[rmolnav]

It gains ... and loses 3.738 x 10^18 joules of kinetic energy due to the slowing of its orbital speed. That assumes I didn't make any kind of calculation error.

I found an equation that predicted orbital velocity based on the distance a satellite is from a planet and the planet's mass. The equation is: orbital velocity (m/s) = square root ((gravitational constant x mass of the planet (kg)) / orbital radius (m)).The orbital velocity I calculated for the Moon's semi-major axis of 384,399,000 meters was 1,018.20824493 meters per second. Then I used Newton's kinetic energy equation to calculate how much kinetic energy the Moon (with a mass of 7.342 x 10^22 kilograms) would have at that velocity. The equation is: kinetic energy (J) = 0.5 x mass (kg) x (velocity (m/s))^2. The result was 38,059,020,182,894,347,263,576,879,000 joules.
Then I repeated the calculations with the Moon being 3.82 centimeters further out (384,399,000.0382 meters), resulting in an orbital velocity of 1,018.20824488 meters per second and a kinetic energy of 38,059,020,179,156,504,796,530,624,000 joules. I subtracted the two to get a kinetic energy decrease of 3,737,842,467,046,255,000 joules per year.

There's an easier way.  The total orbital energy can be found by:
E= -GMm/2a
where a is the semimajor axis.   By using 3.987e14 for GM (we actually know this product more accurately than we know either G or the mass of the Earth) and calling it U, you can find the energy difference by
-Um/2 (1/a1-1/a2)
For your given Moon mass, this works out to 3,783,812,701,003,630,000 Joules.

Sorry, but in those calculations you have considered the further the Moon, the smaller its linear speed ... But, as far as I can understand, that´s not actually the case.
You have made calculations which seem to be correct, but ONLY if Moon were rotating "freely" as a satellite, without experiencing neither other force than gravity towards Earth´s C.G. (or the barycenter of Earth/Moon), nor an external torque ...
But, as I explained in my two last posts, constantly a torque is being exerted on the Moon due to the position of the C.G. of the massive bulge of the Earth´s high tide caused by Moon´s attraction.
Although the other bulge at the antipodes also causes a torque, but opposite, this is smaller due to the facts of being further from the Moon, and a smaller angle out of line between centers of G. (smaller tangential component of a smaller attracting force).
So, Moon´s kinetic energy actually increases with its speed and distance to Earth, at the expense of Earth´s energy.
As the hammer going to be thrown, at the expense of the athlete energy !!
 

[hamdani yusuf (OP)]

I think this video answers the question pretty nicely.

[rmolnav]

I think this video answers the question pretty nicely.

Yes, pretty nicely ... but not completely right !!
Just now I´ve finished drawing and sending following comment to relevant youtube site:
" Your video has many interesting, correct things, some of them rarely mentioned by even physicists. One of them: the interaction between the Moon and Earth´s tidal bulges ...
But you've got a detail wrong. You say:
"The Earth loses a lot of energy due to friction as its insides and the oceans slosh around
during this whole process, which also slows its spin".
Quite right so far. But then you add:
"Since the Earth’s rotation is slowing down, some mass in the Earth-Moon system moves farther from the center to compensate—and that mass, in this case, is the Moon".
And here you are forgetting that that the rotational momentum conservation principle is valid ONLY when in an isolated system, without any external energy input or output ... And tidal friction energy is lost, as heat ...
When you say:
"You know how a spinning figure skater slows down when they extend their arms?"
you are also quite right ... But eventually they would slow down even not extending their arms, due to friction with ice and air !!"