Optimal. Leaf size=27 \[ 2+e^{-2+\frac {4}{\log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}} \]
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Rubi [F] time = 17.92, 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 {\exp \left (\frac {4-2 \log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}{\log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}\right ) \left (-24-4 x-48 x^2-4 x^3-4 x \log (4)+4 x \log (x)\right )}{\left (6 x^3+x^4+\left (-6 x-x^2\right ) \log (4)+\left (6 x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {x^2-\log (4)+\log (x)}{6+x}\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 {\exp \left (\frac {4-2 \log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}{\log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}\right ) \left (-24-48 x^2-4 x^3+x (-4-4 \log (4))+4 x \log (x)\right )}{\left (6 x^3+x^4+\left (-6 x-x^2\right ) \log (4)+\left (6 x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )} \, dx\\ &=\int \frac {4 e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (-6-12 x^2-x^3-x (1+\log (4))+x \log (x)\right )}{x (6+x) \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )} \, dx\\ &=4 \int \frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (-6-12 x^2-x^3-x (1+\log (4))+x \log (x)\right )}{x (6+x) \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )} \, dx\\ &=4 \int \left (\frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (6+12 x^2+x^3+x (1+\log (4))-x \log (x)\right )}{6 (6+x) \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}+\frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (-6-12 x^2-x^3-x (1+\log (4))+x \log (x)\right )}{6 x \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}\right ) \, dx\\ &=\frac {2}{3} \int \frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (6+12 x^2+x^3+x (1+\log (4))-x \log (x)\right )}{(6+x) \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )} \, dx+\frac {2}{3} \int \frac {e^{-2+\frac {4}{\log \left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )}} \left (-6-12 x^2-x^3-x (1+\log (4))+x \log (x)\right )}{x \left (x^2+\log \left (\frac {x}{4}\right )\right ) \log ^2\left (\frac {x^2+\log \left (\frac {x}{4}\right )}{6+x}\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [F] time = 0.28, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {4-2 \log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}{\log \left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )}} \left (-24-4 x-48 x^2-4 x^3-4 x \log (4)+4 x \log (x)\right )}{\left (6 x^3+x^4+\left (-6 x-x^2\right ) \log (4)+\left (6 x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {x^2-\log (4)+\log (x)}{6+x}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 1.07, size = 41, normalized size = 1.52 \begin {gather*} e^{\left (-\frac {2 \, {\left (\log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\right ) - 2\right )}}{\log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\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 {4 \, {\left (x^{3} + 12 \, x^{2} + 2 \, x \log \relax (2) - x \log \relax (x) + x + 6\right )} e^{\left (-\frac {2 \, {\left (\log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\right ) - 2\right )}}{\log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\right )}\right )}}{{\left (x^{4} + 6 \, x^{3} - 2 \, {\left (x^{2} + 6 \, x\right )} \log \relax (2) + {\left (x^{2} + 6 \, x\right )} \log \relax (x)\right )} \log \left (\frac {x^{2} - 2 \, \log \relax (2) + \log \relax (x)}{x + 6}\right )^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 12.63, size = 433, normalized size = 16.04
method | result | size |
risch | \({\mathrm e}^{-\frac {2 \left (i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2} \mathrm {csgn}\left (\frac {i}{x +6}\right )+i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2} \mathrm {csgn}\left (i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )\right )-i \pi \,\mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right ) \mathrm {csgn}\left (\frac {i}{x +6}\right ) \mathrm {csgn}\left (i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )\right )-2 i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2}+2 i \pi -2 \ln \left (x +6\right )+2 \ln \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )+2 \ln \relax (2)-4\right )}{i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2} \mathrm {csgn}\left (\frac {i}{x +6}\right )+i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2} \mathrm {csgn}\left (i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )\right )-i \pi \,\mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right ) \mathrm {csgn}\left (\frac {i}{x +6}\right ) \mathrm {csgn}\left (i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )\right )-2 i \pi \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )}{x +6}\right )^{2}+2 i \pi -2 \ln \left (x +6\right )+2 \ln \left (-\frac {x^{2}}{2}+\ln \relax (2)-\frac {\ln \relax (x )}{2}\right )+2 \ln \relax (2)}}\) | \(433\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.96, size = 384, normalized size = 14.22 \begin {gather*} \frac {x^{3} e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )}}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} + \frac {12 \, x^{2} e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )}}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} + \frac {2 \, x e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )} \log \relax (2)}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} - \frac {x e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )} \log \relax (x)}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} + \frac {x e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )}}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} + \frac {6 \, e^{\left (\frac {4}{\log \left (x^{2} - 2 \, \log \relax (2) + \log \relax (x)\right ) - \log \left (x + 6\right )}\right )}}{x^{3} e^{2} + 12 \, x^{2} e^{2} + x {\left (2 \, \log \relax (2) + 1\right )} e^{2} - x e^{2} \log \relax (x) + 6 \, e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.05, size = 23, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{-2}\,{\mathrm {e}}^{\frac {4}{\ln \left (\frac {\ln \left (\frac {x}{4}\right )+x^2}{x+6}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 62.22, size = 37, normalized size = 1.37 \begin {gather*} e^{\frac {4 - 2 \log {\left (\frac {x^{2} + \log {\relax (x )} - 2 \log {\relax (2 )}}{x + 6} \right )}}{\log {\left (\frac {x^{2} + \log {\relax (x )} - 2 \log {\relax (2 )}}{x + 6} \right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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