Optimal. Leaf size=25 \[ e^{10}-2 x-\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \]
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Rubi [F] time = 1.05, 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 {\left (-10-2 e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )+\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \left (-\frac {1}{2} e^{\frac {e^x}{2}+x} x^2+\left (-10 x-2 e^{\frac {e^x}{2}} x\right ) \log \left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right )}{\left (5+e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^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 \left (-2-\frac {e^{\frac {1}{2} \left (e^x+2 x\right )} x^2 \log ^{-1+x^2}\left (5+e^{\frac {e^x}{2}}\right )}{2 \left (5+e^{\frac {e^x}{2}}\right )}-2 x \log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right ) \, dx\\ &=-2 x-\frac {1}{2} \int \frac {e^{\frac {1}{2} \left (e^x+2 x\right )} x^2 \log ^{-1+x^2}\left (5+e^{\frac {e^x}{2}}\right )}{5+e^{\frac {e^x}{2}}} \, dx-2 \int x \log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.23, size = 22, normalized size = 0.88 \begin {gather*} -2 x-\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.83, size = 21, normalized size = 0.84 \begin {gather*} -2 \, x - \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )^{\left (x^{2}\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^{2} e^{\left (x + e^{\left (x - \log \relax (2)\right )} - \log \relax (2)\right )} + 2 \, {\left (x e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5 \, x\right )} \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right ) \log \left (\log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )\right )\right )} \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )^{\left (x^{2}\right )} + 2 \, {\left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )} \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )}{{\left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )} \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 19, normalized size = 0.76
method | result | size |
risch | \(-2 x -\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{x}}{2}}+5\right )^{x^{2}}\) | \(19\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.73, size = 18, normalized size = 0.72 \begin {gather*} -2 \, x - \log \left (e^{\left (\frac {1}{2} \, e^{x}\right )} + 5\right )^{\left (x^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.15, size = 18, normalized size = 0.72 \begin {gather*} -2\,x-{\ln \left ({\mathrm {e}}^{\frac {{\mathrm {e}}^x}{2}}+5\right )}^{x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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