Evaluating a double integral by using $\int\arctan (1+\sin 2t)\ dt/(1+\sin 2t)$

Let $(\xi,\eta) = (\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}})$. In terms of $(\xi,\eta)$, the integral becomes:

$$\int_0^{\frac{1}{\sqrt{2}}} \frac{2\xi }{1 + 4\xi^4}d\xi + \int_{\frac{1}{\sqrt{2}}}^1 \frac{2\sqrt{1-\xi^2} }{1+4\xi^4}d\xi$$

For the $1^{st}$ integral, let $u = 2\xi^2$, we have:

$$\int_0^{\frac{1}{\sqrt{2}}} \frac{2\xi}{1 + 4\xi^4} d\xi = \frac{1}{2}\int_0^1 \frac{du}{1+u^2} = \frac{\pi}{8}$$

For the $2^{nd}$ integral, let $\cos\theta = \xi$ and then let $t = \tan\theta$, we have:

$$\int_{\frac{1}{\sqrt{2}}}^1 \frac{2\sqrt{1-\xi^2} d\xi}{1+4\xi^4} = \int_0^{\frac{\pi}{4}} \frac{2\sin(\theta)^2 d\theta}{1 + 4\cos(\theta)^4} = \int_0^{1} \frac{2 \left(\frac{t^2}{1+t^2}\right)\left(\frac{dt}{1+t^2}\right) }{1 + 4\left(\frac{1}{1+t^2}\right)^2} = \int_0^1 \frac{2t^2 dt}{t^4+ 2t^2 + 5} $$ Since $$\frac{2t^2}{t^4 + 2t^2 + 5} = \frac{2t^2}{(t^2+1)^2+4} =\frac{t^2}{2i}\left(\frac{1}{t^2+1-2i} - \frac{1}{t^2+1+2i}\right) =\frac{i}{2}\left(\frac{1-2i}{t^2+1-2i} - \frac{1+2i}{t^2+1+2i}\right) $$ We can evaluate the $2^{nd}$ integral and get following ugly expression:

$$\frac{i}{2} \left( \sqrt{1-2i} \tan^{-1}\left(\frac{1}{\sqrt{1-2i}}\right) - \sqrt{1+2i} \tan^{-1}\left(\frac{1}{\sqrt{1+2i}}\right) \right)$$ Since $\sqrt{1-2i} = \frac{\varphi-i}{\sqrt{\varphi}}$ where $\varphi$ is the golden ratio, the original integral can be expressed as: $$\begin{align}&\frac{\pi}{8} + \frac{i}{2\sqrt{\varphi}}\left( (\varphi - i)\tan^{-1}\left(\frac{\varphi + i}{\sqrt{5\varphi}}\right) - (\varphi + i)\tan^{-1}\left(\frac{\varphi - i}{\sqrt{5\varphi}}\right) \right)\\ = & \frac{\pi}{8} + \frac{1}{2\sqrt{\varphi}}\left( \tan^{-1}(\sqrt{\varphi}^3) -\varphi \tanh^{-1}(\frac{1}{\sqrt{\varphi}^3}) \right) \end{align}$$ Numerically, the integral $\sim 0.3926990817+0.1021715030 = 0.4948705847$.