Complicated integral... or is it?

$\begin{aligned} I & = \int_{-\pi}^{\color{#3D99F6}\exp (2\ln \sqrt \pi)} \frac {\sin \left((x^3-\pi x)\ln (x^2+1)\right)}{e^{-x^4+\pi}\cos (3x)} dx & \small \color{#3D99F6} \text{Note that } e^{2\ln \sqrt \pi} = e^{\ln \pi} = \pi \\ & = \int_{-\pi}^{\color{#3D99F6}\pi} \frac {\color{#D61F06}\sin \left((x^3-\pi x)\ln (x^2+1)\right)}{\color{#3D99F6}e^{-x^4+\pi}\cos (3x)} dx & \small \color{#D61F06} \text{Numerator is odd} \color{#3D99F6} \text{ while the denominor is even} \\ & = \color{#D61F06} \boxed 0 & \small \color{#D61F06} \text{the integrand is odd.} \end{aligned}$

The integrand is an odd function, i.e.

$f(x) = \frac{\sin \big( (x^3 - \pi x) \ln(x^2+1) \big)}{e^{-x^4+\pi} \cos(3x)}$

then

$f(-x) = \frac{\sin \big( ((-x)^3 - \pi (-x)) \ln((-x)^2+1) \big)}{e^{-e(-x)^4+\pi} \cos(3(-x))} = - \frac{\sin \big( (x^3 - \pi x) \ln(x^2+1) \big)}{e^{-x^4+\pi} \cos(3x)} = - f(x)$

The integral bounds are $- \pi$ to $e^{2 \ln(\sqrt{\pi})} = e^{ln(\pi)} = \pi$

so

$\int_{-\pi}^{e^{2\ln(\sqrt{\pi})}} \frac{\sin \big( (x^3 - \pi x) \ln(x^2+1) \big)}{e^{-x^4+\pi} \cos(3x)} dx =0$