Optimal. Leaf size=29 \[ e^{\frac {x (x+\log (x))}{-x+\log \left (\left (-3+x+\frac {\log (x)}{e^3 x}\right )^2\right )}} \]
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Rubi [F] time = 120.41, 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 {x^2+x \log (x)}{-x+\log \left (\frac {e^6 \left (9 x^2-6 x^3+x^4\right )+e^3 \left (-6 x+2 x^2\right ) \log (x)+\log ^2(x)}{e^6 x^2}\right )}\right ) \left (-2 x+e^3 \left (3 x^2-x^4\right )+\left (-2+x-x^2-2 e^3 x^2\right ) \log (x)+2 \log ^2(x)+\left (e^3 \left (-3 x-5 x^2+2 x^3\right )+\left (1+2 x+e^3 \left (-3 x+x^2\right )\right ) \log (x)+\log ^2(x)\right ) \log \left (\frac {e^6 \left (9 x^2-6 x^3+x^4\right )+e^3 \left (-6 x+2 x^2\right ) \log (x)+\log ^2(x)}{e^6 x^2}\right )\right )}{e^3 \left (-3 x^3+x^4\right )+x^2 \log (x)+\left (e^3 \left (6 x^2-2 x^3\right )-2 x \log (x)\right ) \log \left (\frac {e^6 \left (9 x^2-6 x^3+x^4\right )+e^3 \left (-6 x+2 x^2\right ) \log (x)+\log ^2(x)}{e^6 x^2}\right )+\left (e^3 \left (-3 x+x^2\right )+\log (x)\right ) \log ^2\left (\frac {e^6 \left (9 x^2-6 x^3+x^4\right )+e^3 \left (-6 x+2 x^2\right ) \log (x)+\log ^2(x)}{e^6 x^2}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-\frac {x^2}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) x^{-\frac {x}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}} \left (\log ^2(x) \left (2+\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )\right )-x \left (2+e^3 x \left (-3+x^2\right )+e^3 \left (3+5 x-2 x^2\right ) \log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )\right )+\log (x) \left (-2+x-x^2-2 e^3 x^2+\left (1+\left (2-3 e^3\right ) x+e^3 x^2\right ) \log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )\right )\right )}{\left (e^3 (-3+x) x+\log (x)\right ) \left (x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )\right )^2} \, dx\\ &=\int \left (\frac {\exp \left (-\frac {x^2}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) x^{-\frac {x}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}} \left (2 x+5 e^3 x^3-e^3 x^4+2 \log (x)-2 x \log (x)-\left (1-5 e^3\right ) x^2 \log (x)-e^3 x^3 \log (x)-2 \log ^2(x)-x \log ^2(x)\right )}{\left (3 e^3 x-e^3 x^2-\log (x)\right ) \left (x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )\right )^2}+\frac {\exp \left (-\frac {x^2}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) x^{-\frac {x}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}} (-1-2 x-\log (x))}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) \, dx\\ &=\int \frac {\exp \left (-\frac {x^2}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) x^{-\frac {x}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}} \left (2 x+5 e^3 x^3-e^3 x^4+2 \log (x)-2 x \log (x)-\left (1-5 e^3\right ) x^2 \log (x)-e^3 x^3 \log (x)-2 \log ^2(x)-x \log ^2(x)\right )}{\left (3 e^3 x-e^3 x^2-\log (x)\right ) \left (x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )\right )^2} \, dx+\int \frac {\exp \left (-\frac {x^2}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) x^{-\frac {x}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}} (-1-2 x-\log (x))}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )} \, dx\\ &=\int \frac {\exp \left (-\frac {x^2}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) x^{-\frac {x}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}} \left (x \left (-2+e^3 (-5+x) x^2\right )+\left (-2+2 x+\left (1-5 e^3\right ) x^2+e^3 x^3\right ) \log (x)+(2+x) \log ^2(x)\right )}{\left (e^3 (-3+x) x+\log (x)\right ) \left (x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )\right )^2} \, dx+\int \left (-\frac {2 \exp \left (-\frac {x^2}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) x^{1-\frac {x}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}}}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}-\frac {\exp \left (-\frac {x^2}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) x^{-\frac {x}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}} \log (x)}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}+\frac {\exp \left (-\frac {x^2}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) x^{-\frac {x}{x-\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}}}{-x+\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [B] time = 0.39, size = 65, normalized size = 2.24 \begin {gather*} e^{\frac {x^2}{-x+\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}} x^{\frac {x}{-x+\log \left (\frac {\left (e^3 (-3+x) x+\log (x)\right )^2}{e^6 x^2}\right )}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.98, size = 59, normalized size = 2.03 \begin {gather*} e^{\left (-\frac {x^{2} + x \log \relax (x)}{x - \log \left (\frac {{\left (2 \, {\left (x^{2} - 3 \, x\right )} e^{3} \log \relax (x) + {\left (x^{4} - 6 \, x^{3} + 9 \, x^{2}\right )} e^{6} + \log \relax (x)^{2}\right )} e^{\left (-6\right )}}{x^{2}}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 456.63, size = 356, normalized size = 12.28
method | result | size |
risch | \({\mathrm e}^{-\frac {2 x \left (x +\ln \relax (x )\right )}{i \pi \mathrm {csgn}\left (i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )^{2}\right )^{3}-2 i \pi \mathrm {csgn}\left (i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )^{2}\right )^{2} \mathrm {csgn}\left (i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )\right )+i \pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )^{2}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )^{2}}{x^{2}}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )^{2}}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right )-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+i \pi \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )^{2}}{x^{2}}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )^{2}}{x^{2}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right )+4 \ln \relax (x )-4 \ln \left (\ln \relax (x )+\left (x^{2}-3 x \right ) {\mathrm e}^{3}\right )+2 x +12}}\) | \(356\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left ({\left (x^{4} - 3 \, x^{2}\right )} e^{3} + {\left (2 \, x^{2} e^{3} + x^{2} - x + 2\right )} \log \relax (x) - 2 \, \log \relax (x)^{2} - {\left ({\left (2 \, x^{3} - 5 \, x^{2} - 3 \, x\right )} e^{3} + {\left ({\left (x^{2} - 3 \, x\right )} e^{3} + 2 \, x + 1\right )} \log \relax (x) + \log \relax (x)^{2}\right )} \log \left (\frac {{\left (2 \, {\left (x^{2} - 3 \, x\right )} e^{3} \log \relax (x) + {\left (x^{4} - 6 \, x^{3} + 9 \, x^{2}\right )} e^{6} + \log \relax (x)^{2}\right )} e^{\left (-6\right )}}{x^{2}}\right ) + 2 \, x\right )} e^{\left (-\frac {x^{2} + x \log \relax (x)}{x - \log \left (\frac {{\left (2 \, {\left (x^{2} - 3 \, x\right )} e^{3} \log \relax (x) + {\left (x^{4} - 6 \, x^{3} + 9 \, x^{2}\right )} e^{6} + \log \relax (x)^{2}\right )} e^{\left (-6\right )}}{x^{2}}\right )}\right )}}{x^{2} \log \relax (x) + {\left ({\left (x^{2} - 3 \, x\right )} e^{3} + \log \relax (x)\right )} \log \left (\frac {{\left (2 \, {\left (x^{2} - 3 \, x\right )} e^{3} \log \relax (x) + {\left (x^{4} - 6 \, x^{3} + 9 \, x^{2}\right )} e^{6} + \log \relax (x)^{2}\right )} e^{\left (-6\right )}}{x^{2}}\right )^{2} + {\left (x^{4} - 3 \, x^{3}\right )} e^{3} - 2 \, {\left ({\left (x^{3} - 3 \, x^{2}\right )} e^{3} + x \log \relax (x)\right )} \log \left (\frac {{\left (2 \, {\left (x^{2} - 3 \, x\right )} e^{3} \log \relax (x) + {\left (x^{4} - 6 \, x^{3} + 9 \, x^{2}\right )} e^{6} + \log \relax (x)^{2}\right )} e^{\left (-6\right )}}{x^{2}}\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.81, size = 109, normalized size = 3.76 \begin {gather*} {\mathrm {e}}^{-\frac {x^2}{x-\ln \left (\frac {x^4-6\,x^3+2\,{\mathrm {e}}^{-3}\,x^2\,\ln \relax (x)+9\,x^2-6\,{\mathrm {e}}^{-3}\,x\,\ln \relax (x)+{\mathrm {e}}^{-6}\,{\ln \relax (x)}^2}{x^2}\right )}}\,{\mathrm {e}}^{-\frac {x\,\ln \relax (x)}{x-\ln \left (\frac {x^4-6\,x^3+2\,{\mathrm {e}}^{-3}\,x^2\,\ln \relax (x)+9\,x^2-6\,{\mathrm {e}}^{-3}\,x\,\ln \relax (x)+{\mathrm {e}}^{-6}\,{\ln \relax (x)}^2}{x^2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 44.42, size = 56, normalized size = 1.93 \begin {gather*} e^{\frac {x^{2} + x \log {\relax (x )}}{- x + \log {\left (\frac {\left (2 x^{2} - 6 x\right ) e^{3} \log {\relax (x )} + \left (x^{4} - 6 x^{3} + 9 x^{2}\right ) e^{6} + \log {\relax (x )}^{2}}{x^{2} e^{6}} \right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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