Solving indefinite integral $\int \frac{dx}{(x^4-1)^3}$ I'm trying to solve next integral, but I can't start. WolframAlpha gives me really terrible answer and can't give any step-by-step instructions, so I simply does not know how to start.
$$\int \frac{dx}{(x^4-1)^3}$$
Please give any hint or start point of solving this integral? Maybe there any way to simplify it?

One step that can simplify the partial fraction decomposition is to first perform an integration by parts with the choice $$u = -(2x)^{-3}, \quad du = \frac{3}{8}x^{-4} \, dx, \quad dv = -\frac{(2x)^3}{(x^4-1)^{3}} \, dx, \quad v = (x^4-1)^{-2},$$ giving $$\int \frac{dx}{(x^4-1)^3} = -\frac{1}{8x^3(x^4-1)^2} - \frac{3}{8} \int \frac{dx}{x^4 (x^4-1)^2}.$$  Partial fraction decomposition on $(x^2 (x^4-1))^{-2}$ results in a simpler integrand with lower-degree polynomial denominator terms:
$$\frac{1}{x^4(x^4-1)^2} = \frac{1}{x^4}+\frac{3}{4 \left(x^2+1\right)}+\frac{1}{4(x^2+1)^2}+\frac{7}{16 (x+1)}+\frac{1}{16 (x+1)^2}-\frac{7}{16(x-1)}+\frac{1}{16 (x-1)^2},$$ all of which are trivially integrable with the exception of the third term, which may be handled with the usual trigonometric substitution $x = \tan \theta$, $dx = \sec^2 \theta \, d\theta$, resulting in the transformed integrand $\frac{1}{4} \cos^2 \theta$, which again is easily handled.
