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

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

I am trying to evaluate the following integral: $$ I:=\iint_D\frac{dy\
dx}{1+(x+y)^4} $$ 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: $$ I = \iint \frac{r\ dr\
d\theta}{1+r^4(\cos\theta+\sin\theta)^4} =
\int_0^{\pi/2}d\theta\int_0^1\frac{d(r^2)}{1+(\cos\theta+\sin\theta)^4(r^2)^2}.
$$ Then, I used the well-known technique to get $$ I =
\int_0^{\pi/2}\frac{\arctan (1+\sin 2\theta)}{1+\sin 2\theta}d\theta. $$
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).