Optimal. Leaf size=25 \[ 5-e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \]
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Rubi [F] time = 9.14, 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^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 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 {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{\left (5-2 e^x-x\right )^2 x} \, dx\\ &=\int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} (-6+x) \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (1-x+x \log \left (\frac {4}{x}\right )\right )}{x \left (-5+2 e^x+x\right )}\right ) \, dx\\ &=-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} (-6+x) \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (1-x+x \log \left (\frac {4}{x}\right )\right )}{x \left (-5+2 e^x+x\right )} \, dx\\ &=-\int \left (-\frac {6 e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2}\right ) \, dx+\int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x}\right ) \, dx\\ &=6 \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x} \, dx\\ &=6 \int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2}\right ) \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x} \, dx-\int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2}\right ) \, dx\\ &=-\left (6 \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{\left (-5+2 e^x+x\right )^2} \, dx\right )+6 \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x}{\left (-5+2 e^x+x\right )^2} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 3.45, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.61, size = 21, normalized size = 0.84 \begin {gather*} -e^{\left (\frac {\log \left (\frac {4}{x}\right ) - 1}{x + 2 \, e^{x} - 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*} \int -\frac {{\left (2 \, {\left (x - 1\right )} e^{x} - {\left (2 \, x e^{x} + x\right )} \log \left (\frac {4}{x}\right ) + 5\right )} e^{\left (\frac {\log \left (\frac {4}{x}\right ) - 1}{x + 2 \, e^{x} - 5}\right )}}{x^{3} - 10 \, x^{2} + 4 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} - 5 \, x\right )} e^{x} + 25 \, x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.20, size = 24, normalized size = 0.96
method | result | size |
risch | \(-{\mathrm e}^{\frac {2 \ln \relax (2)-\ln \relax (x )-1}{2 \,{\mathrm e}^{x}+x -5}}\) | \(24\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.07, size = 41, normalized size = 1.64 \begin {gather*} -e^{\left (\frac {2 \, \log \relax (2)}{x + 2 \, e^{x} - 5} - \frac {\log \relax (x)}{x + 2 \, e^{x} - 5} - \frac {1}{x + 2 \, e^{x} - 5}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.26, size = 29, normalized size = 1.16 \begin {gather*} -{\mathrm {e}}^{-\frac {1}{x+2\,{\mathrm {e}}^x-5}}\,{\left (\frac {4}{x}\right )}^{\frac {1}{x+2\,{\mathrm {e}}^x-5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.93, size = 17, normalized size = 0.68 \begin {gather*} - e^{\frac {\log {\left (\frac {4}{x} \right )} - 1}{x + 2 e^{x} - 5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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