One answer is as follows:
\begin{align*}
\biggl(\frac{1}{x^4+4}\biggr)^{(n)}
&=\sum_{k=0}^n \frac{(-1)^kk!}{(x^4+4)^{k+1}} B_{n,k}\bigl(4x^3,12x^2,24x,24,0,\dotsc,0\bigr)\\
&=\sum_{k=0}^n \frac{(-1)^kk!}{(x^4+4)^{k+1}} x^{4k-n} B_{n,k}\bigl(\langle4\rangle_1, \langle4\rangle_2, \langle4\rangle_3, \langle4\rangle_4, \langle4\rangle_5,\dotsc,\langle4\rangle_{n-k+1}\bigr)\\
&=\sum_{k=0}^n \frac{(-1)^kk!}{(x^4+4)^{k+1}} x^{4k-n} (-1)^k\frac{n!}{k!}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\binom{4\ell}{n}\\
&=\frac{n!}{x^n}\sum_{k=0}^n \frac{x^{4k}}{(x^4+4)^{k+1}} \sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\binom{4\ell}{n},
\end{align*}
where we used the formula
\begin{equation}\label{binom-bell-factorial}
B_{n,k}(\langle\alpha\rangle_1, \langle\alpha\rangle_2, \dotsc,\langle\alpha\rangle_{n-k+1})
=(-1)^k\frac{n!}{k!}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\binom{\alpha\ell}{n}.
\end{equation}
A Reference
- Siqintuya Jin, Bai-Ni Guo, and Feng Qi, Partial Bell polynomials, falling and rising factorials, Stirling numbers, and combinatorial identities, Computer Modeling in Engineering & Sciences 132 (2022), no. 3, 781--799; available online at https://doi.org/10.32604/cmes.2022.019941.