Optimal. Leaf size=28 \[ e^{-x+\frac {e^9}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \]
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Rubi [F] time = 14.80, 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 {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}\right ) \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\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 {\exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}\right ) \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{(-4+x) x \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx\\ &=\int \left (-\exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}\right )+\frac {2 \exp \left (9+\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}\right ) (-x-4 \log (-4+x)+x \log (-4+x))}{(-4+x) x \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}\right ) \, dx\\ &=2 \int \frac {\exp \left (9+\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}\right ) (-x-4 \log (-4+x)+x \log (-4+x))}{(-4+x) x \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx-\int \exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}\right ) \, dx\\ &=2 \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} (x-(-4+x) \log (-4+x))}{(4-x) x \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx-\int e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \, dx\\ &=2 \int \left (\frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} (x+4 \log (-4+x)-x \log (-4+x))}{4 x \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}+\frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} (-x-4 \log (-4+x)+x \log (-4+x))}{4 (-4+x) \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}\right ) \, dx-\int e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \, dx\\ &=\frac {1}{2} \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} (x+4 \log (-4+x)-x \log (-4+x))}{x \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx+\frac {1}{2} \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} (-x-4 \log (-4+x)+x \log (-4+x))}{(-4+x) \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx-\int e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \, dx\\ &=\frac {1}{2} \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} (x-(-4+x) \log (-4+x))}{x \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx+\frac {1}{2} \int \left (-\frac {4 e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{(-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}+\frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} x}{(-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}-\frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} x}{(-4+x) \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}\right ) \, dx-\int e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \, dx\\ &=\frac {1}{2} \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} x}{(-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx-\frac {1}{2} \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} x}{(-4+x) \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx+\frac {1}{2} \int \left (-\frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{\left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}+\frac {4 e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{x \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}+\frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{\log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}\right ) \, dx-2 \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{(-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx-\int e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{\left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx\right )+\frac {1}{2} \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{\log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx+\frac {1}{2} \int \left (\frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{\left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}+\frac {4 e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{(-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}\right ) \, dx-\frac {1}{2} \int \left (\frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{\log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}+\frac {4 e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{(-4+x) \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2}\right ) \, dx-2 \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{(-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx+2 \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{x \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx-\int e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \, dx\\ &=2 \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{x \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx-2 \int \frac {e^{9-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}}}{(-4+x) \log (-4+x) \left (e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )\right )^2} \, dx-\int e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.28, size = 27, normalized size = 0.96 \begin {gather*} e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.61, size = 40, normalized size = 1.43 \begin {gather*} e^{\left (-\frac {x \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right ) - e^{9}}{\log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 13.81, size = 24, normalized size = 0.86 \begin {gather*} e^{\left (-x + \frac {e^{9}}{\log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 417.42, size = 497, normalized size = 17.75
method | result | size |
risch | \({\mathrm e}^{-\frac {-i x \pi \mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{3}+2 i x \pi \mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{2} \mathrm {csgn}\left (i \ln \left (x -4\right )\right )-i x \pi \,\mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \mathrm {csgn}\left (i \ln \left (x -4\right )\right )^{2}+i x \pi \,\mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2}-i x \pi \,\mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right )+i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-i x \pi \mathrm {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{3}+i x \pi \mathrm {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right )+2 \,{\mathrm e}^{2} x -4 x \ln \relax (x )+4 x \ln \left (\ln \left (x -4\right )\right )-2 \,{\mathrm e}^{9}}{-i \pi \mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{3}+2 i \pi \mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{2} \mathrm {csgn}\left (i \ln \left (x -4\right )\right )-i \pi \,\mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \mathrm {csgn}\left (i \ln \left (x -4\right )\right )^{2}+i \pi \,\mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -4\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 \ln \left (x -4\right )^{2}}{x^{2}}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right )+2 \,{\mathrm e}^{2}-4 \ln \relax (x )+4 \ln \left (\ln \left (x -4\right )\right )}}\) | \(497\) |
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^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2} - 2 \, {\left (x - 4\right )} e^{9} \log \left (x - 4\right ) + 2 \, x e^{9}\right )} e^{\left (-\frac {x \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right ) - e^{9}}{\log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )}\right )}}{{\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.71, size = 92, normalized size = 3.29 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^2}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^9}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}\,{\left (x^2\right )}^{\frac {x}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}}{{\left ({\ln \left (x-4\right )}^2\right )}^{\frac {x}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.52, size = 39, normalized size = 1.39 \begin {gather*} e^{\frac {- x \log {\left (\frac {e^{e^{2}} \log {\left (x - 4 \right )}^{2}}{x^{2}} \right )} + e^{9}}{\log {\left (\frac {e^{e^{2}} \log {\left (x - 4 \right )}^{2}}{x^{2}} \right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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