Radial (not Polar!) planimeter theory... I am investigating the theory of a radial (not a polar) planimeter - when tracing a closed contour with the "pole" of the device outside of the contour (normally, the "pole" would be inside the contour).
This has led me to this horrible expression below, whilst trying to evaluate a line integral - and I have been unable to progress beyond it...
$$
\int_0^{2\pi } \frac{r\ldotp \mathrm{cos}\left(t\right)\ldotp \left(a+r\ldotp \mathrm{cos}\left(t\right)\right)}{\sqrt{{\left(a+r\ldotp \mathrm{cos}\left(t\right)\right)}^2 +{\left(b+r\ldotp \mathrm{sin}\left(t\right)\right)}^2 }}+\frac{r\ldotp \mathrm{sin}\left(t\right)\ldotp \left(b+r\ldotp \mathrm{sin}\left(t\right)\right)}{\sqrt{{\left(a+r\ldotp \mathrm{cos}\left(t\right)\right)}^2 +{\left(b+r\ldotp \mathrm{sin}\left(t\right)\right)}^2 }}\mathrm{dt}
$$
Does anyone have any suggestions about how I might proceed with the solution to this? (Wolfram times out and so does the rudimentary symbolic processor which I have access to).  
I therefore reverted to a numerical approach and found that the results appear to be physically plausible (also nicely confirmed by measurements using an actual instrument), but I want to find the analytical solution as this will obviously provide much more insight into the underlying process. Of course, I realise that there may not be an analytical solution... 
Thanks for looking.
 A: Assuming that this is indeed the correct integral you want to evaluate, you can do the following:
Since your "pole" lies outside the contour we can work from
$$
\sqrt{a^2 + b^2} = c > r > 0
$$
where $c$ is the distance from the origin to the point $(a,b)$ and $r$ the radius of the circles of the contour. Expanding the numerators and denominators we get
$$
\int_0^{2 \pi} \frac{
r^2 + a r \cos t + b r \sin t
}{
\sqrt{r^2 + a^2 + b^2 + 2 a r \cos t + 2 b r \sin t}
} d t,
$$
where I already used that $\sin^2 t + \cos^2 t = 1$. Since we also can use that 
$a \cos t + b \sin t = \sqrt{a^2+b^2} \sin(t + \phi) = c\sin(t + \phi) $ for $\phi = \arctan a/b$, the integral becomes simpler
$$
\int_0^{2 \pi} \frac{r^2 + r c \sin (t+\phi)}{\sqrt{r^2 + c^2 + 2 r c \sin (t+\phi)}} d t.
$$
While the integration is over a whole period, the angular shift $\phi$ can be discarded, and we get
$$
\int_0^{2 \pi} \frac{r^2 + r c \sin t}{\sqrt{r^2 + c^2 + 2 r c \sin t}} d t
$$
An alternative way to look at this result is to rotate the frame of reference such that the point $(a,b)$ lies on the $y$-axis at a distance $c$ from the origin.
There are only two integrals to be evaluated:
$$
\int_0^{2 \pi} \frac{1}{\sqrt{1 + q \sin t}} d t
$$
$$
\int_0^{2 \pi} \frac{\sin t}{\sqrt{1 + q \sin t}} d t
$$
with $0 < q = 2 r c/\sqrt{r^2+c^2}<1$. These give rise to the complete elliptic integral functions $E(..)$ and $K(..)$, and will result in
$$
(c-r) \left[ E\left(\frac{-4 c r}{(c-r)^2}\right) - K\left(\frac{4 c r}{(c+r)^2}\right) \right] + (c+r) \left[ E\left(\frac{4 c r}{(c+r)^2}\right) - K\left(\frac{-4 c r}{(c-r)^2}\right) \right]
$$
