Optimal. Leaf size=27 \[ e^{2 \left (-e^3+x+\log \left (-2+x^2+\log \left (2-e^x\right )\right )\right )^2} \]
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Rubi [F] time = 101.95, 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 (2 e^6-4 e^3 x+2 x^2+2 \left (-2 e^3+2 x\right ) \log \left (-2+x^2+\log \left (2-e^x\right )\right )+2 \log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right ) \left (16 x-16 x^2-8 x^3+e^3 \left (-16+16 x+8 x^2\right )+e^x \left (-4 x+8 x^2+4 x^3+e^3 \left (4-8 x-4 x^2\right )\right )+\left (8 e^3-8 x+e^x \left (-4 e^3+4 x\right )\right ) \log \left (2-e^x\right )+\left (16-16 x-8 x^2+e^x \left (-4+8 x+4 x^2\right )+\left (-8+4 e^x\right ) \log \left (2-e^x\right )\right ) \log \left (-2+x^2+\log \left (2-e^x\right )\right )\right )}{4-2 x^2+e^x \left (-2+x^2\right )+\left (-2+e^x\right ) \log \left (2-e^x\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (2 \left (-2+2 x+x^2\right )-e^x \left (-1+2 x+x^2\right )-\left (-2+e^x\right ) \log \left (2-e^x\right )\right ) \left (-e^3+x+\log \left (-2+x^2+\log \left (2-e^x\right )\right )\right )}{2-e^x} \, dx\\ &=4 \int \frac {\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (2 \left (-2+2 x+x^2\right )-e^x \left (-1+2 x+x^2\right )-\left (-2+e^x\right ) \log \left (2-e^x\right )\right ) \left (-e^3+x+\log \left (-2+x^2+\log \left (2-e^x\right )\right )\right )}{2-e^x} \, dx\\ &=4 \int \left (-\frac {2 \exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (e^3-x-\log \left (-2+x^2+\log \left (2-e^x\right )\right )\right )}{-2+e^x}+\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (-1+2 x+x^2+\log \left (2-e^x\right )\right ) \left (-e^3+x+\log \left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \, dx\\ &=4 \int \exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (-1+2 x+x^2+\log \left (2-e^x\right )\right ) \left (-e^3+x+\log \left (-2+x^2+\log \left (2-e^x\right )\right )\right ) \, dx-8 \int \frac {\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (e^3-x-\log \left (-2+x^2+\log \left (2-e^x\right )\right )\right )}{-2+e^x} \, dx\\ &=4 \int \left (-\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (e^3-x\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (-1+2 x+x^2+\log \left (2-e^x\right )\right )+\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (-1+2 x+x^2+\log \left (2-e^x\right )\right ) \log \left (-2+x^2+\log \left (2-e^x\right )\right )\right ) \, dx-8 \int \left (\frac {\exp \left (3+2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x}}{-2+e^x}-\frac {\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) x \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x}}{-2+e^x}-\frac {\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \log \left (-2+x^2+\log \left (2-e^x\right )\right )}{-2+e^x}\right ) \, dx\\ &=-\left (4 \int \exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (e^3-x\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (-1+2 x+x^2+\log \left (2-e^x\right )\right ) \, dx\right )+4 \int \exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \left (-1+2 x+x^2+\log \left (2-e^x\right )\right ) \log \left (-2+x^2+\log \left (2-e^x\right )\right ) \, dx-8 \int \frac {\exp \left (3+2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x}}{-2+e^x} \, dx+8 \int \frac {\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) x \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x}}{-2+e^x} \, dx+8 \int \frac {\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \log \left (-2+x^2+\log \left (2-e^x\right )\right )}{-2+e^x} \, dx\\ &=-\left (4 \int \left (\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (e^3-x\right ) \left (-1+2 x+x^2\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x}+\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (e^3-x\right ) \log \left (2-e^x\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x}\right ) \, dx\right )+4 \int \left (-\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \log \left (-2+x^2+\log \left (2-e^x\right )\right )+2 \exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) x \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \log \left (-2+x^2+\log \left (2-e^x\right )\right )+\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) x^2 \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \log \left (-2+x^2+\log \left (2-e^x\right )\right )+\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \log \left (2-e^x\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \log \left (-2+x^2+\log \left (2-e^x\right )\right )\right ) \, dx-8 \int \frac {\exp \left (3+2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x}}{-2+e^x} \, dx+8 \int \frac {\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) x \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x}}{-2+e^x} \, dx+8 \int \frac {\exp \left (2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )\right ) \left (-2+x^2+\log \left (2-e^x\right )\right )^{-1-4 e^3+4 x} \log \left (-2+x^2+\log \left (2-e^x\right )\right )}{-2+e^x} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.33, size = 54, normalized size = 2.00 \begin {gather*} e^{2 \left (\left (e^3-x\right )^2+\log ^2\left (-2+x^2+\log \left (2-e^x\right )\right )\right )} \left (-2+x^2+\log \left (2-e^x\right )\right )^{-4 e^3+4 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.08, size = 54, normalized size = 2.00 \begin {gather*} e^{\left (2 \, x^{2} - 4 \, x e^{3} + 4 \, {\left (x - e^{3}\right )} \log \left (x^{2} + \log \left (-e^{x} + 2\right ) - 2\right ) + 2 \, \log \left (x^{2} + \log \left (-e^{x} + 2\right ) - 2\right )^{2} + 2 \, e^{6}\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 [B] time = 0.06, size = 58, normalized size = 2.15
method | result | size |
risch | \(\left (\ln \left (-{\mathrm e}^{x}+2\right )+x^{2}-2\right )^{4 x -4 \,{\mathrm e}^{3}} {\mathrm e}^{2 \ln \left (\ln \left (-{\mathrm e}^{x}+2\right )+x^{2}-2\right )^{2}+2 \,{\mathrm e}^{6}-4 x \,{\mathrm e}^{3}+2 x^{2}}\) | \(58\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.02, size = 66, normalized size = 2.44 \begin {gather*} e^{\left (2 \, x^{2} - 4 \, x e^{3} + 4 \, x \log \left (x^{2} + \log \left (-e^{x} + 2\right ) - 2\right ) - 4 \, e^{3} \log \left (x^{2} + \log \left (-e^{x} + 2\right ) - 2\right ) + 2 \, \log \left (x^{2} + \log \left (-e^{x} + 2\right ) - 2\right )^{2} + 2 \, e^{6}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.24, size = 57, normalized size = 2.11 \begin {gather*} {\mathrm {e}}^{2\,{\mathrm {e}}^6}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{-4\,x\,{\mathrm {e}}^3}\,{\mathrm {e}}^{2\,{\ln \left (\ln \left (2-{\mathrm {e}}^x\right )+x^2-2\right )}^2}\,{\left (\ln \left (2-{\mathrm {e}}^x\right )+x^2-2\right )}^{4\,x-4\,{\mathrm {e}}^3} \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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