Optimal. Leaf size=30 \[ x \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right ) \]
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Rubi [F] time = 21.97, 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 e^x x+e^{e^{2 x}+2 e^x x+x^2} \left (-6 e^{2 x} x-6 x^2+e^x \left (-6 x-6 x^2\right )\right )-x \log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )+\left (-e^x x-x \log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )+\left (3 e^x+3 \log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )\right ) \log \left (e^x+\log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )\right )\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )\right )\right )}{-e^x x-x \log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )+\left (3 e^x+3 \log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )\right ) \log \left (e^x+\log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 e^x x-e^{e^{2 x}+2 e^x x+x^2} \left (-6 e^{2 x} x-6 x^2+e^x \left (-6 x-6 x^2\right )\right )+x \log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )-\left (-e^x x-x \log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )+\left (3 e^x+3 \log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )\right ) \log \left (e^x+\log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )\right )\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )\right )\right )}{\left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )} \, dx\\ &=\int \left (\frac {6 e^{\left (e^x+x\right )^2} \left (1+e^x\right ) x \left (e^x+x\right )}{\left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )}+\frac {-2 e^x x+x \log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )+e^x x \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )+x \log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )-3 e^x \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )-3 \log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right ) \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )}{\left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )}\right ) \, dx\\ &=6 \int \frac {e^{\left (e^x+x\right )^2} \left (1+e^x\right ) x \left (e^x+x\right )}{\left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )} \, dx+\int \frac {-2 e^x x+x \log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )+e^x x \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )+x \log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )-3 e^x \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )-3 \log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right ) \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )}{\left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )} \, dx\\ &=6 \int \left (\frac {e^{x+\left (e^x+x\right )^2} x}{x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )}+\frac {e^{\left (e^x+x\right )^2} x \left (1+x-\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )}{x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )}-\frac {e^{\left (e^x+x\right )^2} x \left (x-\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (-1+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )}{\left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )}\right ) \, dx+\int \frac {e^x \left (-2 x+\left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )\right )+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right ) \left (x+\left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )\right )}{\left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )} \, dx\\ &=6 \int \frac {e^{x+\left (e^x+x\right )^2} x}{x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )} \, dx+6 \int \frac {e^{\left (e^x+x\right )^2} x \left (1+x-\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )}{x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )} \, dx-6 \int \frac {e^{\left (e^x+x\right )^2} x \left (x-\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (-1+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )}{\left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )} \, dx+\int \left (\frac {3 x \log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )}{\left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \left (x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )}+\frac {-2 x+x \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right ) \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )\right )}{x-3 \log \left (e^x+\log \left (5 e^{-e^{\left (e^x+x\right )^2}}\right )\right )}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.65, size = 38, normalized size = 1.27 \begin {gather*} x \log \left (-x+3 \log \left (e^x+\log \left (5 e^{-e^{e^{2 x}+2 e^x x+x^2}}\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 31, normalized size = 1.03 \begin {gather*} x \log \left (-x + 3 \, \log \left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} + e^{x} + \log \relax (5)\right )\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 {6 \, {\left (x^{2} + x e^{\left (2 \, x\right )} + {\left (x^{2} + x\right )} e^{x}\right )} e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} - 2 \, x e^{x} + {\left (x e^{x} + x \log \left (5 \, e^{\left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}\right )}\right ) - 3 \, {\left (e^{x} + \log \left (5 \, e^{\left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}\right )}\right )\right )} \log \left (e^{x} + \log \left (5 \, e^{\left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}\right )}\right )\right )\right )} \log \left (-x + 3 \, \log \left (e^{x} + \log \left (5 \, e^{\left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}\right )}\right )\right )\right ) + x \log \left (5 \, e^{\left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}\right )}\right )}{x e^{x} + x \log \left (5 \, e^{\left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}\right )}\right ) - 3 \, {\left (e^{x} + \log \left (5 \, e^{\left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}\right )}\right )\right )} \log \left (e^{x} + \log \left (5 \, e^{\left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}\right )}\right )\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 2.54, size = 34, normalized size = 1.13
method | result | size |
risch | \(\ln \left (3 \ln \left (\ln \relax (5)-\ln \left ({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}}}\right )+{\mathrm e}^{x}\right )-x \right ) x\) | \(34\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.73, size = 31, normalized size = 1.03 \begin {gather*} x \log \left (-x + 3 \, \log \left (-e^{\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} + e^{x} + \log \relax (5)\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.74, size = 189, normalized size = 6.30 \begin {gather*} \frac {x\,\ln \relax (5)\,\ln \left (3\,\ln \left (\ln \relax (5)+{\mathrm {e}}^x-{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\right )-x\right )}{\ln \relax (5)+{\mathrm {e}}^x-{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}}+\frac {x\,{\mathrm {e}}^x\,\ln \left (3\,\ln \left (\ln \relax (5)+{\mathrm {e}}^x-{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\right )-x\right )}{\ln \relax (5)+{\mathrm {e}}^x-{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}}-\frac {x\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\ln \left (3\,\ln \left (\ln \relax (5)+{\mathrm {e}}^x-{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\right )-x\right )}{\ln \relax (5)+{\mathrm {e}}^x-{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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