Optimal. Leaf size=22 \[ 3 \log \left (\log \left (6 \left (-2+e^{-e^x \left (1+x^2\right )} x\right )\right )\right ) \]
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Rubi [F] time = 2.68, 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 {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 \left (-1+e^x x (1+x)^2\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx\\ &=3 \int \frac {-1+e^x x (1+x)^2}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx\\ &=3 \int \left (-\frac {1}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}+\frac {e^x x}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}+\frac {2 e^x x^2}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}+\frac {e^x x^3}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}\right ) \, dx\\ &=-\left (3 \int \frac {1}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx\right )+3 \int \frac {e^x x}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx+3 \int \frac {e^x x^3}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx+6 \int \frac {e^x x^2}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.43, size = 21, normalized size = 0.95 \begin {gather*} 3 \log \left (\log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.89, size = 29, normalized size = 1.32 \begin {gather*} 3 \, \log \left (\log \left (6 \, {\left (x - 2 \, e^{\left ({\left (x^{2} + 1\right )} e^{x}\right )}\right )} e^{\left (-{\left (x^{2} + 1\right )} e^{x}\right )}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.40, size = 33, normalized size = 1.50 \begin {gather*} 3 \, \log \left (\log \left (6 \, {\left (x - 2 \, e^{\left (x^{2} e^{x} + e^{x}\right )}\right )} e^{\left (-x^{2} e^{x} - e^{x}\right )}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.55, size = 232, normalized size = 10.55
method | result | size |
risch | \(3 \ln \left (\ln \left ({\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}\right )+\frac {i \left (\pi \,\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )-\pi \,\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )^{3}+2 i \ln \relax (2)+2 i \ln \relax (3)+2 i \ln \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )}{2}\right )\) | \(232\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.66, size = 37, normalized size = 1.68 \begin {gather*} 3 \, \log \left (i \, \pi - {\left (x^{2} + 1\right )} e^{x} + \log \relax (3) + \log \relax (2) + \log \left (-x + 2 \, e^{\left (x^{2} e^{x} + e^{x}\right )}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.57, size = 22, normalized size = 1.00 \begin {gather*} 3\,\ln \left (\ln \left (6\,x\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}-12\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.88, size = 29, normalized size = 1.32 \begin {gather*} 3 \log {\left (\log {\left (\left (6 x - 12 e^{\left (x^{2} + 1\right ) e^{x}}\right ) e^{- \left (x^{2} + 1\right ) e^{x}} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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