Optimal. Leaf size=27 \[ \left (3-\frac {5}{3} e^{\frac {e^x}{x \left (x+x \log \left (\frac {1}{x}\right )\right )}}\right )^2 \]
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Rubi [A]
time = 4.33, antiderivative size = 42, normalized size of antiderivative = 1.56, number of steps
used = 4, number of rules used = 3, integrand size = 119, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6820, 12,
6840} \begin {gather*} \frac {25}{9} e^{\frac {2 e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}-10 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 6820
Rule 6840
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 e^{x+\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}} \left (9-5 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \left (1-x-(-2+x) \log \left (\frac {1}{x}\right )\right )}{9 x^3 \left (1+\log \left (\frac {1}{x}\right )\right )^2} \, dx\\ &=\frac {10}{9} \int \frac {e^{x+\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}} \left (9-5 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \left (1-x-(-2+x) \log \left (\frac {1}{x}\right )\right )}{x^3 \left (1+\log \left (\frac {1}{x}\right )\right )^2} \, dx\\ &=\frac {2}{9} \text {Subst}\left (\int (9+x) \, dx,x,-5 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right )\\ &=-10 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}+\frac {25}{9} e^{\frac {2 e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 1.14, size = 46, normalized size = 1.70 \begin {gather*} \frac {10}{9} \left (-9 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}+\frac {5}{2} e^{\frac {2 e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.30, size = 34, normalized size = 1.26
method | result | size |
risch | \(\frac {25 \,{\mathrm e}^{-\frac {2 \,{\mathrm e}^{x}}{x^{2} \left (\ln \left (x \right )-1\right )}}}{9}-10 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{x^{2} \left (\ln \left (x \right )-1\right )}}\) | \(34\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.36, size = 43, normalized size = 1.59 \begin {gather*} -\frac {5}{9} \, {\left (18 \, e^{\left (\frac {e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} - 5\right )} e^{\left (-\frac {2 \, e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 42, normalized size = 1.56 \begin {gather*} \frac {25}{9} \, e^{\left (\frac {2 \, e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} - 10 \, e^{\left (\frac {e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.42, size = 39, normalized size = 1.44 \begin {gather*} \frac {25 e^{\frac {2 e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}}}{9} - 10 e^{\frac {e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 43, normalized size = 1.59 \begin {gather*} -10 \, e^{\left (-\frac {e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} + \frac {25}{9} \, e^{\left (-\frac {2 \, e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 6.06, size = 42, normalized size = 1.56 \begin {gather*} \frac {5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}\,\left (5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}-18\right )}{9} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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