long and long int are identical. So are long long and long long int. In both cases, the int is optional.
As to the difference between the two sets, the C++ standard mandates minimum ranges for each, and that long long is at least as wide as long.
The controlling parts of the standard (C++11, but this has been around for a long time) are, for one, 3.9.1 Fundamental types, section 2 (a later section gives similar rules for the unsigned integral types):
There are five standard signed integer types : signed char, short int, int, long int, and long long int. In this list, each type provides at least as much storage as those preceding it in the list.
There’s also a table 9 in 7.1.6.2 Simple type specifiers, which shows the “mappings” of the specifiers to actual types (showing that the int is optional), a section of which is shown below:
Specifier(s) Type ------------- ------------- long long int long long int long long long long int long int long int long long int
Note the distinction there between the specifier and the type. The specifier is how you tell the compiler what the type is but you can use different specifiers to end up at the same type.
Hence long on its own is neither a type nor a modifier as your question posits, it’s simply a specifier for the long int type. Ditto for long long being a specifier for the long long int type.
Although the C++ standard itself doesn’t specify the minimum ranges of integral types, it does cite C99, in 1.2 Normative references, as applying. Hence the minimal ranges as set out in C99 5.2.4.2.1 Sizes of integer types <limits.h> are applicable.
In terms of long double, that’s actually a floating point value rather than an integer. Similarly to the integral types, it’s required to have at least as much precision as a double and to provide a superset of values over that type (meaning at least those values, not necessarily more values).