Integrand size = 49, antiderivative size = 17 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {\log (3)}{x \left (e^{-2-x}+x\right )} \]
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\[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2+x} \left (-1-\left (-1+2 e^{2+x}\right ) x\right ) \log (3)}{x^2 \left (1+e^{2+x} x\right )^2} \, dx \\ & = \log (3) \int \frac {e^{2+x} \left (-1-\left (-1+2 e^{2+x}\right ) x\right )}{x^2 \left (1+e^{2+x} x\right )^2} \, dx \\ & = \log (3) \int \left (\frac {e^{2+x} (1+x)}{x^2 \left (1+e^{2+x} x\right )^2}-\frac {2 e^{2+x}}{x^2 \left (1+e^{2+x} x\right )}\right ) \, dx \\ & = \log (3) \int \frac {e^{2+x} (1+x)}{x^2 \left (1+e^{2+x} x\right )^2} \, dx-(2 \log (3)) \int \frac {e^{2+x}}{x^2 \left (1+e^{2+x} x\right )} \, dx \\ & = \log (3) \int \left (\frac {e^{2+x}}{x^2 \left (1+e^{2+x} x\right )^2}+\frac {e^{2+x}}{x \left (1+e^{2+x} x\right )^2}\right ) \, dx-(2 \log (3)) \int \frac {e^{2+x}}{x^2 \left (1+e^{2+x} x\right )} \, dx \\ & = \log (3) \int \frac {e^{2+x}}{x^2 \left (1+e^{2+x} x\right )^2} \, dx+\log (3) \int \frac {e^{2+x}}{x \left (1+e^{2+x} x\right )^2} \, dx-(2 \log (3)) \int \frac {e^{2+x}}{x^2 \left (1+e^{2+x} x\right )} \, dx \\ \end{align*}
Time = 0.89 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=-\left (\left (-\frac {1}{x^2}+\frac {1}{x^2 \left (1+e^{2+x} x\right )}\right ) \log (3)\right ) \]
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Time = 0.55 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {\ln \left (3\right )}{x \left (x +{\mathrm e}^{-2-x}\right )}\) | \(17\) |
risch | \(\frac {\ln \left (3\right )}{x \left (x +{\mathrm e}^{-2-x}\right )}\) | \(17\) |
parallelrisch | \(\frac {\ln \left (3\right )}{x \left (x +{\mathrm e}^{-2-x}\right )}\) | \(17\) |
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Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {\log \left (3\right )}{x^{2} + x e^{\left (-x - 2\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {\log {\left (3 \right )}}{x^{2} + x e^{- x - 2}} \]
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Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {e^{\left (x + 2\right )} \log \left (3\right )}{x^{2} e^{\left (x + 2\right )} + x} \]
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Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {e^{2} \log \left (3\right )}{x^{2} e^{2} + x e^{\left (-x\right )}} \]
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Time = 12.55 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {\ln \left (3\right )}{x\,{\mathrm {e}}^{-x-2}+x^2} \]
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