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?

Let 
$$I:=\int\frac{dx}{x^4-1}, J:=\int\frac{dx}{(x^4-1)^2}, K:=\int\frac{dx}{(x^4-1)^3}.$$
By parts,
$$I=\frac x{x^4-1}+4\int\frac{x^4\,dx}{(x^4-1)^2}=\frac x{x^4-1}+4(I+J)$$
and
$$J=\frac x{(x^4-1)^2}+8\int\frac{x^4\,dx}{(x^4-1)^3}=\frac x{(x^4-1)^2}+8(J+K).$$
On another hand,
$$I=\int\frac{dx}{x^4-1}=\frac12\int\frac{dx}{x^2-1}-\frac12\int\frac{dx}{x^2+1}=-\frac12(\text{argtanh }x+\arctan x).$$
Then
$$J=-\frac14\left(3I+\frac x{x^4-1}\right)$$
and
$$K=-\frac18\left(7J+\frac x{(x^4-1)^2}\right).$$

The generalization is immediate, and
$$L:=\int\frac{dx}{(x^4-1)^4}=-\frac1{12}\left(11K+\frac x{(x^4-1)^3}\right),$$
$$M:=\int\frac{dx}{(x^4-1)^5}=-\frac1{16}\left(15L+\frac x{(x^4-1)^4}\right)$$
$$\vdots$$
We get linear combinations of $\dfrac x{(x^4-1)^k}$, and $I$, with simple coefficients involving the product of fractions $\dfrac{4k-1}{4k}$.
