I'm not sure it qualifies as "simple", but, using
http://idlastro.gsfc.nasa.gov/ftp/pro/astro/hadec2altaz.pro (and some
additional calculations/simplifications):
$
\begin{array}{|c|c|c|c|}
\hline
\text{Event} & \text{Time} & \phi & Z \\
\hline
\text{Any} & \text{t} & \tan ^{-1}(\cos (\lambda ) \sin (\delta )-\cos
(\delta ) \cos (\alpha -t) \sin (\lambda ),\cos (\delta ) \sin (\alpha
-t)) & \tan ^{-1}\left(\sqrt{(\cos (\lambda ) \sin (\delta )-\cos (\delta
) \cos (\alpha -t) \sin (\lambda ))^2+\cos ^2(\delta ) \sin ^2(\alpha
-t)},\cos (\delta ) \cos (\lambda ) \cos (\alpha -t)+\sin (\delta ) \sin
(\lambda )\right) \\
\hline
\text{Rise} & \alpha -\cos ^{-1}(-\tan (\delta ) \tan (\lambda )) & \tan
^{-1}\left(\sec (\lambda ) \sin (\delta ),\cos (\delta ) \sqrt{1-\tan
^2(\delta ) \tan ^2(\lambda )}\right) & 0 \\
\hline
\text{Transit} & \alpha &
\begin{cases}
\delta >\lambda & 0 \\
\delta =\lambda & \text{Zenith} \\
\delta <\lambda & \pi
\end{cases}
& \frac{\pi }{2}-\left| \delta -\lambda \right| \\
\hline
\text{Set} & \alpha +\cos ^{-1}(-\tan (\delta ) \tan (\lambda )) & \tan
^{-1}\left(\sec (\lambda ) \sin (\delta ),-\cos (\delta ) \sqrt{1-\tan
^2(\delta ) \tan ^2(\lambda )}\right) & 0 \\
\hline
\text{Lowest Point} & \alpha +\pi &
\begin{cases}
\delta >-\lambda & 0 \\
\delta =-\lambda & \text{Nadir} \\
\delta <-\lambda & \pi
\end{cases}
& \left| \delta +\lambda \right|-\frac{\pi }{2} \\
\hline
\end{array}
$
where:
$\phi$ is the azimuth of the object
$Z$ is the altitude of the object above the horizon
$\alpha$ is the right ascension of the object
$\delta$ is the declination of the object
$\lambda$ is the latitude of the observer
$t$ is the current local sidereal time
Note the two-argument form of arctangent is required so that the
results are in the correct quadrant:
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent
Additional caveats:
If $\left| \delta -\lambda \right|>\frac{\pi }{2}$, the object is
always below the horizon, and the equations for rising time and
setting time will not work.
If $\left| \delta +\lambda \right|>\frac{\pi }{2}$, the object is
always above the horizon (circumpolar), and the equations for rising
and setting time will also not work.
The measurements above are in radians. You can convert $\pi \to
180 {}^{\circ}$ for degrees.
Because we use the local sidereal time, the longitude doesn't
appear in any of the formulas above. However, we do need it to find
the local sidereal time, as below.
To find the local sideral time $t$ in radians, we use
http://aa.usno.navy.mil/faq/docs/GAST.php and make some substitions
to get:
$t = 4.894961212735792 + 6.30038809898489 d + \psi$
where $\psi$ is your longitude in radians, and $d$ is the number of
days (including fractional days) since "2000-01-01 12:00:00 UTC". Traditionally, we use $\phi$ for longitude, but I'm already using it in the formulas above for azimuth.
If you combine the formula for local sidereal time and
azimuth/altitude and assume excessive precision, you get my answer to
https://astronomy.stackexchange.com/a/8415/21
Additional computations for these results at:
https://github.com/barrycarter/bcapps/blob/master/STACK/bc-rst.m
I was going to add some graphs to show how the altitude is NOT a sine wave and how the azimuth is NOT a straight line (although you might expect them to be), but they turned out not to be terribly instructive/helpful.
You might also be able to get simpler formulas if you set $t$ to be the "hour angle" (which is $\alpha-t$ in the current setup).