I am trying to perform an integral but the main cause of issues and errors is the following part \begin{equation} -\int_0^1dx\: \log(m^2+Q^2x(x-1)) \end{equation} where there are some other $x$'s multiplied to the front of it but I can get rid of those by integration by parts (if you want to know one of the polynomials multiplied to it is $(20x^2 - 40x + 21)(m^2 + Q^2(x-1)x)$ but I do not know if that would help trying to solve this thing. For reference this integral arose after doing a loop-integral in QFT and I am finally at the end stages of the calculation and this monster appears; this is after performing the momentum integral by $\epsilon$-expansion, keeping terms up to order $\epsilon^1$, and now I am doing the $dx$ integral that came from the Feynman parameter trick of splitting up the denominator of $1/AB = \int_0^1\frac{dx}{Ax + B(1-x)}$.
I don't have a nice integral book on stand by so I am turning to here. Mathematica takes about 10 minutes to do just the $\log$ integral and it spits out \begin{equation} -\frac{2 \sqrt{4 m^2-Q^2} \tan ^{-1}\left(\frac{Q}{\sqrt{4 m^2-Q^2}}\right)}{Q}-\log \left(m^2\right)+2 \end{equation} with the following if condition \begin{equation} \Im(m)^2\leq \Re(m)^2\land \left(\frac{\sqrt{Q^4-4 m^2 Q^2}}{Q^2}\notin \mathbb{R}\lor \Re\left(\frac{\sqrt{Q^4-4 m^2 Q^2}}{Q^2}\right)<-1\lor \Re\left(\frac{\sqrt{Q^4-4 m^2 Q^2}}{Q^2}\right)>1\right)\land \left(\sqrt{1-\frac{4 \Im(m) \Re(m)}{\Im(Q) \Re(Q)}}\notin \mathbb{R}\lor \Re\left(\sqrt{1-\frac{4 \Im(m) \Re(m)}{\Im(Q) \Re(Q)}}\right)\geq 1\lor \Im(m) \left(\frac{\Re(m) \Re(Q)}{\Im(Q)}+\Im(m)\right)\leq \Re(m) \left(\frac{\Im(m) \Im(Q)}{\Re(Q)}+\Re(m)\right)\right)\land \left(\Re\left(\frac{\sqrt{4 m^2-Q^2}}{Q}\right)\neq 0\lor \Im\left(\frac{\sqrt{4 m^2-Q^2}}{Q}\right)<-1\lor \Im\left(\frac{\sqrt{4 m^2-Q^2}}{Q}\right)>1\right) \end{equation}
If I said I knew what that meant I'd be lying.
Any suggestions about how to massage this integral or where to look are very helpful, thanks!
Remember that Mathematica does not know anything about $m$ and $Q$.
If you just type
Integrate[Log[m^2+Q^2*x(x-1)],{x,0,1}, Assumptions -> m>0 && Q>0]Mathematica returns immediately $$2(\log(m)-1)+\frac{2 \sqrt{4 m^2-Q^2} }{Q}\,\,\tan ^{-1}\left(\frac{Q}{\sqrt{4 m^2-Q^2}}\right)\qquad \text{if}\qquad 2m>Q$$
Now, if you write $$\log \left(m^2+Q^2 x(x-1) \right)=\log(Q^2)+\log(x-a)+\log(x-b)$$
where $$(a,b)=\frac{1}{2}\pm\frac{\sqrt{Q^2-4 m^2}}{2 Q}$$ you will get, without any condition $$(1-a) \log (1-a)+a \log (-a)+(1-b) \log (1-b)+b \log (-b)-2$$ which covers all cases.