How do you calculate $\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{1}{2t}\arctan\frac{t^2}{\sqrt{3t^4-4}}dt$? How do you calculate $$\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{1}{2t}\arctan\frac{t^2}{\sqrt{3t^4-4}}dt?$$ Mathematica fails to do it.

Substitute $u=t^4$ to obtain $$\frac{1}{8} \int_{4/3}^2 \frac{\tan^{-1} \sqrt{\frac{u}{3u-4}}}{u} \ \mathrm{d}u$$
Consider $$I(a) = \int_{4/3}^2 \frac{\tan^{-1} \sqrt{\frac{a u}{3u-4}}}{u} \ \mathrm{d}u$$
which, when we take the derivative with respect to $a$, we obtain $$I'(a) = \int_{4/3}^2 \frac{1}{2 \sqrt{\frac{au}{3u-4}} (3u-4) \left(1+\frac{au}{3u-4}\right)}\ \mathrm{d}u$$
The integrand has antiderivative $$\frac{\sqrt{u} \left(\sqrt{a} \tan ^{-1}\left(\frac{\sqrt{a u}}{\sqrt{3
   u-4}}\right)+\sqrt{3} \log \left(3 \sqrt{u}+\sqrt{9 u-12}\right)\right)}{(a+3) \sqrt{3
   u-4} \sqrt{\frac{a u}{3 u-4}}}$$
Substitute $u=2$ to obtain $$\frac{\sqrt{a} \tan ^{-1}\left(\sqrt{a}\right)+\sqrt{3} \log \left(3
   \sqrt{2}+\sqrt{6}\right)}{\sqrt{a} (a+3)}$$
Take the limit as $u \to \frac{4}{3}$ to obtain $$\frac{\frac{\sqrt{3} \log (12)}{\sqrt{a}}+\pi }{2 a+6}$$
The difference of these expressions is $$-\frac{\frac{\sqrt{3} \log \left(2-\sqrt{3}\right)}{\sqrt{a}}-2 \tan
   ^{-1}\left(\sqrt{a}\right)+\pi }{2 a+6}$$
which is the value of $I'(a)$.
Note also that $I(0) = 0$.
Now, we want $I(1)$; the resulting integral is something Mathematica can perform, and it outputs the following:
simp[a_] := -((Pi - 2 ArcTan[Sqrt[a]] + (Sqrt[3] Log[2 - Sqrt[3]])/Sqrt[a])/(6 + 2a))

DSolve[func'[a] == simp[a] && func[0] == 0, func[a], a] /. a -> 1 // FullSimplify

Out[80]= {{func[1] -> 
 1/48 (48 Catalan + 
  I (3 \[Pi]^2 - 4 I \[Pi] Log[729/512 (7 + 4 Sqrt[3])] + 
     24 (PolyLog[2, -2 - Sqrt[3]] - PolyLog[2, I (-2 + Sqrt[3])] +
         PolyLog[2, -2 + Sqrt[3]] - 
        PolyLog[2, -I (2 + Sqrt[3])])))}}

That is, the original integral has value $\frac{1}{8}$ times that, which is (on taking real parts, since I know the value is real) $$\frac{1}{96} \left(12 C+6 \Im\left(\text{Li}_2\left(i
   \left(-2+\sqrt{3}\right)\right)+\text{Li}_2\left(-i
   \left(2+\sqrt{3}\right)\right)\right)+\pi  \log \left(\frac{729}{512} \left(7+4
   \sqrt{3}\right)\right)\right)$$
1/96 (12 Catalan + 
 6 Im[PolyLog[2, I (-2 + Sqrt[3])] + 
  PolyLog[2, -I (2 + Sqrt[3])]] + \[Pi] Log[729/512 (7 + 4 Sqrt[3])])

