Closed form for an almost-elliptic integral Does 
$$\int_0^{2 \pi} \log\left(\frac{1}{2}[1+\sqrt{1-(a \sin\phi)^2}]\right) d\phi $$  have a closed form ?    
An approximation for small $a$ is $2E-\pi$, but it is the exact form that is needed for any $|a|<1$.
The integration standing to Jack is 

$$ -\frac{\pi a^2}{4}\phantom{}_4 F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;a^2\right) $$ 

The question now is: Can it be changed into a finite combination of elementary functions.

For any $b\in(0,1)$ we have:
$$I(b)=\int_{0}^{2\pi}\log\left(1+\sqrt{1-b^2\sin^2 x}\right)\,dx=4\int_{0}^{1}\frac{\log\left(1+\sqrt{1-b^2 t^2}\right)}{\sqrt{1-t^2}}\,dt$$
hence:
$$ I(b)=\frac{1}{b}\int_{0}^{b}\frac{\log(1+\sqrt{1-u^2})}{\sqrt{1-\frac{u^2}{b^2}}}\,du=\frac{1}{b}\int_{0}^{\arcsin b}\frac{\cos\theta}{\sqrt{1-\frac{\sin^2\theta}{b^2}}}\,\log(1+\cos\theta)\,d\theta$$
and the last integral can be evaluated by exploiting the Fourier series of $\log(1+\cos\theta)$ that is pretty well-known. Another possible approach is differentiation under the integral sign. 
We have:
$$ I'(b) = -4\int_{0}^{\pi/2}\frac{b\sin^2 t}{\sqrt{1-b^2\sin^2 t}\left(1+\sqrt{1-b^2\sin^2 t}\right)}\,dt = \frac{2\pi-4K(b^2)}{b}\tag{1}$$
where $K$ is the complete elliptic integral of the first kind. Since $I(0)=2\pi\log 2$, it follows that:

$$ I(b)=2\pi\log 2-\frac{\pi b^2}{4}\phantom{}_4 F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;b^2\right).$$

