Optimal. Leaf size=28 \[ e^{\log ^2\left (2+\frac {1}{2} \left (x^2+\log \left (2-e^x (2-x)\right )\right )\right )} \]
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Rubi [F] time = 13.20, 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 (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \left (8 x+e^x \left (-2-6 x+4 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{8+2 x^2+e^x \left (-8+4 x-2 x^2+x^3\right )+\left (2+e^x (-2+x)\right ) \log \left (2+e^x (-2+x)\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 \exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \left (4 x+e^x \left (-1-3 x+2 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{\left (2+e^x (-2+x)\right ) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )} \, dx\\ &=2 \int \frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \left (4 x+e^x \left (-1-3 x+2 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{\left (2+e^x (-2+x)\right ) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )} \, dx\\ &=2 \int \left (-\frac {2 \exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) (-1+x) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{(-2+x) \left (2-2 e^x+e^x x\right ) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )}+\frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \left (-1-3 x+2 x^2\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{(-2+x) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )}\right ) \, dx\\ &=2 \int \frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \left (-1-3 x+2 x^2\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{(-2+x) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )} \, dx-4 \int \frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) (-1+x) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{(-2+x) \left (2-2 e^x+e^x x\right ) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )} \, dx\\ &=2 \int \left (\frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{4+x^2+\log \left (2+e^x (-2+x)\right )}+\frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{(-2+x) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )}+\frac {2 \exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) x \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{4+x^2+\log \left (2+e^x (-2+x)\right )}\right ) \, dx-4 \int \left (\frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{\left (2-2 e^x+e^x x\right ) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )}+\frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{(-2+x) \left (2-2 e^x+e^x x\right ) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )}\right ) \, dx\\ &=2 \int \frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{4+x^2+\log \left (2+e^x (-2+x)\right )} \, dx+2 \int \frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{(-2+x) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )} \, dx+4 \int \frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) x \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{4+x^2+\log \left (2+e^x (-2+x)\right )} \, dx-4 \int \frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{\left (2-2 e^x+e^x x\right ) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )} \, dx-4 \int \frac {\exp \left (\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{(-2+x) \left (2-2 e^x+e^x x\right ) \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 24, normalized size = 0.86 \begin {gather*} e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 22, normalized size = 0.79 \begin {gather*} e^{\left (\log \left (\frac {1}{2} \, x^{2} + \frac {1}{2} \, \log \left ({\left (x - 2\right )} e^{x} + 2\right ) + 2\right )^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.28, size = 24, normalized size = 0.86 \begin {gather*} e^{\left (\log \left (\frac {1}{2} \, x^{2} + \frac {1}{2} \, \log \left (x e^{x} - 2 \, e^{x} + 2\right ) + 2\right )^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 23, normalized size = 0.82
method | result | size |
risch | \({\mathrm e}^{\ln \left (\frac {\ln \left ({\mathrm e}^{x} \left (x -2\right )+2\right )}{2}+\frac {x^{2}}{2}+2\right )^{2}}\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.67, size = 42, normalized size = 1.50 \begin {gather*} e^{\left (\log \relax (2)^{2} - 2 \, \log \relax (2) \log \left (x^{2} + \log \left ({\left (x - 2\right )} e^{x} + 2\right ) + 4\right ) + \log \left (x^{2} + \log \left ({\left (x - 2\right )} e^{x} + 2\right ) + 4\right )^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.59, size = 24, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^{{\ln \left (\frac {\ln \left (x\,{\mathrm {e}}^x-2\,{\mathrm {e}}^x+2\right )}{2}+\frac {x^2}{2}+2\right )}^2} \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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