Integral with exponent and trigonometric functions

I shall give you the result obtained after a lot of simplifications of the result provided by a CAS. $$J = int e^{-frac{1}{2}cos (x)} cos left( 3x + frac{1}{2} sin (x) right), dx$$ $$J=color{red}{frac{i}{96} left(text{Ei}left(-frac{e^{i x}}{2} right)-text{Ei}left(-frac{e^{-i x}}{2}right)right)}+$$ $$color{blue}{frac{1}{24} left(sin left(x+frac{sin (x)}{2}right)-2 sin left(2x+frac{sin (x)}{2} right)+8 sin left(3x+frac{sin (x)}{2} right)right) times }$$ $$color{blue}{left(cosh left(frac{cos (x)}{2}right)-sinh left(frac{cos (x)}{2}right)right)}$$ Integrated between $0$ and $pi$, the $color{blue}{text{blue term}}$ is obviously zero (because of the sines) and what is left is to evaluate the $color{red}{text{red term}}$. So, for the definite integral, we are left with $$I =color{green}{ int_0^{pi} e^{-frac{1}{2}cos (x)} cos left( 3x + frac{1}{2} sin (x) right) dx=-frac pi{48} }$$

This has been check by numerical integration.