To solve it, you need to factor the lower part and get squares instead of power $4$. The problem is that the solution will be with imaginary numbers, since the denominator cannot be decomposed into factors other than through the roots of negative numbers. This results in the following:
\begin{align} \int\frac{x^2 \sin(x)}{x^4 + a^4}\,dx = {} & \frac1{4a} (-1)^{1/4} (-i \sin(a(-1)^{1/4}) \operatorname{Ci}(x - a (-1)^{1/4}) \\[4pt] & {} - \sin(a (-1)^{3/4}) \operatorname{Ci}(x - a (-1)^{3/4}) \\[4pt] & {} - \sin(a (-1)^{3/4}) \operatorname{Ci}(x + a (-1)^{3/4}) \\[4pt] & {} - i \sin(a (-1)^{1/4}) \operatorname{Ci}(x + a (-1)^{1/4}) \\[4pt] & {} + i \cos(a (-1)^{1/4}) \operatorname{Si}(a (-1)^{1/4} - x) \\[4pt] & {} + \cos(a (-1)^{3/4}) \operatorname{Si}(a (-1)^{3/4} - x) \\[4pt] & {} + i \cos(a (-1)^{1/4}) \operatorname{Si}(x + a (-1)^{1/4}) \\[4pt] & {} + \cos(a (-1)^{3/4}) \operatorname{Si}(x + a (-1)^{3/4})) + \text{constant} \end{align}
$\mathrm{Ci}$ is the cosine integral, $\mathrm{Si}$ is the sine integral, $\mathrm{G}$ is the Meyer G-function.
I went probably through the same steps and,using a CAS, also arrived at $$I(a)=\frac {\sqrt{2\pi}}{4a}\,\, G_{1,5}^{3,1}\left(\left.\frac{a^4}{4^4}\right| \begin{array}{c} \frac{1}{4} \\ \frac{1}{4},\frac{1}{4},\frac{3}{4},0,\frac{1}{2} \end{array} \right)$$ provided that $$a\neq 0\land ((\Im(a)\neq \Re(a)\land \Im(a)+\Re(a)\neq 0)\lor \Re(a)=0)$$ If $a$ is a real, this function is defined everywhere (except for $a=0$).
It just cancels just above $a=3$ (more exactly for $a=3.01106$); is is negative for larger values, goes through a minimum value for $a=4$, and tends asymptotically to $0^-$.