Does $\int_0^{2\pi} \sqrt{(1-a\cos x)^2 + b} \mathrm{d}x$ evaluate in terms of elliptic integrals? Consider the integral
$$\int_0^{2\pi} \mathrm{d}x \sqrt{(1-a\cos x)^2 + b}$$
with $a,b>0$. I believe I can somehow evaluate this integral by incomplete elliptic integrals. This would be obvious, for example, if $a=1$. But, in the general case, this integral appears rather challenging. Mathematica, and SymPy have not been useful. And the Weierstrass substitution returns polynomials of 4th degree.
 A: Here is a step-by-step solution where each step is not too hard.
Step 1: Instead of doing a Weierstrass substitution, substitute $t=\cos x$ and take factors out to get
$$I=2a\int_{-1}^1\frac{\sqrt{(t-1/a)^2+b/a^2}}{\sqrt{1-t^2}}\,dt$$
The roots of the upper quadratic are always complex since $a,b>0$: $c=\frac{1+\sqrt{-b}}a$ and $c^*$.
$$I=2a\int_{-1}^1\frac{\sqrt{(t-c)(t-c^*)}}{\sqrt{1-t^2}}\,dt$$
Step 2: Perform a linear fractional transformation that keeps the integrand's poles at $\pm1$ but shifts the zeros at $c$ and $c^*$ to the imaginary axis, so as to leave only even powers of the polynomial under root. This is the most involved part. Define$\newcommand{Re}{\operatorname{Re}}$
$$A=\frac{1+|c|^2+|c^2-1|}{2\Re(c)}>1$$
$$A_1=A^2-2A\Re(c)+|c|^2>0\qquad A_2=A^2|c|^2-2A\Re(c)+1>0\qquad B=\frac{A_2}{A_1}>0$$
The actual substitution is $u=\frac{At+1}{t+A}$. After taking out more factors
$$I=2a\sqrt{A_1(A^2-1)}\int_{-1}^1\frac1{(u+A)^2}\sqrt{\frac{u^2+B}{1-u^2}}\,du$$
Define $g=2a\sqrt{A_1(A^2-1)}$ and move on.
Step 3: Multiply top and bottom by $(u-A)^2\sqrt{u^2+B}$ to get
$$I=g\int_{-1}^1\frac{(u-A)^2(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du$$
$$=g\left(\int_{-1}^1\frac{(u^2+A^2)(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du-A\int_{-1}^1\frac{(2u)(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du\right)$$
The two integrands are even and odd about zero respectively, so
$$I=2g\int_0^1\frac{(u^2+A^2)(u^2+B)}{(u^2-A^2)^2}\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du$$
We perform a partial fraction decomposition of the rational part of the integrand:
$$I=2g\int_0^1\left(1-\frac{3A^2+B}{A^2-u^2}+\frac{2A^2(A^2+B)}{(A^2-u^2)^2}\right)\frac1{\sqrt{(u^2+B)(1-u^2)}}\,du$$
Step 4: Finally we use elliptic integrals (all arguments here follow Mathematica/mpmath conventions). Byrd and Friedman 213.11 gives$\newcommand{sn}{\operatorname{sn}}$
$$I=\frac{2g}{\sqrt{1+B}}\left(V_0-\frac{3A^2+B}{A^2-1}V_1+\frac{2A^2(A^2+B)}{(A^2-1)^2}V_2\right)$$
where
$$V_k=\int_0^{K(m)}\frac1{(1-n\sn^2u)^k}\,du\qquad m=\frac1{B+1}\qquad n=\frac1{1-A^2}$$
Step 5: B&F 336.00, .01, .02 gives solutions for the $V_k$.
$$V_0=K(m)\qquad V_1=\Pi(n,m)$$
$$V_2=\frac{nE(m) + (m-n)K(m) + (2nm+2n-n^2-3m)\Pi(n,m)}{2(n-1)(m-n)}$$
Putting everything together and simplifying a lot we get the final answer.
$$I=4a\sqrt{-n(A_1+A_2)}(E(m)-K(m)+(1-n)\Pi(n,m))$$
A: If you make the $z=tan(x/2)$ substitution, Mathematica quickly evaluates the indefinite integral, with this horrible expression:
$$\frac{1}{2} z \sqrt{\frac{\left(a
   \left(z^2-1\right)+z^2+1\right)^2}{\left(z^2+1\right)^2}+b}-\frac{i \left(z^2+1\right)
   \sqrt{1-\frac{z^2 \left(a-i \sqrt{b}+1\right)}{a+i \sqrt{b}-1}} \sqrt{1-\frac{z^2
   \left(a+i \sqrt{b}+1\right)}{a-i \sqrt{b}-1}} \sqrt{\frac{\left(a
   \left(z^2-1\right)+z^2+1\right)^2}{\left(z^2+1\right)^2}+b} \left(2 \left(-i a
   \sqrt{b}+a+b+1\right) F\left(i \sinh ^{-1}\left(\sqrt{\frac{a+i \sqrt{b}+1}{-a+i
   \sqrt{b}+1}} z\right)|\frac{-a^2+2 i \sqrt{b} a+b+1}{-a^2-2 i \sqrt{b}
   a+b+1}\right)+\left(a^2+2 i a \sqrt{b}-b-1\right) E\left(i \sinh
   ^{-1}\left(\sqrt{\frac{a+i \sqrt{b}+1}{-a+i \sqrt{b}+1}} z\right)|\frac{-a^2+2 i
   \sqrt{b} a+b+1}{-a^2-2 i \sqrt{b} a+b+1}\right)-4 a \Pi \left(\frac{-a+i
   \sqrt{b}+1}{a+i \sqrt{b}+1};i \sinh ^{-1}\left(\sqrt{\frac{a+i \sqrt{b}+1}{-a+i
   \sqrt{b}+1}} z\right)|\frac{-a^2+2 i \sqrt{b} a+b+1}{-a^2-2 i \sqrt{b}
   a+b+1}\right)\right)}{2 \sqrt{\frac{a+i \sqrt{b}+1}{-a+i \sqrt{b}+1}} \left(a^2
   \left(z^2-1\right)^2+2 a \left(z^4-1\right)+(b+1) \left(z^2+1\right)^2\right)}$$
