Optimal. Leaf size=32 \[ \log \left (e^{-2+x}+\frac {1-\frac {5}{x}+\log (x)}{-e^{-1+\frac {3}{x}}+x}\right ) \]
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Rubi [F] time = 21.71, 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 {-10 x^2+e^{\frac {3-x}{x}} \left (-15+8 x+x^2\right )+e^{-2+x} \left (-e^{\frac {2 (3-x)}{x}} x^3+2 e^{\frac {3-x}{x}} x^4-x^5\right )+\left (3 e^{\frac {3-x}{x}} x+x^3\right ) \log (x)}{5 x^3-x^4+e^{\frac {3-x}{x}} \left (-5 x^2+x^3\right )+e^{-2+x} \left (-e^{\frac {2 (3-x)}{x}} x^3+2 e^{\frac {3-x}{x}} x^4-x^5\right )+\left (e^{\frac {3-x}{x}} x^3-x^4\right ) \log (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 {-10 e^4 x^2-e^{\frac {6}{x}+x} x^3+2 e^{1+\frac {3}{x}+x} x^4-e^{2+x} x^5+e^{3+\frac {3}{x}} \left (-15+8 x+x^2\right )+e^3 x \left (3 e^{3/x}+e x^2\right ) \log (x)}{x^2 \left (e^{3/x}-e x\right ) \left (e^3 (-5+x)-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )} \, dx\\ &=\int \left (1-\frac {e^3 \left (-15 e^{3/x}+8 e^{3/x} x-10 e x^2+6 e^{3/x} x^2-5 e x^3-e^{3/x} x^3+e x^4+3 e^{3/x} x \log (x)+e x^3 \log (x)-e^{3/x} x^3 \log (x)+e x^4 \log (x)\right )}{x^2 \left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )}\right ) \, dx\\ &=x-e^3 \int \frac {-15 e^{3/x}+8 e^{3/x} x-10 e x^2+6 e^{3/x} x^2-5 e x^3-e^{3/x} x^3+e x^4+3 e^{3/x} x \log (x)+e x^3 \log (x)-e^{3/x} x^3 \log (x)+e x^4 \log (x)}{x^2 \left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )} \, dx\\ &=x-e^3 \int \frac {-e x^2 \left (-10-5 x+x^2\right )+e^{3/x} \left (15-8 x-6 x^2+x^3\right )-\left (e x^3 (1+x)-e^{3/x} x \left (-3+x^2\right )\right ) \log (x)}{x^2 \left (e^{3/x}-e x\right ) \left (e^3 (-5+x)-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )} \, dx\\ &=x-e^3 \int \left (\frac {6 e^{3/x}}{\left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )}-\frac {15 e^{3/x}}{x^2 \left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )}+\frac {8 e^{3/x}}{x \left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )}-\frac {e^{3/x} x}{\left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )}+\frac {3 e^{3/x} \log (x)}{x \left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )}-\frac {e^{3/x} x \log (x)}{\left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )}-\frac {10 e}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )}-\frac {5 e x}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )}+\frac {e x^2}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )}+\frac {e x \log (x)}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )}+\frac {e x^2 \log (x)}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )}\right ) \, dx\\ &=x+e^3 \int \frac {e^{3/x} x}{\left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )} \, dx+e^3 \int \frac {e^{3/x} x \log (x)}{\left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )} \, dx-\left (3 e^3\right ) \int \frac {e^{3/x} \log (x)}{x \left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )} \, dx-\left (6 e^3\right ) \int \frac {e^{3/x}}{\left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )} \, dx-\left (8 e^3\right ) \int \frac {e^{3/x}}{x \left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )} \, dx+\left (15 e^3\right ) \int \frac {e^{3/x}}{x^2 \left (e^{3/x}-e x\right ) \left (5 e^3-e^3 x+e^{\frac {3}{x}+x} x-e^{1+x} x^2-e^3 x \log (x)\right )} \, dx-e^4 \int \frac {x^2}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )} \, dx-e^4 \int \frac {x \log (x)}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )} \, dx-e^4 \int \frac {x^2 \log (x)}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )} \, dx+\left (5 e^4\right ) \int \frac {x}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )} \, dx+\left (10 e^4\right ) \int \frac {1}{\left (-e^{3/x}+e x\right ) \left (-5 e^3+e^3 x-e^{\frac {3}{x}+x} x+e^{1+x} x^2+e^3 x \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 0.63, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-10 x^2+e^{\frac {3-x}{x}} \left (-15+8 x+x^2\right )+e^{-2+x} \left (-e^{\frac {2 (3-x)}{x}} x^3+2 e^{\frac {3-x}{x}} x^4-x^5\right )+\left (3 e^{\frac {3-x}{x}} x+x^3\right ) \log (x)}{5 x^3-x^4+e^{\frac {3-x}{x}} \left (-5 x^2+x^3\right )+e^{-2+x} \left (-e^{\frac {2 (3-x)}{x}} x^3+2 e^{\frac {3-x}{x}} x^4-x^5\right )+\left (e^{\frac {3-x}{x}} x^3-x^4\right ) \log (x)} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.52, size = 50, normalized size = 1.56 \begin {gather*} -\log \left (-x + e^{\left (-\frac {x - 3}{x}\right )}\right ) + \log \left (\frac {{\left (x^{2} - x e^{\left (-\frac {x - 3}{x}\right )}\right )} e^{\left (x - 2\right )} + x \log \relax (x) + x - 5}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [A] time = 0.06, size = 54, normalized size = 1.69
method | result | size |
risch | \(-\ln \left ({\mathrm e}^{-\frac {x -3}{x}}-x \right )-1+\ln \left (\ln \relax (x )+\frac {x^{2} {\mathrm e}^{x -2}-x \,{\mathrm e}^{\frac {x^{2}-3 x +3}{x}}+x -5}{x}\right )\) | \(54\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 61, normalized size = 1.91 \begin {gather*} x - \log \left (-x e + e^{\frac {3}{x}}\right ) + \log \left (-\frac {{\left (x^{2} e^{\left (x + 1\right )} + x e^{3} \log \relax (x) + x e^{3} - x e^{\left (x + \frac {3}{x}\right )} - 5 \, e^{3}\right )} e^{\left (-x\right )}}{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 = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{x-2}\,\left (x^3\,{\mathrm {e}}^{-\frac {2\,\left (x-3\right )}{x}}-2\,x^4\,{\mathrm {e}}^{-\frac {x-3}{x}}+x^5\right )-\ln \relax (x)\,\left (3\,x\,{\mathrm {e}}^{-\frac {x-3}{x}}+x^3\right )+10\,x^2-{\mathrm {e}}^{-\frac {x-3}{x}}\,\left (x^2+8\,x-15\right )}{{\mathrm {e}}^{x-2}\,\left (x^3\,{\mathrm {e}}^{-\frac {2\,\left (x-3\right )}{x}}-2\,x^4\,{\mathrm {e}}^{-\frac {x-3}{x}}+x^5\right )-\ln \relax (x)\,\left (x^3\,{\mathrm {e}}^{-\frac {x-3}{x}}-x^4\right )+{\mathrm {e}}^{-\frac {x-3}{x}}\,\left (5\,x^2-x^3\right )-5\,x^3+x^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.27, size = 26, normalized size = 0.81 \begin {gather*} \log {\left (e^{x - 2} + \frac {x \log {\relax (x )} + x - 5}{x^{2} - x e^{\frac {3 - x}{x}}} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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