Integral of the squared distance between an arc and a circle Assume a circle of radius $r_a$ is centered at the origin. Assume there is a circle arc, with center $(c_x,c_y)$ and radius $r_b$. I am interested in the integral of the (squared) distance between the arc and the circle, i.e.
$$
\int_{\theta_a}^{\theta_b}\left( \sqrt{\left(c_x + r_b \cos{\left(\theta \right)}\right)^{2} + \left(c_y + r_b \sin{\left(\theta \right)}\right)^{2}}- r_a \right)^{2} d\theta
$$
But I cannot solve the integral, neither prove it has no closed form.
 A: Hoping that you do not mind, I shall change notations and consider
$$I=\int \Big[\sqrt{(a+b \cos (t))^2+(c+b \sin (t))^2}-d\Big]^2\,dt$$
Expanding the integrand, we have
$$\Big[\sqrt{(a+b \cos (t))^2+(c+b \sin (t))^2}-d\Big]^2=2 a b \cos (t)+2 b c \sin (t)+(a^2+b^2+c^2+d^2)-$$
$$2d \sqrt{(a^2+b^2+c^2)+2 a b \cos (t)+2 b c \sin (t)}$$ The $\color{red}{\large\text{monster}}$ comes from the last term which is
$$\int \sqrt{1+A \cos(t)+B \sin(t)}\,dt$$ No problem if one of the $(A,B)$ terms is equal to $0$ (that is to say if $a$ or $c$ is equal to $0$
If this is not the case, a CAS finds a very long antiderivative where appear several Appell hypergeometric functions of two variables  with extremely complicated arguments.
Edit
Formula $2.617.5$ (page $211$) of "Table of Integrals, Series and Products" (seventh edition) by I.S. Gradshteyn and I.M. Ryzhik gives a formula for
$$\sqrt{a+b \sin(x)+c\cos(x)}\,dt$$ in terms of  incomplete elliptic integrals.
But, as said in comments,
$$\int \sqrt{1+c \cos (t+\varphi )}\,d\phi=2 \sqrt{c+1} \,\,E\left(\frac{t+\varphi }{2}|\frac{2 c}{c+1}\right)$$
