Habitable Moon Day Cycle

B : Rotation movement of the secondary (Moon)

C : Star rays

How to calculate your day cycle :

1. Construct your star and gas giant. If you want a very earth like planet (>0.5 earth mass) your planet need to be at least 8'000 times this mass, which means 4000/318 = 12.578616352 jovian masses, meaning the upper limit of planets (13 jovian masses), so this mean that a earth like satelite would be impossible and would be found only around small brown dwarves. However, since it's all speculation (we only know of the solar system's gas giants moons systems) you can be less conservative. But still chose a superjovian planet when doing it. It should be at least 8 jovian masses to have still something somewhat believable.
EDIT : It seems I was mistaken, 10'000 is only an order of scale. in fact since Titan is 1/4000 the mass of Saturn it is probable that the higher limit is something like 1/3000. With that, you can have a moon of 0.5 Earth masses and a primary of 1500-2000 earth masses, meaning ~4.7-6.3 jovian masses. Also, for an even more Earth like moon, if you give the moon a mass of 75% that of Earth, your planet can be between 7-10 jovian masses easily. However take this with caution, it's possible that the bigger the planet is, the more it sucks the materials around it and so maybe the smaller proportionally a moon can be. But, good news : it may apply only to Jovian objects and not to Neptunian objects and only to the satellites that where formed around the planet. A kind of migrating inward super Neptune (intermediary between Neptune and Saturn) could capture Mars like objects for example. Take this all with a grain of salt, as long as we don't observe extrasolar moons nothing of all of that is confirmed.
Here some further reading :
www.universetoday.com/100/gas-…
www.reddit.com/r/askscience/co…
www.reddit.com/r/worldbuilding…

You can go on the great Artifexian's channel to build both star and gas giant :
www.youtube.com/watch?v=80oQBG…
To build its orbit :
www.youtube.com/watch?v=Ig_Dzm…
The star :

www.youtube.com/watch?v=x55nxx…

Yes, I know that his method has some issues, especially in surface temperature and size of stars. However, for the habitable range of masses, it's pretty accurate from experience. Just remember he calculates everything in relation of the Sun (multiply by the caracteristics of the Sun to obtain it in international units), his methode of gloldilock zone calculation is a bit confusing though, cause it's in relation of Earth's and not the Sun : he calculates where a planet needs to be in UA to have practically the same conditions as Earth and then multiply this number by 0.95 and 1.37, just multiply by 149'600'000 to obtain it in kilometers.

Your primary should be somewhere in the habitable zone, but contrarily to artifexian, i consider it's possible to put it slightly over it thank to tidal heat.

www.youtube.com/watch?v=Evq7n2…
Here a more detailed explanation.

2. Calculate the orbit of your moon (use the build orbit video to do so). Here we'll take the case of a moon that has a not too huge eccentricity. It's orbit need to be in the range 10-20 radii of your primary. 

Now we want to calculate the orbital period, I use this formula :

P[♁days] = √(((4π²(a)³)/G(M+m)))/864000

Where a is half great axis of the moon,
G is the gravitational constant,
M is mass of primary
m is mass of the moon, it's optional cause negligeable

Please note that you need to put the masses in kilogram and the half great axis in meters. If you have it in Earth masses multiply it by 5.97*10²⁴.

Now, to calculate the solar day of this tidally locked moon, we will need to use a formula that take into account the fact that each time the moon does a revolution, the planet will slightly move. You can calculate the revolution of the planet around the star in Earth years by using this formula :

A[♁years] = √(a³/M)

Where a is half great axis of the planet in UA,
M is mass of the sun in solar masses.

To obtain it in Earth days multiply by 365.25.=A[♁days]

Now we will need the average angular speed of both the primary and secondary. ^^
The angular speed of the primary will be :
Ω[rad/d] = 2π/A
And thus, that of the moon :
ω[rad/d] = 2π/P

Now, the formula for calculating the solar day :
D[♁days] = P/(1-(Ω/ω))

Since Ω/ω = (2π/A)/(2π/P) = (2π/A)(P/2π) = P/A,
you can use this slightly simpler equation and so not calculate Ω and ω :
D[♁days] = P/(1-(P/A))

3. calculate the eclipse periode :
Now you can ask yourself of the duration of the eclipses when the moon is going through the planet's penumbra and umbra. Note that here we're only considering the case where the moon is more or less in the same plan as its planet. In other cases, the eclipse will not be at each turn, however, since the planet is very close (proportionally) and very big, it will be apparently way bigger than our moon in our sky. You can by the way calculate it's apparent size with this formula :
App = 2arcsin(r/D)

r is the radius of the planet/star,
D is it's distance.

So for the eclipse period, what will it be? first we need to calculate what i call the tangential distance (if you have a better term tell it to me). It's a distance that is obtained by connecting a tangent on the surface of the planet to the center of the moon. We will consider the mean distance, so we will take the half great axis :
d_t = √(a²-r²)

a is the half great axis of the orbit of the moon
r is the radius of the planet
The units need to be all the same (be it planet radii (which i strongly advise), moon radii, earth radii, UA or KM)

With that we can determine with some trigonometry the angles that is formed by the arc formed by the intersection of the orbit with umbra/penumbra. This angle (that i advise to calculate in radiants) is given for the penumbra :

ψ = 2arctan((Ar+d_t(R+r))/(Rd_t-r(R+r)))

For the umbra :

υ = 2arctan((Ar+d_t(r-R))/(Rd_t-r(R+r)))

A is the half great axis of the planet,
r is the radius of the planet,
R is the radius of the star,
d_t is the tangential distance (as previously defined) from the planet to the moon.

Be sure to verify that the moon is indeed going through the umbra. For that the back focal point (the point where the planet and the star have the exact same apparent sizes) needs to be further from the planet than the moon is. To calculate the distance from the planet to the focal point you can use this equation :

d_f = Ar/(R-r)

At last you need to calculate the mean duration of the eclipse period. For that you just need to use the orbital speeds from earlier. I'll not adjust for the movement of the planet around the star cause it's rarely needed cause of the short period of time it represents.

T_pen = ψ/ω
T_umbra = υ/ω

/!\ Disclaimer/!\ I'll verify this later /!\ but if you want to take into account the planetary movement, use these formulas :
P_pen = ψ/(ω-Ω)
P_umbra = υ/(ω-Ω)

I'll show what we can achieve with that soon