Integrand size = 68, antiderivative size = 33 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=x+\frac {e^{-x} \left (1+\log (2)+\log \left (\frac {2}{x}\right )\right )}{(1-x) \log \left (5 e^2\right )} \]
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Time = 1.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52, number of steps used = 20, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {6, 12, 1608, 27, 6874, 2208, 2209, 2228, 2634} \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=x+\frac {e^{-x} \log \left (\frac {2}{x}\right )}{(1-x) (2+\log (5))}+\frac {e^{-x} (1+\log (2))}{(1-x) (2+\log (5))} \]
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Rule 6
Rule 12
Rule 27
Rule 1608
Rule 2208
Rule 2209
Rule 2228
Rule 2634
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (-1+x+x^2 (1+\log (2))+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx \\ & = \frac {\int \frac {e^{-x} \left (-1+x+x^2 (1+\log (2))+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{x-2 x^2+x^3} \, dx}{2+\log (5)} \\ & = \frac {\int \frac {e^{-x} \left (-1+x+x^2 (1+\log (2))+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{x \left (1-2 x+x^2\right )} \, dx}{2+\log (5)} \\ & = \frac {\int \frac {e^{-x} \left (-1+x+x^2 (1+\log (2))+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{(-1+x)^2 x} \, dx}{2+\log (5)} \\ & = \frac {\int \left (2+\frac {e^{-x}}{(-1+x)^2}-\frac {e^{-x}}{(-1+x)^2 x}+\frac {e^{-x} x (1+\log (2))}{(-1+x)^2}+\log (5)+\frac {e^{-x} x \log \left (\frac {2}{x}\right )}{(-1+x)^2}\right ) \, dx}{2+\log (5)} \\ & = x+\frac {\int \frac {e^{-x}}{(-1+x)^2} \, dx}{2+\log (5)}-\frac {\int \frac {e^{-x}}{(-1+x)^2 x} \, dx}{2+\log (5)}+\frac {\int \frac {e^{-x} x \log \left (\frac {2}{x}\right )}{(-1+x)^2} \, dx}{2+\log (5)}+\frac {(1+\log (2)) \int \frac {e^{-x} x}{(-1+x)^2} \, dx}{2+\log (5)} \\ & = x+\frac {e^{-x}}{(1-x) (2+\log (5))}+\frac {e^{-x} (1+\log (2))}{(1-x) (2+\log (5))}+\frac {e^{-x} \log \left (\frac {2}{x}\right )}{(1-x) (2+\log (5))}-\frac {\int \left (\frac {e^{-x}}{1-x}+\frac {e^{-x}}{(-1+x)^2}+\frac {e^{-x}}{x}\right ) \, dx}{2+\log (5)}-\frac {\int \frac {e^{-x}}{-1+x} \, dx}{2+\log (5)}+\frac {\int \frac {e^{-x}}{(1-x) x} \, dx}{2+\log (5)} \\ & = x+\frac {e^{-x}}{(1-x) (2+\log (5))}-\frac {\text {Ei}(1-x)}{e (2+\log (5))}+\frac {e^{-x} (1+\log (2))}{(1-x) (2+\log (5))}+\frac {e^{-x} \log \left (\frac {2}{x}\right )}{(1-x) (2+\log (5))}+\frac {\int \left (\frac {e^{-x}}{1-x}+\frac {e^{-x}}{x}\right ) \, dx}{2+\log (5)}-\frac {\int \frac {e^{-x}}{1-x} \, dx}{2+\log (5)}-\frac {\int \frac {e^{-x}}{(-1+x)^2} \, dx}{2+\log (5)}-\frac {\int \frac {e^{-x}}{x} \, dx}{2+\log (5)} \\ & = x-\frac {\text {Ei}(-x)}{2+\log (5)}+\frac {e^{-x} (1+\log (2))}{(1-x) (2+\log (5))}+\frac {e^{-x} \log \left (\frac {2}{x}\right )}{(1-x) (2+\log (5))}+\frac {\int \frac {e^{-x}}{1-x} \, dx}{2+\log (5)}+\frac {\int \frac {e^{-x}}{-1+x} \, dx}{2+\log (5)}+\frac {\int \frac {e^{-x}}{x} \, dx}{2+\log (5)} \\ & = x+\frac {e^{-x} (1+\log (2))}{(1-x) (2+\log (5))}+\frac {e^{-x} \log \left (\frac {2}{x}\right )}{(1-x) (2+\log (5))} \\ \end{align*}
Time = 5.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\frac {e^{-x} \left (-1+e^x (-1+x) x (2+\log (5))-\log \left (\frac {4}{x}\right )\right )}{(-1+x) (2+\log (5))} \]
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Time = 0.43 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24
method | result | size |
default | \(\frac {\frac {\left (-1-\ln \left (2\right )-\ln \left (\frac {2}{x}\right )\right ) {\mathrm e}^{-x}}{-1+x}+x \ln \left (5 \,{\mathrm e}^{2}\right )}{\ln \left (5 \,{\mathrm e}^{2}\right )}\) | \(41\) |
parts | \(x +\frac {\left (\frac {\ln \left (x \right )}{\ln \left (5 \,{\mathrm e}^{2}\right )}-\frac {\ln \left (2\right )+\ln \left (\frac {2}{x}\right )+\ln \left (x \right )+1}{\ln \left (5 \,{\mathrm e}^{2}\right )}\right ) {\mathrm e}^{-x}}{-1+x}\) | \(45\) |
norman | \(\frac {\left (-{\mathrm e}^{x}+{\mathrm e}^{x} x^{2}-\frac {1+\ln \left (2\right )}{2+\ln \left (5\right )}-\frac {\ln \left (\frac {2}{x}\right )}{2+\ln \left (5\right )}\right ) {\mathrm e}^{-x}}{-1+x}\) | \(48\) |
parallelrisch | \(\frac {\left (\ln \left (5 \,{\mathrm e}^{2}\right ) {\mathrm e}^{x} x^{2}-1-{\mathrm e}^{x} \ln \left (5 \,{\mathrm e}^{2}\right )-\ln \left (2\right )-\ln \left (\frac {2}{x}\right )\right ) {\mathrm e}^{-x}}{\ln \left (5 \,{\mathrm e}^{2}\right ) \left (-1+x \right )}\) | \(52\) |
risch | \(\frac {{\mathrm e}^{-x} \ln \left (x \right )}{\left (2+\ln \left (5\right )\right ) \left (-1+x \right )}-\frac {\left (2-2 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+2 x \,{\mathrm e}^{x} \ln \left (5\right )-4 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x +4 \ln \left (2\right )\right ) {\mathrm e}^{-x}}{2 \left (2+\ln \left (5\right )\right ) \left (-1+x \right )}\) | \(71\) |
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Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\frac {{\left ({\left (2 \, x^{2} + {\left (x^{2} - x\right )} \log \left (5\right ) - 2 \, x\right )} e^{x} - \log \left (2\right ) - \log \left (\frac {2}{x}\right ) - 1\right )} e^{\left (-x\right )}}{{\left (x - 1\right )} \log \left (5\right ) + 2 \, x - 2} \]
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Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=x + \frac {\left (- \log {\left (\frac {2}{x} \right )} - 1 - \log {\left (2 \right )}\right ) e^{- x}}{x \log {\left (5 \right )} + 2 x - 2 - \log {\left (5 \right )}} \]
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Time = 0.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\frac {x^{2} {\left (\log \left (5\right ) + 2\right )} - x {\left (\log \left (5\right ) + 2\right )} - {\left (2 \, \log \left (2\right ) - \log \left (x\right ) + 1\right )} e^{\left (-x\right )}}{{\left (x - 1\right )} \log \left (5 \, e^{2}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\frac {x^{2} \log \left (5\right ) + 2 \, x^{2} - x \log \left (5\right ) - 2 \, e^{\left (-x\right )} \log \left (2\right ) + e^{\left (-x\right )} \log \left (x\right ) - 2 \, x - e^{\left (-x\right )}}{{\left (x - 1\right )} \log \left (5 \, e^{2}\right )} \]
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Timed out. \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\int \frac {{\mathrm {e}}^{-x}\,\left (x+x^2\,\ln \left (2\right )+x^2+x^2\,\ln \left (\frac {2}{x}\right )+\ln \left (5\,{\mathrm {e}}^2\right )\,{\mathrm {e}}^x\,\left (x^3-2\,x^2+x\right )-1\right )}{\ln \left (5\,{\mathrm {e}}^2\right )\,\left (x^3-2\,x^2+x\right )} \,d x \]
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