TL;DR: you need to compute solar declination given your axial tilt, current true anomaly and the true anomaly of the winter solstice. You can feed that and your latitude into the sunrise equation.
Here's a complete worked example though, mostly for my own edification but others might find it helpul or useful.
(and for future readers, the date it was written was the date used to compute various numbers, and was 2020-02-08)
1) For a given orbit day (elapsed planetary days since perihelion, for simplicity) calculate the true anomaly.
Earth's perihelion in 2020 was on the 5th of January, so we're on day 34 of our current orbit. If we say the year length is 365 days and perihelion was precisely at midnight, that makes the current Mean Anomaly 33.5° (something like wolfram alpha will give you a more accurate value, but this'll do for an example).
If Earth had a perfectly circular orbit, the true anomaly would be exactly the same as the mean anomaly. Alas, real life is irrational and unhelpful, and so we do not have a nice tractable circular orbit.
We can compute the True Anomaly via this nice simple equation:
$$\nu = M + \left(2e - \frac{1}{4} e^3\right) \sin M + \frac{5}{4} e^2 \sin 2M + \frac{13}{12} e^3 \sin 3M + \cdots$$
where $e$ is the eccentricity of the orbit, which for Earth is ~0.0167, and $M$ is the mean anomaly we computed above. Using just these first three terms of the series expansion, we get a true anomaly $\nu$ of ~35.63° (and if you wanted more terms, you can have a read of this). Again, a slightly more reputable source than "some person on the internet" will give you a better value, but we're still close enough to see that this simple(ish) example isn't totally wrong.
2) From the true anomaly, calculate the orbital angular velocity.
The orbital velocity of a body changes as it procedes around its orbit... it will be fastest at perihelion, and slowest at aphelion. It is the rate of change of the true anomaly.
For a perfectly circular orbit, it would be simple: about .986° per day, or ~1.1416x10-5 degrees per second.
As before, ellipses ruin everything. You get the the specific relative angular momentum of an orbit where the orbiting body masses much less than the orbited body (as is the case with Earth and the Sun, for example) via this equation:
$$h = \sqrt{GM_sa(1-e^2)}$$
where $M_s$ is the mass of the Sun and $a$ is the semi-major axis of the planet and $e$ is still its orbital eccentricity. Courtesy of this handy answer on physics.SE, you can see that angular velocity $\omega$, the rate of change of true anomaly, can be obtained from $h = \omega r^2$.
You can get $r$ from $\nu$ like this:
$$r = \frac{a(1-e^2)}{1 + e \cos(\nu)}$$
So, today's value of $r$ is approximately 1.4755x1011m, giving us a current angular velocity of about 1.1724x10-5 degrees per second. As expected, this is a little faster than the circular equivalent, because we're closer to Earth's perihelion than the aphelion and so our orbital speed is a little higher than average.
3) From the orbital angular velocity and the rotation angular velocity, calculate the mean angular velocity of the sun across the sky.
In a circular orbit, if the rotational angular velocity were equal to the orbital angular velocty the world would be tidally locked and the sun would never appear to move. That would make the question a little too easy to answer, though.
The rotational period of the earth (the sidereal day) is a little shorter than the average 24 hour day (the solar day), which is the length of time between the sun reaching its zenith point in successive cycles. There's a handy answer on this very site for computing the solar day length: How to calculate the solar day from sidereal day and sidereal orbital period?
This of course gives you an average solar day length, which isn't quite right as the day length changes slightly due to orbital and rotational inconveniences. I'm going to skip handling the equation of time for now, and cheat by assuming an average 24 day which gives us the mean angular velocity of ~0.0042°/s. I might revisit this later, but don't hold your breath.
3) From the latitude, axial tilt and [true anomaly-solstice anomaly], calculate the angular length of sun's path in the sky at the required latitude.
The solstice anomaly mentioned here is presumably the true anomaly of the planet when it was last at a solstice, that being the point at which a pole is closest to (or further from) the sun. Again, we live on an inconvenient planet where the solstices do not coincide with the apsides (though for various reasons the gap between them changes over time in multi-millenia cycles which I shall ignore entirely. They've coincided in the past, will in the future, and could coincide for your fictional worlds, too). The last winter solstice was roughly at day 350 of last year, and you can compute its true anomaly using the method in step (1), giving $\nu_w$ of approximately 343.98°.
Solar declination declination is the angle between the sun's current zenith, and its zenith during the equinoxes, and you can compute it from your planet's axial tilt and the length of time since the last solstice:
$$\delta_\odot = \theta_a \cdot \cos(\nu - \nu_w)$$
where $\theta_a$ is the Earth's axial tilt, about -23.44°. Today's declination is therefore approximately -14.55°.
My latitude $\Phi$ is about 52° north. You can use the sunrise equation to find the hour angle of sunrise and sunset:
$$\pm \cos \omega_0 = -\tan \Phi \tan \delta_\odot$$
Where sunrise has the positive hour angle and sunset has the negative. The day length is then the sunrise angle minus the sunset angle... in this case about 141 degrees.
(This does assume that the Sun is a point source of light instead of a disc, and atmospheric refraction of light from an over-the-horizon sun is also ignored. You can use a more generalised equation which has an additional term to take these things into account)
Note that when $\Phi$ becomes large enough you will find that sunrise and sunset times become no longer defined. This is a sign that the latitude you're looking at is in a period of 24 hour night or day, where it will remain until $\delta_\odot$ has moved a bit closer to zero. 75.4N is roughly the limit for a sunrise at the moment, which is why places like Svalbard don't manage a proper daytime right now.
If $\delta_\odot$ is zero, then the day has the same length regardless of latitude. This happens on the equinoxes. If axial tilt is zero then on every day of the year the day length will be the same regardless of latitude (though one day might have a slightly different length than the next, depending on your planet's orbital eccentricity).
5) From 3) and 4), derive the day length.
Divide the day length angle from (4) by the angular velocity in (3). In my case, this ands up being approximately 33890 seconds, or a bit over 9 hours and 24 minutes. This is within a few minutes of the actual day length according to timeanddate.com, which is a nice result.
Easy as that!
Note: Handling the difference between civil twilight, nautical twilight, astronomical twilight and night will be left as an exercise for the reader. The additional complexity is minimal ;-)
