Integrand size = 95, antiderivative size = 24 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\log \left (e^{1+e^{-x} x}+\log \left (\log \left (125 e^2 \log ^2(x)\right )\right )\right ) \]
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Time = 0.95 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6873, 6816} \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\log \left (e^{e^{-x} x+1}+\log \left (\log \left (125 \log ^2(x)\right )+2\right )\right ) \]
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Rule 6816
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )\right )}{x \log (x) \left (2 \left (1+\frac {3 \log (5)}{2}\right )+\log \left (\log ^2(x)\right )\right ) \left (e^{1+e^{-x} x}+\log \left (2+\log \left (125 \log ^2(x)\right )\right )\right )} \, dx \\ & = \log \left (e^{1+e^{-x} x}+\log \left (2+\log \left (125 \log ^2(x)\right )\right )\right ) \\ \end{align*}
\[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.70 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.29
\[\ln \left ({\mathrm e}^{1+x \,{\mathrm e}^{-x}}+\ln \left (3 \ln \left (5\right )+2+2 \ln \left (\ln \left (x \right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )+\operatorname {csgn}\left (i \ln \left (x \right )\right )\right )}^{2}}{2}\right )\right )\]
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none
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\log \left ({\left (e^{x} \log \left (\log \left (125 \, e^{2} \log \left (x\right )^{2}\right )\right ) + e^{\left ({\left ({\left (x + 1\right )} e^{x} + x\right )} e^{\left (-x\right )}\right )}\right )} e^{\left (-x\right )}\right ) \]
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Time = 1.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\log {\left (e^{x e^{- x}} + \frac {\log {\left (\log {\left (125 e^{2} \log {\left (x \right )}^{2} \right )} \right )}}{e} \right )} \]
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Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\log \left (e^{\left (x e^{\left (-x\right )} + 1\right )} + \log \left (3 \, \log \left (5\right ) + 2 \, \log \left (\log \left (x\right )\right ) + 2\right )\right ) \]
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Exception generated. \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\text {Exception raised: TypeError} \]
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Time = 12.43 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\ln \left ({\mathrm {e}}^{x\,{\mathrm {e}}^{-x}+1}+\ln \left (\ln \left (125\,{\mathrm {e}}^2\,{\ln \left (x\right )}^2\right )\right )\right ) \]
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