Approach ideas for the integral $\int\frac{dx}{(x^4-16)^2}$ Well, the title sums it up pretty well. I'm in search for some smart approach ideas for solving this indefinite integral:
$$\int\frac{dx}{(x^4-16)^2}$$
I know one that would work for sure, namely partial fraction decomposition, but it gets really heavy when regrouping the coefficients for the powers of $x$ and then solving an $8\times8$ system of linear equations. It will eventually work, but I suspect there is something more ingenuine behind this problem.
I also tried all sorts of trigonometric substitutions and formulations, but that added square power really is a bummer to it all.
I'm generally open to any exchange on the topic and would be glad to hear some advice in such situations. Many thanks in advance!

As $x^4-2^4=(x^2-2^2)(x^2+2^2)$ and $(x^2+2^2)-(x^2-2^2)=8$
$$\dfrac{8^2}{(x^4-16)^2}=\dfrac{(x^2+4-(x^2-4))^2}{(x^2-4)^2(x^2+4)^2}=\dfrac1{(x^2-4)^2}+\dfrac1{(x^2+4)^2}-\dfrac2{(x^2-4)(x^2+4)}$$
For the first, write the numerator as $\dfrac{(x+2-(x-2))^2}{16}$  and expand
For the second either $x=2\tan t$
or integrate by parts $$\int\dfrac1x\cdot\dfrac x{(x^2+4)^2}\ dx$$
Again for the last, $\dfrac{(x^2+4)-(x^2-4)}{(x^2-4)(x^2+4)}$
