Optimal. Leaf size=25 \[ \frac {x^2}{1-e^x+\log \left (\frac {x^2}{\log (\log (5 x))}\right )} \]
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Rubi [F] time = 3.83, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {x+e^x \left (-2 x+x^2\right ) \log (5 x) \log (\log (5 x))+2 x \log (5 x) \log (\log (5 x)) \log \left (\frac {x^2}{\log (\log (5 x))}\right )}{\left (1-2 e^x+e^{2 x}\right ) \log (5 x) \log (\log (5 x))+\left (2-2 e^x\right ) \log (5 x) \log (\log (5 x)) \log \left (\frac {x^2}{\log (\log (5 x))}\right )+\log (5 x) \log (\log (5 x)) \log ^2\left (\frac {x^2}{\log (\log (5 x))}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x+x \log (5 x) \log (\log (5 x)) \left (e^x (-2+x)+2 \log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )}{\log (5 x) \log (\log (5 x)) \left (1-e^x+\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2} \, dx\\ &=\int \left (\frac {(-2+x) x}{-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )}+\frac {x \left (1-2 \log (5 x) \log (\log (5 x))+x \log (5 x) \log (\log (5 x))+x \log (5 x) \log (\log (5 x)) \log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )}{\log (5 x) \log (\log (5 x)) \left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2}\right ) \, dx\\ &=\int \frac {(-2+x) x}{-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )} \, dx+\int \frac {x \left (1-2 \log (5 x) \log (\log (5 x))+x \log (5 x) \log (\log (5 x))+x \log (5 x) \log (\log (5 x)) \log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )}{\log (5 x) \log (\log (5 x)) \left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2} \, dx\\ &=\int \left (-\frac {2 x}{-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )}+\frac {x^2}{-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )}\right ) \, dx+\int \frac {x+x \log (5 x) \log (\log (5 x)) \left (-2+x+x \log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )}{\log (5 x) \log (\log (5 x)) \left (1-e^x+\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2} \, dx\\ &=-\left (2 \int \frac {x}{-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )} \, dx\right )+\int \frac {x^2}{-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )} \, dx+\int \left (-\frac {2 x}{\left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2}+\frac {x^2}{\left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2}+\frac {x}{\log (5 x) \log (\log (5 x)) \left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2}+\frac {x^2 \log \left (\frac {x^2}{\log (\log (5 x))}\right )}{\left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {x}{\left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2} \, dx\right )-2 \int \frac {x}{-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )} \, dx+\int \frac {x^2}{\left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2} \, dx+\int \frac {x}{\log (5 x) \log (\log (5 x)) \left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2} \, dx+\int \frac {x^2}{-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )} \, dx+\int \frac {x^2 \log \left (\frac {x^2}{\log (\log (5 x))}\right )}{\left (-1+e^x-\log \left (\frac {x^2}{\log (\log (5 x))}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 25, normalized size = 1.00 \begin {gather*} \frac {x^2}{1-e^x+\log \left (\frac {x^2}{\log (\log (5 x))}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 25, normalized size = 1.00 \begin {gather*} -\frac {x^{2}}{e^{x} - \log \left (\frac {x^{2}}{\log \left (\log \left (5 \, x\right )\right )}\right ) - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.73, size = 397, normalized size = 15.88 \begin {gather*} \frac {x^{3} e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) + x^{3} e^{x} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) - 2 \, x^{2} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) - 2 \, x^{2} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) + x^{2} \log \left (5 \, x\right )}{2 \, x e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) + 2 \, x e^{x} \log \left (5 \, x\right ) \log \relax (x)^{2} \log \left (\log \left (5 \, x\right )\right ) - x e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) - x e^{x} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) - x e^{\left (2 \, x\right )} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) + x e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) - x e^{\left (2 \, x\right )} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) + x e^{x} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) + 2 \, e^{x} \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) + 2 \, e^{x} \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) - 4 \, \log \relax (5) \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) - 4 \, \log \left (5 \, x\right ) \log \relax (x)^{2} \log \left (\log \left (5 \, x\right )\right ) + 2 \, \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) + 2 \, \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) - 2 \, \log \relax (5) \log \left (5 \, x\right ) \log \left (\log \left (5 \, x\right )\right ) - 2 \, \log \left (5 \, x\right ) \log \relax (x) \log \left (\log \left (5 \, x\right )\right ) - e^{x} \log \relax (5) - e^{x} \log \relax (x) + 2 \, \log \relax (5) \log \relax (x) + 2 \, \log \relax (x)^{2} - \log \relax (5) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) - \log \relax (x) \log \left (\log \left (\log \left (5 \, x\right )\right )\right ) + \log \relax (5) + \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {2 x \ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right ) \ln \left (\frac {x^{2}}{\ln \left (\ln \left (5 x \right )\right )}\right )+\left (x^{2}-2 x \right ) {\mathrm e}^{x} \ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right )+x}{\ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right ) \ln \left (\frac {x^{2}}{\ln \left (\ln \left (5 x \right )\right )}\right )^{2}+\left (-2 \,{\mathrm e}^{x}+2\right ) \ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right ) \ln \left (\frac {x^{2}}{\ln \left (\ln \left (5 x \right )\right )}\right )+\left ({\mathrm e}^{2 x}-2 \,{\mathrm e}^{x}+1\right ) \ln \left (5 x \right ) \ln \left (\ln \left (5 x \right )\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.59, size = 22, normalized size = 0.88 \begin {gather*} -\frac {x^{2}}{e^{x} - 2 \, \log \relax (x) + \log \left (\log \left (\log \relax (5) + \log \relax (x)\right )\right ) - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {x+2\,x\,\ln \left (5\,x\right )\,\ln \left (\ln \left (5\,x\right )\right )\,\ln \left (\frac {x^2}{\ln \left (\ln \left (5\,x\right )\right )}\right )-\ln \left (5\,x\right )\,{\mathrm {e}}^x\,\ln \left (\ln \left (5\,x\right )\right )\,\left (2\,x-x^2\right )}{\ln \left (5\,x\right )\,\ln \left (\ln \left (5\,x\right )\right )\,{\ln \left (\frac {x^2}{\ln \left (\ln \left (5\,x\right )\right )}\right )}^2-\ln \left (5\,x\right )\,\ln \left (\ln \left (5\,x\right )\right )\,\left (2\,{\mathrm {e}}^x-2\right )\,\ln \left (\frac {x^2}{\ln \left (\ln \left (5\,x\right )\right )}\right )+\ln \left (5\,x\right )\,\ln \left (\ln \left (5\,x\right )\right )\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.54, size = 20, normalized size = 0.80 \begin {gather*} - \frac {x^{2}}{e^{x} - \log {\left (\frac {x^{2}}{\log {\left (\log {\left (5 x \right )} \right )}} \right )} - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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