$\int_0^{2\pi} \sqrt{a^2 +b^2+2ab \cos\varphi}\,\mathrm{d}\varphi$ $$\int_0^{2\pi} \sqrt{a^2 +b^2+2ab \cos\varphi}\,\mathrm{d}\varphi$$
Where $a$ and $b$ are constants.
I had to find the distance travelled by a point at a distance of $b$ from the centre of a rolling disc with radius $a$ in one full rotation. I got the final answer as this integral, but I was not able to solve this integral.

First consider the integrand
\begin{align}
\sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)}
\end{align}
which can be seen as
\begin{align}
\sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} &= \sqrt{(a+b)^{2} - 4ab \sin^{2}\left(\frac{\phi}{2}\right) } \\
&= (a+b) \sqrt{ 1 - \left( \frac{4ab}{(a+b)^{2}} \right) \, \sin^{2}\left( \frac{\phi}{2}\right) }.
\end{align}
Now,
\begin{align}
I &= \int_{0}^{2\pi} \sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} \, d\phi \\
&= (a+b) \, \int_{0}^{2\pi} \sqrt{ 1 - \left( \frac{4ab}{(a+b)^{2}} \right) \, \sin^{2}\left( \frac{\phi}{2}\right) } \, d\phi \\
&= 2(a+b) \int_{0}^{\pi} \sqrt{1- k^{2} \sin^{2}(x) } \, dx
\end{align}
where $(a+b)^{2} k^{2} = 4ab$. The integral remaining if the complete elliptical integral of the second kind. The "solution" of the integral is
\begin{align}
I = 4(a+b) E\left(\frac{4ab}{(a+b)^{2}}\right). 
\end{align}
Hence,
\begin{align}
\int_{0}^{2\pi} \sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} \, d\phi = 4(a+b) E\left(\frac{4ab}{(a+b)^{2}}\right).
\end{align}
