Evaluating a double integral by using $\int\arctan (1+\sin 2t)\ dt/(1+\sin 2t)$ I am trying to evaluate the following integral:
$$
I:=\iint_D\frac{\mathrm dy\ \mathrm dx}{1+(x+y)^4}
$$
where $D := \{(x,y)\mid x^2+y^2\le1,x\ge0,y\ge0\}$.
I tried to evaluate it in the following way: first I rewrite it in terms of the polar coordinate:
$$
I = \iint \frac{r\ \mathrm dr\ \mathrm d\theta}{1+r^4(\cos\theta+\sin\theta)^4} = \int_0^{\pi/2}\mathrm d\theta\int_0^1\frac{\mathrm d(r^2)}{1+(\cos\theta+\sin\theta)^4(r^2)^2}.
$$
Then, I used the well-known technique to get
$$
I = \int_0^{\pi/2}\frac{\arctan (1+\sin 2\theta)}{1+\sin 2\theta}\mathrm d\theta.
$$
The problem is how to evaluate this.  I could get a simpler formula by substituting $1+\sin 2\theta$, but I would not proceed further.
I would be grateful if you give a clue (not necessarily a complete solution).

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 $\sim 0.3926990817+0.1021715030 = 0.4948705847$.
