Optimal. Leaf size=27 \[ -5+\frac {\log \left (10 \left (e^{3 e^x}+\frac {4}{x}\right )+x\right )}{\log (-2+x)} \]
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Rubi [F] time = 3.80, 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 {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {\left (-40+\left (1+30 e^{3 e^x+x}\right ) x^2\right ) \log (-2+x)}{x \left (40+10 e^{3 e^x} x+x^2\right )}-\frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{-2+x}}{\log ^2(-2+x)} \, dx\\ &=\int \left (\frac {30 e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)}+\frac {80 \log (-2+x)-40 x \log (-2+x)-2 x^2 \log (-2+x)+x^3 \log (-2+x)-40 x \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )-10 e^{3 e^x} x^2 \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )-x^3 \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) x \left (40+10 e^{3 e^x} x+x^2\right ) \log ^2(-2+x)}\right ) \, dx\\ &=30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \frac {80 \log (-2+x)-40 x \log (-2+x)-2 x^2 \log (-2+x)+x^3 \log (-2+x)-40 x \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )-10 e^{3 e^x} x^2 \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )-x^3 \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) x \left (40+10 e^{3 e^x} x+x^2\right ) \log ^2(-2+x)} \, dx\\ &=30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \frac {-\frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)}{x \left (40+10 e^{3 e^x} x+x^2\right )}+\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(2-x) \log ^2(-2+x)} \, dx\\ &=30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \left (\frac {-40+x^2}{x \left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)}-\frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) \log ^2(-2+x)}\right ) \, dx\\ &=30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \frac {-40+x^2}{x \left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx-\int \frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) \log ^2(-2+x)} \, dx\\ &=30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \left (-\frac {40}{x \left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)}+\frac {x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)}\right ) \, dx-\int \frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) \log ^2(-2+x)} \, dx\\ &=30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx-40 \int \frac {1}{x \left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \frac {x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx-\int \frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) \log ^2(-2+x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.64, size = 24, normalized size = 0.89 \begin {gather*} \frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{\log (-2+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 35, normalized size = 1.30 \begin {gather*} \frac {\log \left (\frac {{\left (10 \, x e^{\left (x + 3 \, e^{x}\right )} + {\left (x^{2} + 40\right )} e^{x}\right )} e^{\left (-x\right )}}{x}\right )}{\log \left (x - 2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 26, normalized size = 0.96 \begin {gather*} \frac {\log \left (x^{2} + 10 \, x e^{\left (3 \, e^{x}\right )} + 40\right ) - \log \relax (x)}{\log \left (x - 2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.16, size = 186, normalized size = 6.89
method | result | size |
risch | \(\frac {\ln \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{\ln \left (x -2\right )}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )\right ) \mathrm {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )\right ) \mathrm {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )^{3}+2 \ln \relax (x )}{2 \ln \left (x -2\right )}\) | \(186\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 26, normalized size = 0.96 \begin {gather*} \frac {\log \left (x^{2} + 10 \, x e^{\left (3 \, e^{x}\right )} + 40\right ) - \log \relax (x)}{\log \left (x - 2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.68, size = 25, normalized size = 0.93 \begin {gather*} \frac {\ln \left (\frac {10\,x\,{\mathrm {e}}^{3\,{\mathrm {e}}^x}+x^2+40}{x}\right )}{\ln \left (x-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 4.83, size = 22, normalized size = 0.81 \begin {gather*} \frac {\log {\left (\frac {x^{2} + 10 x e^{3 e^{x}} + 40}{x} \right )}}{\log {\left (x - 2 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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