Simple expression for the integral $ \int_0^{\infty} \frac{dt}{e^{xt}+\sin t}, \hspace{0.5cm} \operatorname{Re}(x)>0 $

Question

I'm wondering if there exists any nice expression (in terms of special functions or simple looking infinite series) for the integral $$ \int_0^{\infty} \frac{dt}{e^{xt}+\sin t}, \hspace{0.5cm} \operatorname{Re}(x)>0 $$ So far this is all I've been able to think of: $$ \int_0^{\infty} \frac{dt}{e^{xt}+\sin t} = \int_0^{\infty} e^{-xt} \frac{dt}{1+ \sin t \, e^{-xt}} = \int_0^{\infty} \sum_{n=0}^{\infty} (-1)^n \sin^nt \, e^{-x(n+1)t}\, dt \\ = \sum_{n=0}^{\infty} \int_0^{\infty} \sin^{2n}t \, e^{-x(2n+1)t} \, dt - \sum_{n=0}^{\infty} \int_0^{\infty} \sin^{2n+1}t \, e^{-x(2n+2)t} \, dt $$ Using the two formulas $$ \int_0^{\infty} \sin^{2n}t \, e^{-at} \, dt = \frac{(2n)!}{a(2^2+a^2)(4^2+a^2) \cdots ((2n)^2+a^2)} $$ $$ \int_0^{\infty} \sin^{2n+1}t \, e^{-at} \, dt = \frac{(2n+1)!}{(1^2+a^2)(3^2+a^2) \cdots ((2n+1)^2+a^2)} $$ we get $$ \int_0^{\infty} \frac{dt}{e^{xt}+\sin t} = \sum_{n=0}^{\infty} \frac{(2n)!}{(2n+1)x} \prod_{k=1}^n \left((2k)^2+(2n+1)^2x^2 \right)^{-1} \\ \hspace{3.5cm} - \sum_{n=0}^{\infty} (2n+1)! \prod_{k=0}^n \left((2k+1)^2+(2n+2)^2x^2 \right) $$ but the resulting expression doesn't look nice at all. Any help is appreciated!