\[ \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx \]
Optimal antiderivative \[ -{\mathrm e}^{\ln \left (x^{2}\right )+\frac {\ln \left (\left (-\ln \left (3+x \right )+16\right ) {\mathrm e}^{-1}\right )}{x}-x} \]
command
Int[(E^((-x^2 + x*Log[x^2] + Log[(16 - Log[3 + x])/E])/x)*(95*x - 16*x^2 - 16*x^3 + (-6*x + x^2 + x^3)*Log[3 + x] + (-48 - 16*x + (3 + x)*Log[3 + x])*Log[(16 - Log[3 + x])/E]))/(-48*x^2 - 16*x^3 + (3*x^2 + x^3)*Log[3 + x]),x]
Rubi 4.17.3 under Mathematica 13.3.1 output
\[ -e^{-x-\frac {1}{x}} x^2 (16-\log (x+3))^{\frac {1}{x}} \]
Rubi 4.16.1 under Mathematica 13.3.1 output
\[ \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx \]________________________________________________________________________________________