Integral with fourth power in denominator How can I find the antiderivate of $$\int \frac{1}{(x-a)^4+(x-b)^4} dx$$ with $$a<b$$
My try was to rewrite the integral as $$\int \frac{\frac{1}{(x-a)^2(x-b)^2}}{\frac{(x-a)^2}{(x-b)^2}+\frac{(x-b)^2}{(x-a)^2}} dx$$ which is equal to $$\int \frac{\frac{1}{(x-a)^2(x-b)^2}}{(\frac{(x-a)}{(x-b)}+\frac{(x-b)}{(x-a)})^2-2} dx $$now since $$\frac{d}{dx}\left ( \frac{(x-a)}{(x-b)}+\frac{(x-b)}{(x-a)} \right )$$ have $$(x-a)^2(x-b)^2$$ in the denominator i suspect it might have a nice form in the end, but I fail to finish it.

Let $d=\frac{a+b}{2}$ and substitute $x=u+d$ to make the integrand
$$\frac{1}{(u+c)^4+(u-c)^4}.$$
Where $c=(b-a)/2.$  Factor out a $c^4$:
$$\frac{1}{c^4((\frac{u}{c}+1)^4 + (\frac{u}{c}-1)^4)}.$$
Substitute $v=\frac{u}{c}$ to get 
$$\frac{c}{c^4((v+1^4)+(v-1)^4)}.$$
Ignoring the $c$'s, we need to integrate
$$\frac{1}{(v+1^4)+(v-1)^4} = \frac{1}{2v^4+12v^2+2}.$$
The biquadratic is easily factored and partial fractions give
$$\frac{1}{16}\left( \frac{\sqrt{2}}{v^2+3-2\sqrt{2}} - \frac{\sqrt{2}}{v^2+3+2\sqrt{2}}\right).$$
Complete the square in both denominators and trig sub in the usual way to get:
$$\frac{1}{8} \left(\frac{\tan^{-1}\left(\frac{v}{\sqrt{2}-1}\right)}{2-\sqrt{2}} - \frac{\tan^{-1}\left(\frac{v}{\sqrt{2}-1}\right)}{2+\sqrt{2}} \right).$$ 
Backsubstitute and recall that there's a $c^3$ in the denominator.
