What is the general form of $f(t)$ in the inverse Laplace transform $$\mathcal{L}^{-1}\left\{\frac{p'(s)}{p(s)}\right\}=f(t),$$ for some given polynomial $p(s)$ ?
Put another way, what $f(t)$ satisfies the Laplace transform $$\mathcal{L}\{f(t)\} = \frac{p'(s)}{p(s)},$$ where the polynomial $p(s)$ is given.
Let $n=\text{deg}(p)$ be the degree of the polynomial $p$. Then by partial fractions we may write $$\frac{p'(s)}{p(s)}=\sum_{k=1}^n\frac{A_k}{s-\sigma_k},$$ for some $A_k$, where $p(\sigma_k)=0$ for all $1\leq k\leq n$. But since $\frac{1}{s-a}=\mathcal{L}\left\{e^{-at}\right\}$, then $$\sum_{k=1}^n\frac{A_k}{s-\sigma_k}=\sum_{k=1}^n A_k\mathcal{L}\{e^{-\sigma_k t}\}=\mathcal{L}\left\{\sum_{k=1}^n A_ke^{-\sigma_kt}\right\}.$$ Hence, $$\mathcal{L}^{-1}\left\{\frac{p'(s)}{p(s)}\right\}=\sum_{k=1}^n A_ke^{-\sigma_kt}.$$ It may be that the right hand side has closed form due to the exponential forms of the sine and cosine functions, depending on our choice of $p$.
Note - this generalises the original question since $p'/p\equiv q/p$, where $\text{deg}(q)<\text{deg}(p)$.
However, in the original case for $p'/p$, it appears that $A_k=1$ for all $k$, in which case there will indeed be closed fom in terms of sines, cosines, and their hyperbolic equivalents.