Difficult integral $\int_{\nu+\epsilon}^{2\nu}\frac{x^{2}}{(x-\nu)^{4}+b}dx$ As part of breaking up a larger integral to solve, I have ended up with the following integral:
\begin{equation*} \int_{\nu+\epsilon}^{2\nu}\frac{x^{2}}{(x-\nu)^{4}+b}dx \end{equation*}
where $\nu,b>0$ are fixed constants and $\epsilon\in (0,\nu)$ is small as I want. I have tried everything I can think of (i.e. various substitutions, breaking the integral apart even more, wolframalpha, etc..) Maybe I have missed something or maybe there is some way to do this with partial fraction decomposition, but I am not sure.
I have a solution to this integral if I assume instead that $b=0$, but in this case letting $\epsilon\to 0$ makes the answer blow up and I need it to stay bounded. In any case, I would be very grateful for any help/hints in solving the integral if $b>0$.
Edit: If an explicit solution to this integral is difficult, it would suffice for my purposes to show that the integral is bounded above by a positive constant that is independent of $\nu$ and $\epsilon$.
 A: Translate the integral
$$\int_\epsilon^\nu \frac{x^2+\nu^2}{x^4+b}dx+\int_\epsilon^\nu \frac{2x\nu}{x^4+b}dx$$
The second integral evaluates to
$$\frac{\nu}{\sqrt{b}}\left(\tan^{-1}\left(\frac{\nu^2}{\sqrt{b}}\right)-\tan^{-1}\left(\frac{\epsilon^2}{\sqrt{b}}\right)\right) = \frac{\nu}{\sqrt{b}}\tan^{-1}\left(\frac{\sqrt{b}(\nu^2-\epsilon^2)}{b+\nu^2\epsilon^2}\right)$$
For the second use the substitution $x = b^{\frac{1}{4}}t$, factor the denominator as $$t^4+1 = (t^2-\sqrt{2}t+1)(t^2+\sqrt{2}t+1)$$ and decompose with partial fractions. Can you take it from here?
A: Rescale the integral with $t= \frac{x-v}{b^{1/4}}$, $a=\frac{v}{b^{1/4}}$ and $\delta=\frac{\epsilon}{b^{1/4}}$
\begin{align*}
& \int_{v +\epsilon}^{2v}\frac{x^{2}}{(x-v)^{4}+b}dx 
=\frac1{b^{1/4}} \int_{\delta}^{a}\frac{t^{2}+2at +a^2}{t^{4}+1}dt\\= &\frac1{b^{1/4}} \int_{\delta}^{a}\left( \frac{1+a^2}2 \frac{t^{2}+1}{t^{4}+1}
 + \frac{1-a^2}2 \frac{t^{2}-1}{t^{4}+1}+ \frac{2at}{t^{4}+1}  \right)dt \\= &\frac{1+a^2}{2\sqrt2 b^{1/4}} \left( \tan^{-1}\frac{a^2-1}{\sqrt2a} - \tan^{-1}\frac{\delta^2-1}{\sqrt2\delta} \right) \\
&\hspace{2mm} -  \frac{1-a^2}{2\sqrt2 b^{1/4}} \left( \coth^{-1}\frac{a^2+1}{\sqrt2a} - \coth^{-1}\frac{\delta^2+1}{\sqrt2\delta} \right) 
+ \frac a{b^{1/4}} \tan^{-1} \frac{a^2-\delta^2}{1+a^2\delta^2}
\end{align*}
