Optimal. Leaf size=25 \[ e^{5+e^x+x} x \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \]
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Rubi [F] time = 2.12, 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 {e^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+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 e^{5+e^x+x} \left (2 \left (2+x+e^x x\right )-\frac {2 x}{(-2+x) \log (-2+x)}-\left (1+x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \, dx\\ &=\int \left (2 e^{5+e^x+x} \left (2+x+e^x x\right )-\frac {2 e^{5+e^x+x} x}{(-2+x) \log (-2+x)}-e^{5+e^x+x} \left (1+x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \, dx\\ &=2 \int e^{5+e^x+x} \left (2+x+e^x x\right ) \, dx-2 \int \frac {e^{5+e^x+x} x}{(-2+x) \log (-2+x)} \, dx-\int e^{5+e^x+x} \left (1+x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right ) \, dx\\ &=-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int \left (2 e^{5+e^x+x}+e^{5+e^x+x} x+e^{5+e^x+2 x} x\right ) \, dx-2 \int \left (\frac {e^{5+e^x+x}}{\log (-2+x)}+\frac {2 e^{5+e^x+x}}{(-2+x) \log (-2+x)}\right ) \, dx+\int 2 e^{5+e^x+x} \left (-1+\frac {x}{(-2+x) \log (-2+x)}\right ) \, dx\\ &=-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx+2 \int e^{5+e^x+x} \left (-1+\frac {x}{(-2+x) \log (-2+x)}\right ) \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx+4 \int e^{5+e^x+x} \, dx-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx\\ &=-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx+2 \int \left (-e^{5+e^x+x}+\frac {e^{5+e^x+x} x}{(-2+x) \log (-2+x)}\right ) \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx+4 \operatorname {Subst}\left (\int e^{5+x} \, dx,x,e^x\right )\\ &=4 e^{5+e^x}-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}-2 \int e^{5+e^x+x} \, dx+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx+2 \int \frac {e^{5+e^x+x} x}{(-2+x) \log (-2+x)} \, dx-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx\\ &=4 e^{5+e^x}-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx+2 \int \left (\frac {e^{5+e^x+x}}{\log (-2+x)}+\frac {2 e^{5+e^x+x}}{(-2+x) \log (-2+x)}\right ) \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx-2 \operatorname {Subst}\left (\int e^{5+x} \, dx,x,e^x\right )-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx\\ &=2 e^{5+e^x}-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.67, size = 24, normalized size = 0.96 \begin {gather*} -e^{5+e^x+x} x \left (-2+\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.84, size = 24, normalized size = 0.96 \begin {gather*} x e^{\left (x + e^{x} + \log \left (-\log \left (\frac {\log \left (x - 2\right )^{2}}{x^{2}}\right ) + 2\right ) + 5\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*} \mathit {undef} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.38, size = 627, normalized size = 25.08
method | result | size |
risch | \(\frac {2 i \left (-\frac {i \pi x \,\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (i \ln \left (x -2\right )\right )^{2} \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )}{2}-i \pi x \,\mathrm {csgn}\left (i \ln \left (x -2\right )\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{2}-\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}+i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{3}}{2}+\frac {i \pi x \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{3}}{2}-\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+2 x +\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )}{2}+2 x \ln \relax (x )-2 x \ln \left (\ln \left (x -2\right )\right )\right ) \left (2 \ln \relax (x )-2 \ln \left (\ln \left (x -2\right )\right )+\frac {i \pi \,\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \left (-\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )+\mathrm {csgn}\left (i \ln \left (x -2\right )\right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}+\frac {i \pi \,\mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right ) \left (-\mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )+\mathrm {csgn}\left (\frac {i}{x^{2}}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )+\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )\right )}{2}+2\right ) {\mathrm e}^{{\mathrm e}^{x}+5+x}}{4 i \ln \relax (x )+4 i-4 i \ln \left (\ln \left (x -2\right )\right )-\pi \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{3}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )-\pi \mathrm {csgn}\left (i \ln \left (x -2\right )\right )^{2} \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )+2 \pi \,\mathrm {csgn}\left (i \ln \left (x -2\right )\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{2}+\pi \,\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{2}}\) | \(627\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.74, size = 33, normalized size = 1.32 \begin {gather*} -2 \, x e^{\left (x + e^{x} + 5\right )} \log \left (\log \left (x - 2\right )\right ) + 2 \, {\left (x e^{5} \log \relax (x) + x e^{5}\right )} e^{\left (x + e^{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {{\mathrm {e}}^{x+\ln \left (2-\ln \left (\frac {{\ln \left (x-2\right )}^2}{x^2}\right )\right )+{\mathrm {e}}^x+5}\,\left (2\,x+\ln \left (x-2\right )\,\left ({\mathrm {e}}^x\,\left (4\,x-2\,x^2\right )-2\,x^2+8\right )-\ln \left (x-2\right )\,\ln \left (\frac {{\ln \left (x-2\right )}^2}{x^2}\right )\,\left (x+{\mathrm {e}}^x\,\left (2\,x-x^2\right )-x^2+2\right )\right )}{\ln \left (x-2\right )\,\left (2\,x-4\right )-\ln \left (x-2\right )\,\ln \left (\frac {{\ln \left (x-2\right )}^2}{x^2}\right )\,\left (x-2\right )} \,d 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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