Optimal. Leaf size=26 \[ \frac {x^2}{\log ^2\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \]
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Rubi [F] time = 0.74, 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 {-2 x+2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\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 {2 x \left (-1+\log (x) \log (\log (x)) \log \left (\frac {5 \log (\log (x))}{8 e^5-4 e^9}\right )\right )}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \, dx\\ &=2 \int \frac {x \left (-1+\log (x) \log (\log (x)) \log \left (\frac {5 \log (\log (x))}{8 e^5-4 e^9}\right )\right )}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \, dx\\ &=2 \int \left (-\frac {x}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}+\frac {x}{\log ^2\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {x}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \, dx\right )+2 \int \frac {x}{\log ^2\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 25, normalized size = 0.96 \begin {gather*} \frac {x^2}{\log ^2\left (\frac {5 \log (\log (x))}{8 e^5-4 e^9}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.73, size = 21, normalized size = 0.81 \begin {gather*} \frac {x^{2}}{\log \left (-\frac {5 \, \log \left (\log \relax (x)\right )}{4 \, {\left (e^{9} - 2 \, e^{5}\right )}}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.39, size = 80, normalized size = 3.08 \begin {gather*} \frac {x^{2}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) \log \left (-5 \, \log \left (\log \relax (x)\right )\right ) + \log \left (-5 \, \log \left (\log \relax (x)\right )\right )^{2} + 4 \, \log \relax (2) \log \left (e^{4} - 2\right ) - 2 \, \log \left (-5 \, \log \left (\log \relax (x)\right )\right ) \log \left (e^{4} - 2\right ) + \log \left (e^{4} - 2\right )^{2} + 20 \, \log \relax (2) - 10 \, \log \left (-5 \, \log \left (\log \relax (x)\right )\right ) + 10 \, \log \left (e^{4} - 2\right ) + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 23, normalized size = 0.88
method | result | size |
risch | \(\frac {x^{2}}{\ln \left (-\frac {5 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{-5}}{4 \,{\mathrm e}^{4}-8}\right )^{2}}\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.65, size = 102, normalized size = 3.92 \begin {gather*} -\frac {x^{2}}{2 \, {\left (\log \left (-e^{4} + 2\right ) + 5\right )} \log \relax (5) - \log \relax (5)^{2} + 4 \, {\left (\log \relax (5) - \log \left (-e^{4} + 2\right ) - 5\right )} \log \relax (2) - 4 \, \log \relax (2)^{2} - \log \left (-e^{4} + 2\right )^{2} - 2 \, {\left (\log \relax (5) - 2 \, \log \relax (2) - \log \left (-e^{4} + 2\right ) - 5\right )} \log \left (\log \left (\log \relax (x)\right )\right ) - \log \left (\log \left (\log \relax (x)\right )\right )^{2} - 10 \, \log \left (-e^{4} + 2\right ) - 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.55, size = 149, normalized size = 5.73 \begin {gather*} {\ln \left (\ln \relax (x)\right )}^2\,\left (2\,x^2\,{\ln \relax (x)}^2+x^2\,\ln \relax (x)\right )+\frac {x^2-x^2\,\ln \left (-\frac {5\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)}{{\ln \left (-\frac {5\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )}^2}+\frac {x^2\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)-x\,\ln \left (-\frac {5\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (x+x\,\ln \left (\ln \relax (x)\right )+2\,x\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\right )}{\ln \left (-\frac {5\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )}+x^2\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.35, size = 24, normalized size = 0.92 \begin {gather*} \frac {x^{2}}{\log {\left (- \frac {5 \log {\left (\log {\relax (x )} \right )}}{\left (-8 + 4 e^{4}\right ) e^{5}} \right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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