How can we take the inverse Laplace transform of
$$f_1(s)X(s)+f_2(s)\left(e^{i\phi}X(s-i\alpha_1)+e^{-i\phi}X(s+i\alpha_1)\right)= f_0(s)$$
Where $ f_1(s)$ is in the form of $ \frac{f^2(s)}{f^3(s)}$, $f_2(s)=k_0 s$, and $f_0(s)$ is in the form of $ k_1+ \frac{f^4(s)}{s (s^2+\alpha_0^2) f^2(s)}$
My main concern is how we can handle $X(s)$ and $X(s\pm i\alpha_1)$ which are wrapped around with complicated polynomials of $s$
Hint.
Anti-transforming
$$ e^{i\phi}X(s-i\alpha_1)+e^{-i\phi}X(s+i\alpha_1)= G_0(s)+G_1(s)X(s) $$
we get at
$$ \cos(\alpha_1\,t + \phi) \, x_3(t) = \int_0^tg_1(\tau)x_3(t-\tau)d\tau+g_0(t) $$
an integral equation for $x_3(t)$