Optimal. Leaf size=15 \[ e^{x^{e^{1-\frac {e^x}{2}}}} \]
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Rubi [F] time = 1.15, 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 {1}{2} e^{\frac {1}{2} \left (2-e^x\right )+x^{e^{\frac {1}{2} \left (2-e^x\right )}}} x^{-1+e^{\frac {1}{2} \left (2-e^x\right )}} \left (2-e^x x \log (x)\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int e^{\frac {1}{2} \left (2-e^x\right )+x^{e^{\frac {1}{2} \left (2-e^x\right )}}} x^{-1+e^{\frac {1}{2} \left (2-e^x\right )}} \left (2-e^x x \log (x)\right ) \, dx\\ &=\frac {1}{2} \int e^{\frac {1}{2} \left (2-e^x+2 x^{e^{1-\frac {e^x}{2}}}\right )} x^{-1+e^{\frac {1}{2} \left (2-e^x\right )}} \left (2-e^x x \log (x)\right ) \, dx\\ &=\frac {1}{2} \int \left (2 e^{\frac {1}{2} \left (2-e^x+2 x^{e^{1-\frac {e^x}{2}}}\right )} x^{-1+e^{\frac {1}{2} \left (2-e^x\right )}}-e^{x+\frac {1}{2} \left (2-e^x+2 x^{e^{1-\frac {e^x}{2}}}\right )} x^{e^{\frac {1}{2} \left (2-e^x\right )}} \log (x)\right ) \, dx\\ &=-\left (\frac {1}{2} \int e^{x+\frac {1}{2} \left (2-e^x+2 x^{e^{1-\frac {e^x}{2}}}\right )} x^{e^{\frac {1}{2} \left (2-e^x\right )}} \log (x) \, dx\right )+\int e^{\frac {1}{2} \left (2-e^x+2 x^{e^{1-\frac {e^x}{2}}}\right )} x^{-1+e^{\frac {1}{2} \left (2-e^x\right )}} \, dx\\ &=\frac {1}{2} \int \frac {\int e^{1-\frac {e^x}{2}+x+x^{e^{1-\frac {e^x}{2}}}} x^{e^{\frac {1}{2} \left (2-e^x\right )}} \, dx}{x} \, dx-\frac {1}{2} \log (x) \int e^{\frac {1}{2} \left (2-e^x+2 x+2 x^{e^{1-\frac {e^x}{2}}}\right )} x^{e^{\frac {1}{2} \left (2-e^x\right )}} \, dx+\int e^{\frac {1}{2} \left (2-e^x+2 x^{e^{1-\frac {e^x}{2}}}\right )} x^{-1+e^{\frac {1}{2} \left (2-e^x\right )}} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.28, size = 15, normalized size = 1.00 \begin {gather*} e^{x^{e^{1-\frac {e^x}{2}}}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 10, normalized size = 0.67 \begin {gather*} e^{\left (x^{e^{\left (-\frac {1}{2} \, e^{x} + 1\right )}}\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 (x e^{x} \log \relax (x) - 2\right )} x^{e^{\left (-\frac {1}{2} \, e^{x} + 1\right )}} e^{\left (x^{e^{\left (-\frac {1}{2} \, e^{x} + 1\right )}} - \frac {1}{2} \, e^{x} + 1\right )}}{2 \, x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 11, normalized size = 0.73
method | result | size |
risch | \({\mathrm e}^{x^{{\mathrm e}^{1-\frac {{\mathrm e}^{x}}{2}}}}\) | \(11\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.51, size = 10, normalized size = 0.67 \begin {gather*} e^{\left (x^{e^{\left (-\frac {1}{2} \, e^{x} + 1\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.46, size = 11, normalized size = 0.73 \begin {gather*} {\mathrm {e}}^{x^{\frac {\mathrm {e}}{\sqrt {{\mathrm {e}}^{{\mathrm {e}}^x}}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 41.79, size = 14, normalized size = 0.93 \begin {gather*} e^{e^{e^{1 - \frac {e^{x}}{2}} \log {\relax (x )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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