Trigonometric integral: $\int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$ Is it possible to evaluate the following in a closed form?
$$\int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$$

I found the above definite integral at I&S but the solution is not given. I have tried various methods but none of them lead me to the solution, I am honestly out of ideas for this one. 
Any help is appreciated. Thanks! 
 A: We have to compute, 
$$\int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$$
$$=$$
$$\int_{25\pi/4}^{32\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx + \int_{32\pi/4}^{39\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx + \int_{39 \pi/4}^{46\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx +\int_{46\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx = I_1+I_2+I_3+I_4 $$
Now, in $I_2$ substitute, $u=x-\frac{7 \pi}{4}$ , in $I_3$ substitute, $u = x=\frac{7\pi}{2}$ and in $I_4$ substitute, $u=x-\frac{21\pi}{4}$ .
So after substitution limits in each will integral will become same and we can directly add them. So,
We have 
$$\int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx + \frac{1}{(1+2^{\sin x+7\pi/2})(1+2^{\cos x+7\pi/2})}\,dx+ \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{(1+2^{\sin x+7\pi/4})(1+2^{\cos x+7\pi/4})}\,dx + \frac{1}{(1+2^{\sin x+21\pi/4})(1+2^{\cos x+21\pi/4})}\,dx  = J_1+J_2$$
Notice that, $ \sin(a+7\pi/2) = -\cos(a) $ and $\cos(a+7\pi/2)=\sin(a) $
Using this for $J_1$ and $J_2$  we get 
$$J_1 = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{\sin(x)}}$$
And $$ J_2 = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{\sin(x+7\pi/4)}} $$
Now, using well known result, that is 
$$\int_{a}^{b} f(x) dx  = \int_{a}^{b} f(a+b-x) dx$$
$$ I  = J_1+J_2 = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{\sin(x)}} +  \frac{1}{1+2^{\sin(x+7\pi/4)}}  $$
Using that result, $I$ also equals
$$ I = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} \frac{1}{1+2^{-\sin(x)}} +  \frac{1}{1+2^{-\sin(x+7\pi/4)}}  $$
$$ 2I = \int_{\frac{25\pi}{4}}^{\frac{32\pi}{4}} 2 dx $$ 
$$ I = \frac{7\pi}{4} $$

Note:: 
In this solution, if i have written 
$$ \int f(x) + \int g(x) + \cdots = I_1+I_2+\cdots $$
Then this means $$ I_1 = \int f(x) , I_2 = \int g(x) \cdots $$
A: $$I_{a,b}=\int_{a\pi}^{b\pi} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx$$
$$I_{25/4,53/4}=I_{0,7}=3I_{0,2}+I_{0,1}$$
