tl;dr The change in time between moonrise and moonset is different because of the changing declination of the Moon. When the declination is increasing, the time between moonrise is shorter than normal, but the time between moonset is longer than normal. As shown in Figure 1, the line connecting the Moon's position when rising is shallow to the horizon, so the change between rising is smaller. The same line when setting is steep to the horizon, so the change between setting is larger. Figure 1: Moon was at position 1 when rising on the first day (although shown some time after moonrise) and at position 3 when rising the next day. The Moon is at position 2 when setting on the first day and position 4 when setting on the next day (although shown some time before moonset).
Long story: The mathematical demonstration that the rise and set times change by different values is based on the following figure and definitions. Note that my numbers are based on a point source, no atmosphere, and the northern hemisphere. Those assumptions would change the actual numbers for how the rate changes, but it does not change the conclusion. Figure 2: Nomenclature used in the following equations.
* decl. is the declination of the object; the angle of the object above or below the celestial equation.
* RA is the Right Ascension; the angle of the object along the equator measured from the March equinox. 1h RA = 15 degrees.
* Meridian is the line from due south to the zenith (the point overhead) to due north.
* LST is the Local Sidereal Time; the LST equals the right ascension (RA) of an object on the meridian. 797
LST increases at a uniform rate of approximately 1.0027 hours of RA per hour of time. (24 hours of RA in 23 h 56 m 4.091 s of time; the sidereal day)
* HA is the hour angle; the angle (in degrees or hours of time) between the meridian and the object, measured along the equator. That is, $HA = RA-LST$ when the object is east of the meridian, and $HA = LST-RA$ when the object is west of the meridian.
* For an object on the equator (decl.=0), the HA is 6 hours when it rises or sets. For an object with a positive declination, the HA is longer than 6 hours. For an object with a negative declination, the HA is less than 6 hours.
When rising lets begin with the Moon on the celestial equation so that $$HA = 6h = RA_1 - LST_1 $$ when ignoring the diameter of the Moon and refraction. Subscript "1" indicates the moonrise on the first day. On the next day, the Moon has moved eastward to position 3, so its RA has increased. Just to assign an approximate number, assume the Moon's RA increases at 13 degrees per day = $\frac{13}{360}t$ hours of RA per hour of time t, and the new HA on the next day has increased by $h(decl._3)$ hours. (h(decl.) is based on some function of the declination. Both the Moon's changing declination and h(decl.) are beyond the details here!) On the next day we have $$\begin{align} 6h+h(decl._3) & = RA_3-LST_3 \ & = (RA_1+\frac{13}{360}t)-(LST_1+1.0027t-24d) \ & = (RA_1-LST_1) +\frac{13}{360}t-1.0027t+24d \ & = (6h) +\frac{13}{360}t-1.0027t+24d \end{align} $$ where d=1 for the next day, d=2 for the second day after, and so on, 24d puts the LST back in the range of 0 to 24 hours RA, and t is the number of hours that have elapsed since moonrise on the first day. Rearranging the terms enables the elapsed time t to be calculated from $$ 1.0027t-\frac{13}{360}t = 24d-h(decl._3) $$ When the declination increases, h(decl.) is a positive number. Therefore, the time between moonrises is smaller than if h(decl.) were zero.
When setting on the first day $$HA = 6h+h(decl._2) = LST_2-RA_2 $$ On the next day, the Moon has moved eastward to position 4 $$\begin{align} 6h+h(decl. 371