Optimal. Leaf size=30 \[ e^{\frac {5+e^{5+\frac {x+2 x^2}{1+x}}+x-\log (x)}{x}} \]
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Rubi [F] time = 25.81, 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 {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2+2 x^3+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^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2 \left (1+2 x+x^2\right )} \, dx\\ &=\int \frac {e^{\frac {5+e^{\frac {5+6 x+2 x^2}{1+x}}+x-\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2 (1+x)^2} \, dx\\ &=\int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} \left (-6-12 x-6 x^2+e^{\frac {5+6 x+2 x^2}{1+x}} \left (-1-x+3 x^2+2 x^3\right )+\left (1+2 x+x^2\right ) \log (x)\right )}{x^2 (1+x)^2} \, dx\\ &=\int \left (\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right ) (1+2 x) \left (-1+x+x^2\right )}{x^2 (1+x)^2}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} (-6+\log (x))}{x^2}\right ) \, dx\\ &=\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right ) (1+2 x) \left (-1+x+x^2\right )}{x^2 (1+x)^2} \, dx+\int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} (-6+\log (x))}{x^2} \, dx\\ &=\int \left (-\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{x^2}+\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{x}+\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{(1+x)^2}+\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{1+x}\right ) \, dx+\int \left (-\frac {6 e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}}}{x^2}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} \log (x)}{x^2}\right ) \, dx\\ &=-\left (6 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}}}{x^2} \, dx\right )-\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{x^2} \, dx+\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{x} \, dx+\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{(1+x)^2} \, dx+\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}+\frac {5+6 x+2 x^2}{1+x}-\frac {\log (x)}{x}\right )}{1+x} \, dx+\int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {5+6 x+2 x^2}{1+x}}}{x}-\frac {\log (x)}{x}} \log (x)}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.24, size = 29, normalized size = 0.97 \begin {gather*} e^{\frac {5+e^{4+2 x+\frac {1}{1+x}}+x}{x}} x^{-1/x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 29, normalized size = 0.97 \begin {gather*} e^{\left (\frac {x + e^{\left (\frac {2 \, x^{2} + 6 \, x + 5}{x + 1}\right )} - \log \relax (x) + 5}{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.45, size = 46, normalized size = 1.53 \begin {gather*} e^{\left (\frac {e^{\left (\frac {2 \, x^{2}}{x + 1} + \frac {6 \, x}{x + 1} + \frac {5}{x + 1}\right )}}{x} - \frac {\log \relax (x)}{x} + \frac {5}{x} + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 33, normalized size = 1.10
method | result | size |
risch | \({\mathrm e}^{-\frac {\ln \relax (x )-{\mathrm e}^{\frac {2 x^{2}+6 x +5}{x +1}}-5-x}{x}}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.59, size = 30, normalized size = 1.00 \begin {gather*} e^{\left (\frac {e^{\left (2 \, x + \frac {1}{x + 1} + 4\right )}}{x} - \frac {\log \relax (x)}{x} + \frac {5}{x} + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.91, size = 49, normalized size = 1.63 \begin {gather*} \frac {\mathrm {e}\,{\mathrm {e}}^{5/x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {6\,x}{x+1}}\,{\mathrm {e}}^{\frac {2\,x^2}{x+1}}\,{\mathrm {e}}^{\frac {5}{x+1}}}{x}}}{x^{1/x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.13, size = 24, normalized size = 0.80 \begin {gather*} e^{\frac {x + e^{\frac {2 x^{2} + 6 x + 5}{x + 1}} - \log {\relax (x )} + 5}{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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