How can we take the inverse Laplace transform of

$${f}_{1}(s)X(s)+{f}_{2}(s)({e}^{i\varphi}X(s-i{\alpha}_{1})+{e}^{-i\varphi}X(s+i{\alpha}_{1}))={f}_{0}(s)$$

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 ${k}_{1}+\frac{{f}^{4}(s)}{s({s}^{2}+{\alpha}_{0}^{2}){f}^{2}(s)}$

$${f}_{1}(s)X(s)+{f}_{2}(s)({e}^{i\varphi}X(s-i{\alpha}_{1})+{e}^{-i\varphi}X(s+i{\alpha}_{1}))={f}_{0}(s)$$

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 ${k}_{1}+\frac{{f}^{4}(s)}{s({s}^{2}+{\alpha}_{0}^{2}){f}^{2}(s)}$