EDIT: Instead of expressing the integral as the imaginary part of another integral, I instead expanded $\sin^{3}(x)$ in terms of complex exponentials and I don't run into problems anymore.
\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx \\ &= \frac{1}{2} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{x^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{(x-i \epsilon)^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i} (e^{3ix}-3e^{ix})}{(x-i \epsilon)^{5}} + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{3e^{-ix}-e^{-3ix}}{(x-i \epsilon)^{5}} \ dx \end{align}
Then I integrated $ f(z) = \frac{z^{3}+ \frac{1}{8i}(e^{3iz}-3e^{iz})}{(z-i \epsilon)^{5}}$ around the upper half of $|z|=R$ and $ g(z) = \frac{3e^{-iz}-e^{-3iz}}{(z-i \epsilon)^{5}}$ around the lower half of $|z|=R$ and applied Jordan's lemma.
\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}x}{x^{5}} \ dx &= \frac{1}{2} \lim_{\epsilon \to 0^{+}}2 \pi i \ \text{Res}[f(z),i \epsilon] + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} 2 \pi i (0) \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \frac{2 \pi i}{4!} \lim_{z \to i \epsilon} \frac{d^{4}}{dz^{4}} \Big(z^{3}+\frac{1}{8i}e^{3iz}-\frac{3}{8i}e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \ \lim_{z \to i \epsilon}\Big( \frac{1}{8i}(3i)^{4}e^{3iz}- \frac{3}{8i} (i)^{4} e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \Big( \frac{81}{8i}e^{- 3\epsilon} - \frac{3}{8i}e^{- \epsilon} \Big) \\ &= \frac{\pi i}{24} \Big(\frac{81}{8i}-\frac{3}{8i} \Big) \\ &= \frac{13 \pi}{32} \end{align}
I like to calculate this integral as follows:
Let us note that
$$\frac{1}{x^5}=\frac{1}{4!}\int_0^\infty t^4e^{-xt}dt$$ So
$$I=\frac{1}{4!}\int_{0}^{\infty}(x^{3}-\sin^{3}x)\int_0^\infty t^4e^{-xt}\;dt\;dx$$
$$=\frac{1}{4!}\int_{0}^{\infty}t^4\int_{0}^{\infty}(x^{3}-\sin^{3}x)e^{-xt}\;dx\;dt$$
$$=\frac{1}{4!} \int_{0}^{\infty}t^4\left [\frac{6}{t^4}-\frac{6}{(t^2+1)(t^2+9)}\right ]dt$$
$$=\frac{1}{4}\int_{0}^{\infty}\frac{10t^2+9}{(t^2+1)(t^2+9)}dt=\frac{13\pi}{32}$$