+e^x\right ) \log \left (3+e^x\right )\right )} \, dx=\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^2}{x+\ln \left ({\mathrm {e}}^x+3\right )+46} \]
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