Let's say I've got the integral:
$\int [ 2 Q^4 - 5 Q^2 + 3 ]^{-1} dQ$
This integral evaluates to:
$\int [ 2 Q^4 - 5 Q^2 + 3 ]^{-1} dQ = \tanh^{-1}\left( Q \right) + \sqrt{ \frac{2}{3} } \tanh^{-1}\left( \sqrt{ \frac{2}{3} } Q \right)$
I don't wish to have the above written as a sum of two $\tanh^{-1}$ functions. Ideally, it would be nice to write the above as $\tanh^{-1}(f(Q))$ (as a single $\tanh^{-1}$). I have been unable to do this though.
So my question is, is there some transformation I can take $\ Q \to \widetilde{Q}$, where the above integral can be written more compactly, as $\ f(Q)=F(\widetilde{Q})$? I'm sorry if this is a vague question.
EDIT: If it helps to have some context, the integral falls out of a DE I have, which is $\frac{dQ}{dx}= 2 Q^4 - 5 Q^2 + 3$, so really we have $x = \int [ 2 Q^4 - 5 Q^2 + 3 ]^{-1} dQ$ and I'm $trying$ to invert the above expression to get a function $Q(x)$.
Hint:
$$\frac{1}{2Q^4-5Q^2+3}=\frac{1}{(2Q^2-3)(Q^2-1)}=\frac{1}{(\sqrt{2}Q-\sqrt{3})(\sqrt{2}Q+\sqrt{3})(Q-1)(Q+1)}$$