Optimal. Leaf size=36 \[ e^x \left (-\frac {1}{2} e^{2 e^{-1/x}}+\frac {-x+\log \left (\frac {x}{4}\right )}{-2+x}\right ) \]
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Rubi [F] time = 2.55, 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^{-1/x} \left (e^{\frac {1}{x}+x} \left (-4 x+6 x^2+4 x^3-2 x^4\right )+e^{2 e^{-1/x}+x} \left (-8+8 x-2 x^2+e^{\frac {1}{x}} \left (-4 x^2+4 x^3-x^4\right )\right )+e^{\frac {1}{x}+x} \left (-6 x^2+2 x^3\right ) \log \left (\frac {x}{4}\right )\right )}{8 x^2-8 x^3+2 x^4} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-1/x} \left (e^{\frac {1}{x}+x} \left (-4 x+6 x^2+4 x^3-2 x^4\right )+e^{2 e^{-1/x}+x} \left (-8+8 x-2 x^2+e^{\frac {1}{x}} \left (-4 x^2+4 x^3-x^4\right )\right )+e^{\frac {1}{x}+x} \left (-6 x^2+2 x^3\right ) \log \left (\frac {x}{4}\right )\right )}{x^2 \left (8-8 x+2 x^2\right )} \, dx\\ &=\int \frac {e^{-1/x} \left (e^{\frac {1}{x}+x} \left (-4 x+6 x^2+4 x^3-2 x^4\right )+e^{2 e^{-1/x}+x} \left (-8+8 x-2 x^2+e^{\frac {1}{x}} \left (-4 x^2+4 x^3-x^4\right )\right )+e^{\frac {1}{x}+x} \left (-6 x^2+2 x^3\right ) \log \left (\frac {x}{4}\right )\right )}{2 (-2+x)^2 x^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{-1/x} \left (e^{\frac {1}{x}+x} \left (-4 x+6 x^2+4 x^3-2 x^4\right )+e^{2 e^{-1/x}+x} \left (-8+8 x-2 x^2+e^{\frac {1}{x}} \left (-4 x^2+4 x^3-x^4\right )\right )+e^{\frac {1}{x}+x} \left (-6 x^2+2 x^3\right ) \log \left (\frac {x}{4}\right )\right )}{(-2+x)^2 x^2} \, dx\\ &=\frac {1}{2} \int e^x \left (-e^{2 e^{-1/x}}-\frac {2 e^{2 e^{-1/x}-\frac {1}{x}}}{x^2}-\frac {2 \left (2-3 x-2 x^2+x^3\right )}{(-2+x)^2 x}+\frac {2 (-3+x) \log \left (\frac {x}{4}\right )}{(-2+x)^2}\right ) \, dx\\ &=\frac {1}{2} \int \left (-e^{2 e^{-1/x}+x}-\frac {2 e^{2 e^{-1/x}-\frac {1}{x}+x}}{x^2}-\frac {2 e^x \left (2-3 x-2 x^2+x^3\right )}{(-2+x)^2 x}+\frac {2 e^x (-3+x) \log \left (\frac {x}{4}\right )}{(-2+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int e^{2 e^{-1/x}+x} \, dx\right )-\int \frac {e^{2 e^{-1/x}-\frac {1}{x}+x}}{x^2} \, dx-\int \frac {e^x \left (2-3 x-2 x^2+x^3\right )}{(-2+x)^2 x} \, dx+\int \frac {e^x (-3+x) \log \left (\frac {x}{4}\right )}{(-2+x)^2} \, dx\\ &=-\frac {e^x \log \left (\frac {x}{4}\right )}{2-x}-\frac {1}{2} \int e^{2 e^{-1/x}+x} \, dx-\int \left (e^x-\frac {2 e^x}{(-2+x)^2}+\frac {3 e^x}{2 (-2+x)}+\frac {e^x}{2 x}\right ) \, dx-\int \frac {e^{2 e^{-1/x}-\frac {1}{x}+x}}{x^2} \, dx+\int \frac {e^x}{(2-x) x} \, dx\\ &=-\frac {e^x \log \left (\frac {x}{4}\right )}{2-x}-\frac {1}{2} \int e^{2 e^{-1/x}+x} \, dx-\frac {1}{2} \int \frac {e^x}{x} \, dx-\frac {3}{2} \int \frac {e^x}{-2+x} \, dx+2 \int \frac {e^x}{(-2+x)^2} \, dx-\int e^x \, dx+\int \left (-\frac {e^x}{2 (-2+x)}+\frac {e^x}{2 x}\right ) \, dx-\int \frac {e^{2 e^{-1/x}-\frac {1}{x}+x}}{x^2} \, dx\\ &=-e^x+\frac {2 e^x}{2-x}-\frac {3}{2} e^2 \text {Ei}(-2+x)-\frac {\text {Ei}(x)}{2}-\frac {e^x \log \left (\frac {x}{4}\right )}{2-x}-\frac {1}{2} \int e^{2 e^{-1/x}+x} \, dx-\frac {1}{2} \int \frac {e^x}{-2+x} \, dx+\frac {1}{2} \int \frac {e^x}{x} \, dx+2 \int \frac {e^x}{-2+x} \, dx-\int \frac {e^{2 e^{-1/x}-\frac {1}{x}+x}}{x^2} \, dx\\ &=-e^x+\frac {2 e^x}{2-x}-\frac {e^x \log \left (\frac {x}{4}\right )}{2-x}-\frac {1}{2} \int e^{2 e^{-1/x}+x} \, dx-\int \frac {e^{2 e^{-1/x}-\frac {1}{x}+x}}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.04, size = 37, normalized size = 1.03 \begin {gather*} -\frac {e^x \left (e^{2 e^{-1/x}} (-2+x)+2 x+\log (16)-2 \log (x)\right )}{2 (-2+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.61, size = 67, normalized size = 1.86 \begin {gather*} -\frac {{\left ({\left (x - 2\right )} e^{\left ({\left (x e^{\frac {1}{x}} + 2\right )} e^{\left (-\frac {1}{x}\right )} + \frac {1}{x}\right )} + 2 \, x e^{\left (\frac {x^{2} + 1}{x}\right )} - 2 \, e^{\left (\frac {x^{2} + 1}{x}\right )} \log \left (\frac {1}{4} \, x\right )\right )} e^{\left (-\frac {1}{x}\right )}}{2 \, {\left (x - 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 38, normalized size = 1.06 \begin {gather*} -\frac {x e^{x} + 2 \, e^{x} \log \relax (2) - e^{x} \log \relax (x)}{x - 2} - \frac {1}{2} \, e^{\left (x + 2 \, e^{\left (-\frac {1}{x}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (-x^{4}+4 x^{3}-4 x^{2}\right ) {\mathrm e}^{\frac {1}{x}}-2 x^{2}+8 x -8\right ) {\mathrm e}^{x} {\mathrm e}^{2 \,{\mathrm e}^{-\frac {1}{x}}}+\left (2 x^{3}-6 x^{2}\right ) {\mathrm e}^{\frac {1}{x}} {\mathrm e}^{x} \ln \left (\frac {x}{4}\right )+\left (-2 x^{4}+4 x^{3}+6 x^{2}-4 x \right ) {\mathrm e}^{\frac {1}{x}} {\mathrm e}^{x}\right ) {\mathrm e}^{-\frac {1}{x}}}{2 x^{4}-8 x^{3}+8 x^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.86, size = 37, normalized size = 1.03 \begin {gather*} -\frac {{\left (x - 2\right )} e^{\left (x + 2 \, e^{\left (-\frac {1}{x}\right )}\right )} + 2 \, {\left (x + 2 \, \log \relax (2) - \log \relax (x)\right )} e^{x}}{2 \, {\left (x - 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.51, size = 44, normalized size = 1.22 \begin {gather*} -\frac {{\mathrm {e}}^{x+2\,{\mathrm {e}}^{-\frac {1}{x}}}}{2}-\frac {x\,{\mathrm {e}}^x}{x-2}-\frac {x\,\ln \left (\frac {x}{4}\right )\,{\mathrm {e}}^x}{2\,x-x^2} \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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