So ugly, only a mother could love it

We can evaluate this one using tables of infinite integrals. We note that $\prod_{n=1}^\infty \frac{n^2(16n^4 + 24n^2 + 9)}{16(n^6 + 2n^3 + 1)} \; = \; \displaystyle \prod_{n=1}^\infty \frac{n^2(4n^2 + 3)^2}{16(n^3 + 1)^2} \; = \; \prod_{n=1}^\infty \frac{\big(1 + \tfrac34n^{-2}\big)^2}{\big(1 + n^{-3}\big)^2} \; = \; \left(\frac{A}{B}\right)^2$ where $A \; = \; \prod_{n=1}^\infty \big(1 + \tfrac34n^{-2}\big) \qquad \qquad B \; = \; \prod_{n=1}^\infty \big(1 + n^{-3}\big) \;.$ Lists of infinite products (Mathworld has just about enough) give $A \; = \; \tfrac{2}{\pi\sqrt{3}}\sinh\big(\tfrac12\pi\sqrt{3}\big) \qquad \qquad B \; = \; \tfrac{1}{\pi}\cosh\big(\tfrac12\pi\sqrt{3}\big) \;,$ and so the product is $\left(\frac{A}{B}\right)^2 \; = \; \tfrac43\tanh^2\big(\tfrac12\pi\sqrt{3}\big) \;,$ making the answer $4 \times 3 \times 2 = \boxed{24}$ .