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?
 A: Hint. According to the Partial Fraction Decomposition Theorem,
$$\frac{1}{(x^4-1)^3}=\frac{1}{(x+1)^3(x-1)^3(x^2+1)^3}=
\sum_{k=1}^3\frac{A_k}{(x-1)^k}+\sum_{k=1}^3\frac{B_k}{(x+1)^k}
+\sum_{k=1}^3\frac{C_kx+D_k}{(x^2+1)^k}.$$
It looks terrible indeed!
A: Another one ...
Power series (I used the negative of the original, to get positive coefficients),
$$
\frac{1}{(1-x^4)^3} = \sum_{k=0}^\infty \frac{(k+1)(k+2)}{2}\;x^{4k}
$$
valid for $|x|<1$.  Integrate,
$$
\int \frac{dx}{(1-x^4)^3} = \sum_{k=0}^\infty \frac{(k+1)(k+2)}{2(4k+1)}\;x^{4k+1} + C
$$
So now all (haha) you have to do is sum that hypergeometric series:
$$
\frac{x(11-7x^4)}{32(1-x^4)^2} + \frac{21\big(\log(1+x)-\log(1-x)+2\arctan x\big)}{128} + C
$$
A: 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.
A: 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 $\dfrac{4k-1}{4k}$.
