Optimal. Leaf size=34 \[ e^{4+\frac {x+\frac {-4+\frac {x}{2}-\log (x)}{\log (4-2 x-\log (5))}}{x}} \]
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Rubi [F] time = 8.41, 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 {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (2 x^3+x^2 (-4+\log (5))\right ) \log ^2(4-2 x-\log (5))} \, dx\\ &=\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{x^2 (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx\\ &=\int \left (\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (3+\log (x))}{x^2 \log (4-2 x-\log (5))}\right ) \, dx\\ &=\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (3+\log (x))}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=\int \left (\frac {2 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (-8+x-2 \log (x))}{(-4+\log (5)) (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x (-4+\log (5)) \log ^2(4-2 x-\log (5))}\right ) \, dx+\int \left (\frac {3 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))}\right ) \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (-8+x-2 \log (x))}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x \log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \left (-\frac {8 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) x}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}-\frac {2 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}\right ) \, dx}{4-\log (5)}+\frac {\int \left (-\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))}+\frac {8 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))}+\frac {2 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))}\right ) \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) x}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {8 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \left (\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{2 \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (4-\log (5))}{2 (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}\right ) \, dx}{4-\log (5)}-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {8 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {8 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}-\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 30, normalized size = 0.88 \begin {gather*} e^{5+\frac {-8+x-2 \log (x)}{2 x \log (4-2 x-\log (5))}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 38, normalized size = 1.12 \begin {gather*} e^{\left (\frac {10 \, x \log \left (-2 \, x - \log \relax (5) + 4\right ) + x - 2 \, \log \relax (x) - 8}{2 \, x \log \left (-2 \, x - \log \relax (5) + 4\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.81, size = 53, normalized size = 1.56 \begin {gather*} e^{\left (-\frac {\log \relax (x)}{x \log \left (-2 \, x - \log \relax (5) + 4\right )} + \frac {1}{2 \, \log \left (-2 \, x - \log \relax (5) + 4\right )} - \frac {4}{x \log \left (-2 \, x - \log \relax (5) + 4\right )} + 5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 41, normalized size = 1.21
method | result | size |
risch | \({\mathrm e}^{-\frac {-10 x \ln \left (-\ln \relax (5)+4-2 x \right )+2 \ln \relax (x )-x +8}{2 x \ln \left (-\ln \relax (5)+4-2 x \right )}}\) | \(41\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.85, size = 56, normalized size = 1.65 \begin {gather*} \frac {{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {4}{x\,\ln \left (4-\ln \relax (5)-2\,x\right )}}\,{\mathrm {e}}^{\frac {1}{2\,\ln \left (4-\ln \relax (5)-2\,x\right )}}}{x^{\frac {1}{x\,\ln \left (4-\ln \relax (5)-2\,x\right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.69, size = 34, normalized size = 1.00 \begin {gather*} e^{\frac {5 x \log {\left (- 2 x - \log {\relax (5 )} + 4 \right )} + \frac {x}{2} - \log {\relax (x )} - 4}{x \log {\left (- 2 x - \log {\relax (5 )} + 4 \right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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