How does an IIR filter change depending on the approximation used?
A simple Wikipedia tour (B,CH,E) will tell you everything, but fear not, I can copy and paste some information that is already stated there.

This is how the different filters behaves like:



Image was grabbed from the Wikipedia page. 

Given the same order of a filter, the following information holds true:

The Butterworth filter:


Has the lowest ripple in the passband and stopband
Has longest transition band
Requires a higher number of order to get a similar short transition band like the elliptic filter.


The CH1 filter:


Has ripples in passband
Has nearly no ripples in stopband
Has shorter transition band than Butterworth
Has longer transition band than Elliptic


The CH2 filter:


Has nearly no ripples in passband
Has ripples in stopband
Has shorter transition band than Butterworth
Has longer transition band than Elliptic


The Elliptic filter:


Has ripples in passband
Has ripples in stopband
Has the shortest transition band for any given order. 




A low order in analog domain means few number of components. Adding more components, like inductors, capacitors can cost a lot. 

A low order in digital domain means a few number of additions, multiplication and time delays. Performing an addition, multiplication or a time delay does not cost that much.



You use the elliptic filter if you are okay with ripples in the passband and stopband. This depends on whatever you are working with. 

If you want what Butterworth offer but don't want to pay the full price, then use CH2 or CH1, depending on what that matters the most. 

If you are dealing with high quality equipment, and/or you are working with digital filters which means a higher order filter is very cheap, then you can easily use a butterworth filter. The best kind of filter, and the most expensive order wise.