The call type(expr, arctrig) returns true if expr is a function and the function name is one of the inverse trigonometric or hyperbolic functions:

The call type(expr, arctrig(x)) checks, in addition, that the argument to the inverse trigonometric function contains the variable name x.

The call type(expr, arctrigh) returns true if expr is a function and the function name is one of the inverse hyperbolic functions:

The call type(expr, arctrigh(x)) checks, in addition, that the argument to the inverse hyperbolic function contains the variable name x.

Note that the arctrigh type is a subset of the arctrig type, in the sense that an expression of type arctrigh is also of type arctrig. However, the converse is not true: the arcsin function is of type arctrig but not of type arctrigh.

$\mathrm{type}\left(\arcsin \left(x\right)\,'\mathrm{arctrig}'\right)$

$\mathrm{true}$

$\mathrm{type}\left(\arcsin \left(x\right)\,'\mathrm{arctrigh}'\right)$

$\mathrm{false}$

$\mathrm{type}\left(\mathrm{arcsinh}\left(x\right)\,'\mathrm{arctrig}'\right)$

$\mathrm{true}$

$\mathrm{type}\left(\mathrm{arcsinh}\left(x\right)\,'\mathrm{arctrigh}'\right)$

$\mathrm{true}$

$\mathrm{type}\left(\exp \left(x\right)\,'\mathrm{arctrig}'\right)$

$\mathrm{false}$

$\mathrm{type}\left(\exp \left(x\right)\,'\mathrm{arctrigh}'\right)$

$\mathrm{false}$

$\mathrm{type}\left(\arcsin \left(x\right)+\arccos \left(x\right)\,'\mathrm{arctrig}'\right)$

$\mathrm{false}$

$\mathrm{type}\left(\arcsin \left(x\right)+\arccos \left(x\right)\,'\mathrm{arctrigh}'\right)$

$\mathrm{false}$

$\mathrm{type}\left(\arcsin \left(1\right)\,'\mathrm{arctrig}\left(x\right)'\right)$

$\mathrm{false}$

$\mathrm{type}\left(\arcsin \left(1\right)\,'\mathrm{arctrigh}\left(x\right)'\right)$

$\mathrm{false}$

$\mathrm{type}\left(\mathrm{arctanh}\left(3x-1\right)\,'\mathrm{arctrig}\left(x\right)'\right)$

$\mathrm{true}$

$\mathrm{type}\left(\mathrm{arctanh}\left(3x-1\right)\,'\mathrm{arctrigh}\left(x\right)'\right)$

$\mathrm{true}$