Optimal. Leaf size=27 \[ 3 e^{\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}} \]
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Rubi [F] time = 17.38, 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 {\exp \left (\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {\exp \left (\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx\\ &=\frac {1}{2} \int \frac {3 \exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4)) \left (-2 (x+\log (4))+x \log \left (\frac {2 x^2}{5}\right )\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx\\ &=\frac {3}{2} \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4)) \left (-2 (x+\log (4))+x \log \left (\frac {2 x^2}{5}\right )\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx\\ &=\frac {3}{2} \int \left (-\frac {2 \exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4))^2}{x \log ^3\left (\frac {2 x^2}{5}\right )}+\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4))}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \, dx\\ &=\frac {3}{2} \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4))}{\log ^2\left (\frac {2 x^2}{5}\right )} \, dx-3 \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) (x+\log (4))^2}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx\\ &=\frac {3}{2} \int \left (\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) x}{\log ^2\left (\frac {2 x^2}{5}\right )}+\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \log (4)}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \, dx-3 \int \left (\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) x}{\log ^3\left (\frac {2 x^2}{5}\right )}+\frac {2 \exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \log (4)}{\log ^3\left (\frac {2 x^2}{5}\right )}+\frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) \log ^2(4)}{x \log ^3\left (\frac {2 x^2}{5}\right )}\right ) \, dx\\ &=\frac {3}{2} \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) x}{\log ^2\left (\frac {2 x^2}{5}\right )} \, dx-3 \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right ) x}{\log ^3\left (\frac {2 x^2}{5}\right )} \, dx+\frac {1}{2} (3 \log (4)) \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right )}{\log ^2\left (\frac {2 x^2}{5}\right )} \, dx-(6 \log (4)) \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right )}{\log ^3\left (\frac {2 x^2}{5}\right )} \, dx-\left (3 \log ^2(4)\right ) \int \frac {\exp \left (\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 4.43, size = 27, normalized size = 1.00 \begin {gather*} 3 e^{\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.71, size = 90, normalized size = 3.33 \begin {gather*} 3 \, e^{\left (\frac {e^{\left (\frac {x^{2} + 4 \, x \log \relax (2) + 4 \, \log \relax (2)^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}}\right )} \log \left (\frac {2}{5} \, x^{2}\right )^{2} + 4 \, x^{2} + 16 \, x \log \relax (2) + 16 \, \log \relax (2)^{2}}{4 \, \log \left (\frac {2}{5} \, x^{2}\right )^{2}} - \frac {x^{2} + 4 \, x \log \relax (2) + 4 \, \log \relax (2)^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}}\right )} \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 {3 \, {\left (2 \, x^{2} + 8 \, x \log \relax (2) + 8 \, \log \relax (2)^{2} - {\left (x^{2} + 2 \, x \log \relax (2)\right )} \log \left (\frac {2}{5} \, x^{2}\right )\right )} e^{\left (\frac {x^{2} + 4 \, x \log \relax (2) + 4 \, \log \relax (2)^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}} + \frac {1}{4} \, e^{\left (\frac {x^{2} + 4 \, x \log \relax (2) + 4 \, \log \relax (2)^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}}\right )}\right )}}{2 \, x \log \left (\frac {2}{5} \, x^{2}\right )^{3}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.41, size = 24, normalized size = 0.89
method | result | size |
risch | \(3 \,{\mathrm e}^{\frac {{\mathrm e}^{\frac {\left (x +2 \ln \relax (2)\right )^{2}}{\ln \left (\frac {2 x^{2}}{5}\right )^{2}}}}{4}}\) | \(24\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.82, size = 124, normalized size = 4.59 \begin {gather*} 3 \, e^{\left (\frac {1}{4} \, e^{\left (\frac {x^{2}}{\log \relax (5)^{2} - 2 \, \log \relax (5) \log \relax (2) + \log \relax (2)^{2} - 4 \, {\left (\log \relax (5) - \log \relax (2)\right )} \log \relax (x) + 4 \, \log \relax (x)^{2}} + \frac {4 \, x \log \relax (2)}{\log \relax (5)^{2} - 2 \, \log \relax (5) \log \relax (2) + \log \relax (2)^{2} - 4 \, {\left (\log \relax (5) - \log \relax (2)\right )} \log \relax (x) + 4 \, \log \relax (x)^{2}} + \frac {4 \, \log \relax (2)^{2}}{\log \relax (5)^{2} - 2 \, \log \relax (5) \log \relax (2) + \log \relax (2)^{2} - 4 \, {\left (\log \relax (5) - \log \relax (2)\right )} \log \relax (x) + 4 \, \log \relax (x)^{2}}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.80, size = 139, normalized size = 5.15 \begin {gather*} 3\,{\mathrm {e}}^{\frac {2^{\frac {4\,x}{2\,\ln \left (x^2\right )\,\ln \relax (2)-2\,\ln \left (x^2\right )\,\ln \relax (5)-2\,\ln \relax (2)\,\ln \relax (5)+{\ln \left (x^2\right )}^2+{\ln \relax (2)}^2+{\ln \relax (5)}^2}}\,{\mathrm {e}}^{\frac {4\,{\ln \relax (2)}^2}{2\,\ln \left (x^2\right )\,\ln \relax (2)-2\,\ln \left (x^2\right )\,\ln \relax (5)-2\,\ln \relax (2)\,\ln \relax (5)+{\ln \left (x^2\right )}^2+{\ln \relax (2)}^2+{\ln \relax (5)}^2}}\,{\mathrm {e}}^{\frac {x^2}{2\,\ln \left (x^2\right )\,\ln \relax (2)-2\,\ln \left (x^2\right )\,\ln \relax (5)-2\,\ln \relax (2)\,\ln \relax (5)+{\ln \left (x^2\right )}^2+{\ln \relax (2)}^2+{\ln \relax (5)}^2}}}{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.68, size = 32, normalized size = 1.19 \begin {gather*} 3 e^{\frac {e^{\frac {x^{2} + 4 x \log {\relax (2 )} + 4 \log {\relax (2 )}^{2}}{\log {\left (\frac {2 x^{2}}{5} \right )}^{2}}}}{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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