Optimal. Leaf size=29 \[ \frac {e^{-6+2 \log (x) \log (5 x)} \left (-1+e^{\frac {x}{e^5}}\right ) x^2}{\log (x)} \]
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Rubi [B] time = 0.44, antiderivative size = 83, normalized size of antiderivative = 2.86, number of steps used = 1, number of rules used = 1, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {2288} \begin {gather*} -\frac {x^{2 \log (5 x)} \left (\left (e^5 x-e^{\frac {x}{e^5}+5} x\right ) \log ^2(x)+\left (e^5 x-e^{\frac {x}{e^5}+5} x\right ) \log (5 x) \log (x)\right )}{e^{11} \log ^2(x) \left (\frac {\log (x)}{x}+\frac {\log (5 x)}{x}\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 2288
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=-\frac {x^{2 \log (5 x)} \left (\left (e^5 x-e^{5+\frac {x}{e^5}} x\right ) \log ^2(x)+\left (e^5 x-e^{5+\frac {x}{e^5}} x\right ) \log (x) \log (5 x)\right )}{e^{11} \log ^2(x) \left (\frac {\log (x)}{x}+\frac {\log (5 x)}{x}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.47, size = 30, normalized size = 1.03 \begin {gather*} \frac {e^{-6+2 \log ^2(x)} \left (-1+e^{\frac {x}{e^5}}\right ) x^{2+\log (25)}}{\log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.82, size = 43, normalized size = 1.48 \begin {gather*} -\frac {{\left (x^{2} e^{5} - x^{2} e^{\left ({\left (x + 5 \, e^{5}\right )} e^{\left (-5\right )}\right )}\right )} e^{\left (2 \, \log \relax (5) \log \relax (x) + 2 \, \log \relax (x)^{2} - 11\right )}}{\log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (2 \, {\left (x e^{5} - x e^{\left (x e^{\left (-5\right )} + 5\right )}\right )} \log \left (5 \, x\right ) \log \relax (x) + 2 \, {\left (x e^{5} - x e^{\left (x e^{\left (-5\right )} + 5\right )}\right )} \log \relax (x)^{2} - x e^{5} + x e^{\left (x e^{\left (-5\right )} + 5\right )} + {\left (2 \, x e^{5} - {\left (x^{2} + 2 \, x e^{5}\right )} e^{\left (x e^{\left (-5\right )}\right )}\right )} \log \relax (x)\right )} e^{\left (2 \, \log \left (5 \, x\right ) \log \relax (x) - 11\right )}}{\log \relax (x)^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 27, normalized size = 0.93
method | result | size |
risch | \(\frac {x^{2} \left ({\mathrm e}^{{\mathrm e}^{-5} x}-1\right ) x^{2 \ln \left (5 x \right )} {\mathrm e}^{-6}}{\ln \relax (x )}\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.62, size = 49, normalized size = 1.69 \begin {gather*} \frac {{\left (x^{2} e^{\left (x e^{\left (-5\right )} + 2 \, \log \relax (5) \log \relax (x) + 2 \, \log \relax (x)^{2}\right )} - x^{2} e^{\left (2 \, \log \relax (5) \log \relax (x) + 2 \, \log \relax (x)^{2}\right )}\right )} e^{\left (-6\right )}}{\log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.93, size = 29, normalized size = 1.00 \begin {gather*} \frac {x^{2\,\ln \relax (5)+2}\,{\mathrm {e}}^{2\,{\ln \relax (x)}^2-6}\,\left ({\mathrm {e}}^{x\,{\mathrm {e}}^{-5}}-1\right )}{\ln \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.46, size = 31, normalized size = 1.07 \begin {gather*} \frac {\left (x^{2} e^{\frac {x}{e^{5}}} - x^{2}\right ) e^{2 \left (\log {\relax (x )} + \log {\relax (5 )}\right ) \log {\relax (x )} - 6}}{\log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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