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.
 A: Let $I$ represent the original integral and
$$ I(a)=\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{1}{2t}\arctan\frac{at^2}{\sqrt{3t^4-4}}dt.$$
Clearly $I(0)=0$ and $I(1)=I$.
Note
\begin{eqnarray}
I'(a)&=&\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{\partial}{\partial a}\frac{1}{2t}\arctan\frac{at^2}{\sqrt{3t^4-4}}dt\\
&=&\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{t \sqrt{3 t^4-4}}{\left(a^2+3\right) t^4-4}dt\\
&=&\frac14\int_{(4/3)^{1/2}}^{2^{1/2}}\frac{\sqrt{3 t^2-4}}{\left(a^2+3\right) t^2-4}dt\\
&=&\frac{a \arctan\left(\frac{a t}{\sqrt{3 t^2-4}}\right)+\sqrt{3} \log
   \left(\sqrt{9 t^2-12}+3 t\right)}{4 \left(a^2+3\right)}\bigg|_{(4/3)^{1/2}}^{2^{1/2}}\\
&=&-\frac{\pi  a-2 a \arctan(a)+\sqrt{3} \left(\log (6)-2 \log
   \left(3+\sqrt{3}\right)\right)}{8 \left(a^2+3\right)}.
\end{eqnarray}
So
\begin{eqnarray}
I(a)
&=&-\int_0^1\frac{\pi  a-2 a \arctan(a)+\sqrt{3} \left(\log (6)-2 \log
   \left(3+\sqrt{3}\right)\right)}{8 \left(a^2+3\right)}da\\
&=&\frac{1}{48} \pi  \log \left(\frac{27}{64} \left(2+\sqrt{3}\right)\right)+\frac14\int_0^1\frac{a\arctan(a)}{a^2+3}da.
\end{eqnarray}
Now we solve
\begin{eqnarray}
\int_0^1\frac{a\arctan(a)}{a^2+3}da.
\end{eqnarray}
Let
$$ J(b)=\int_0^1\frac{a\arctan(ab)}{a^2+3}da.$$
Then
$$ J'(b)=\int_0^1\frac{a^2}{(a^2+3)(1+a^2b^2)}da=\frac{6 \arctan (b)-\sqrt{3} \pi  b}{6 \left(b-3 b^3\right)}.$$
So
\begin{eqnarray}
\int_0^1\frac{a\arctan(a)}{a^2+3}da&=&\int_0^1\frac{6 \arctan (b)-\sqrt{3} \pi  b}{6 \left(b-3 b^3\right)}db\\
&=&\frac{1}{48} \left(8 C+\pi  \log \left(64 \left(97-56
   \sqrt{3}\right)\right)\right)
\end{eqnarray}
which is obtained from Mathematica. Here $C$ is Catalan constant. Thus
$$ I=\frac{1}{192} \left(8 C+\pi  \left(\log \left(64 \left(97-56
   \sqrt{3}\right)\right)+4 \log \left(\frac{27}{64}
   \left(2+\sqrt{3}\right)\right)\right)\right)$$
A: 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])])

A: Let $I$ denote the integral. Then Mathematica gives
$$I = \frac{1}{24}G - \frac{3\pi}{32}\log 2 + \frac{\pi}{16}\log 3 \approx 0.0497285555762 \cdots, $$
where $G$ is the Catalan's constant. Currently I obtained the following representation
$$ I = \frac{\pi}{16}\log(3/2) - \frac{1}{4} \int_{0}^{1} \frac{w \arctan w}{3-w^2} \, dw, $$
though I am not sure if this will lead me to a correct way. I will update my answer when I find a full solution.
