Closed form of distance travelled by each corner of a rectangle that is simultaneously rotated and translated Motivation, Background, and Research
I'm a programmer for an FRC robotics team. We use a holonomic drivetrain called Swerve, wherein each wheel is independently rotated and driven. During our journey through this season, we've come across several interesting math problems, mostly rooted in trigonometry. This is a question we've managed to sidestep so far, by changing the way we've approached some problems. However, I'm still curious about it, even if we never end up needing to know. Inverse kinematics and other information for the Swerve drivetrain can be found here.
TL;DR: I might be overthinking this.

I know there are formulas for setting wheel velocities and headings in the white paper (derivation for rectangular case here) at the linked page. However, I can't tell if they produce the desired behavior exactly. It may be as simple as integrating the velocities over the necessary time, but I'm not sure. We use a combination of these formulas and path tracking that currently uses the robot's center position, calculated by averaging the recorded positions of each wheel module. We can calculate where we've been, but I'd like a closed-form solution for any translation and rotation. This would allow us to follow per-wheel paths more precisely.

I'm taking Calculus II right now, and frankly I expect that the answer(s) will be over my head. That's fine with me—I'll just keep learning until I understand. I have done some searching about loci, arc length, parametric functions, and translations, but I haven't been able to find anything about this specific problem.
Problem
A note on the angles used in this question: For whatever reason, the angles used in Swerve kinematics are wonky from a math standpoint. They're similar to the headings on a compass, though: $0^\circ$ in normal math corresponds to $90^\circ$ in Swerve math, and vice versa. Positive angles are clockwise.
Consider a rectangle with a center point $R$, initial rotation $\alpha$, width $W$, and length $L$. We want to translate the rectangle by a vector $V = \left<\Delta x,\Delta y\right>$ and rotate it by an angle $\Delta\alpha$ about its center. Both transformations take place over the same period of time and at constant rates. That is, if it takes 10 seconds to perform this motion, then at 5 seconds, the rectangle's center is halfway to its final destination, and its rotation is halfway between where it was and where it will be. The center $R_i$ will always lie on the segment $\overline{RR'}$. I've built a Geogebra model here.
Question
What is the distance travelled by each vertex of the rectangle?
 A: This answer might be unnecessarily complicated, I will try to simplify it a bit later (or someone will post simpler one meanwhile), but it should work as it is.
Additionally to your established notation, let the position of rectangle center be $R=(R_x,R_y)$ and initial position of the corner under investigation $C_0=(C_x,C_y)$ (here notice that initial angle $\alpha$, width $W$ and length $L$ are already incorporated in thes positions, and they are not important in further calculations). Also for simplicity let assume the time $t\in [0,1]$ (you can always achieve that by simple substitution). Let $C(t)$ be position of a selected corner at time $t$, it can be seen that it is simply
$$
C(t)=Rot(C_0+tV,R+tV,t\Delta\alpha)
$$
where $Rot(X,S,\Delta\alpha)$ is rotation of point $X$ around point $S$ by angle $\Delta\alpha$ (clockwise, as given in your model). Basically it says that you translate the point $C_0$ to $C_0+tV$ and then rotate it around the translated center $R+tV$ by angle $\Delta\alpha$.
It is well known that
$$
Rot(X,S,\Delta\alpha) = 
\begin{bmatrix}
    \cos\Delta\alpha &\sin\Delta\alpha \\
    -\sin\Delta\alpha &\cos\Delta\alpha \\
\end{bmatrix}
(X-S)+S.
$$
Denoting $C(t)=(x(t),y(t))$, after some simplification we have trajectory of the corner point given as a parametric curve
\begin{align}
x(t)&=(C_x-R_x)\cos(\Delta\alpha t)+(C_y-R_y)\sin(\Delta\alpha t)+\Delta x\cdot t+R_x\\
y(t)&=(R_x-C_x)\sin(\Delta\alpha t)+(C_y-R_y)\cos(\Delta\alpha t)+\Delta y\cdot t+R_y.\\
\end{align}
for $t\in[0,1]$.
Now the length of such curve is known to be
$$l=\int_0^1 \sqrt{x'(t)^2+y'(t)^2}\, dt$$
which due to many parameters can get quite ugly, so I will just substitute the constants along the way to keep things simple. So let 
\begin{align}
u&=C_x-R_x\\
v&=C_y-R_y,
\end{align}
then
\begin{align}
x(t)&=u\cos(\Delta\alpha t)+v\sin(\Delta\alpha t)+\Delta x \cdot t+R_x\\
y(t)&=-u\sin(\Delta\alpha t)+v\cos(\Delta\alpha t)+\Delta y \cdot t+R_y\\
\end{align}
and so
\begin{align}
x'(t)&=-u\Delta\alpha\sin(\Delta\alpha t)+v\Delta\alpha\cos(\Delta\alpha t)+\Delta x \\
y'(t)&=-u\Delta\alpha\cos(\Delta\alpha t)-v\Delta\alpha\sin(\Delta\alpha t)+\Delta y .\\
\end{align}
Let's further put 
\begin{align}
a&=&u\Delta\alpha&=&(C_x-R_x)\Delta\alpha\\
b&=&v\Delta\alpha&=&(C_y-R_y)\Delta\alpha
\end{align}
so that
\begin{align}
x'(t)&=-a\sin(\Delta\alpha t)+b\cos(\Delta\alpha t)+\Delta x \\
y'(t)&=-a\cos(\Delta\alpha t)-b\sin(\Delta\alpha t)+\Delta y .\\
\end{align}
Now using basic algebra we arrive at
\begin{align}
x'(t)^2+y'(t)^2 &= a^2+b^2+\Delta x ^2+\Delta y ^2\\
&\ \ +(-2a\Delta x -2b\Delta y )\sin(\Delta\alpha t)+(2b\Delta x -2a\Delta y )\cos(\Delta\alpha t).
\end{align}
Again to simplify, let 
\begin{align}
p&=&a^2+b^2+\Delta x ^2+\Delta y ^2&=&((C_x-R_x)\Delta\alpha)^2+((C_y-R_y)\Delta\alpha)^2+\Delta x ^2+\Delta y ^2\\
q&=&(-2a\Delta x -2b\Delta y )&=&(-2((C_x-R_x)\Delta\alpha)\Delta x -2((C_y-R_y)\Delta\alpha)\Delta y )\\
r&=&(2b\Delta x -2a\Delta y )&=&(2((C_y-R_y)\Delta\alpha)\Delta x -2((C_x-R_x)\Delta\alpha)\Delta y )
\end{align}
so finally path length is
\begin{align}
l=\int_0^1\sqrt{x'(t)^2+y'(t)^2}\, dt &= \int_0^1\sqrt{p+q\sin(\Delta\alpha t)+r\cos(\Delta\alpha t)}\, dt
\end{align}
Now this integral does not seem to lead to anything nice (or I am unable to find it), but in your case it might be more practical to approximate it using numerical integration methods (see for example https://en.wikipedia.org/wiki/Numerical_integration). So for example using trapezoidal rule we might approximate
\begin{align}
l\approx \frac{1}{n}\left(\frac{f(0)}{2}+\sum_{k=1}^{n-1}f\left(\frac{k}{n}\right)+\frac{f(1)}{2}\right)
\end{align}
where $f(t)=\sqrt{p+q\sin(\Delta\alpha t)+r\cos(\Delta\alpha t)}$, $p,q,r$ are as defined above and $n$ is number of intervals in the approximation (bigger $n$ gives more accurate approximation).
You can do some sanity checks to see this works. For example choose $C_0=R$, then the expected length is $\sqrt{\Delta x ^2+\Delta y ^2}$ since that point only moves along the line segment. If you plug this $C_0$ to the equation above, you will find out that $q=r=0$ and $p=\Delta x ^2+\Delta y ^2$ and the integral above evaluates to $\sqrt{\Delta x ^2+\Delta y ^2}$ as expected (actually even the approximation mentioned above gives exact result in this case for any $n$). You can check that same is true also for $\Delta\alpha = 0$.
However let's apply it to the example you linked in the geogebra above, let's say for the upper left corner:


  We have
  \begin{align}
R&=(15.3, 15.7)\\
C_0&=(11.2, 28.1)\\
\Delta\alpha&=\frac{23}{45}\pi\\
V&=(25,-3)
\end{align}
  and
  \begin{align}
p&=1072.78\\
q&=448.63\\
r&=953.04
\end{align}
  And so we wish to evaluate $\int_0^1\sqrt{1072.78+448.63\sin(\frac{23}{45}\pi t)+953.04\cos(\frac{23}{45}\pi t)}\, dt$
  which by the method above gives us
  \begin{array}{c|c}
n &l\approx \\\hline
1 &41.791054 \\
10 &44.140258 \\
100 &44.163487 \\
1000 &44.163719 \\
\end{array}

