5. Let $f(x)=x^{3}+3\left(x^{2}+x+\sin \pi x\right)$, then $f(1-\pi)+f(\pi-3)=$
According to the problem, $f(x)=x^{3}+3 x^{2}+3 x+3 \sin \pi x=(x+1)^{3}+3 \sin \pi x-1$, thus $f(-2-x)=(-x-1)^{3}+3 \sin \pi(-x)-1=-(x+1)^{3}-3 \sin \pi x-1$, which means $f(x)+f(-2-x)=-2$. Noting that $1-\pi+\pi-3=-2$, so $f(1-\pi)+f(\pi-3)=-2$.