An expression expr is of type radfun (radical function) if it is a rational function in the variable(s) V over the domain K extended by radicals or the number I.

If no variables were specified, all the indeterminates of expr which are names are used, so expr must be an algebraic function in all of its variables.

If no domain is specified, the default domain 'constant' is used.

$\mathrm{type}\left(\frac{x}{1-x}\,\mathrm{radfun}\left(\mathrm{integer}\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(\mathrm{sqrt}\left(x-y\right)\,\mathrm{radfun}\left(\mathrm{rational}\,\left[x\,y\right]\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(\mathrm{sqrt}\left(x-\sin \left(y\right)\right)\,\mathrm{radfun}\left(\mathrm{anything}\,x\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(\mathrm{sqrt}\left(x-\sin \left(y\right)\right)\,\mathrm{radfun}\left(\mathrm{anything}\,y\right)\right)$

$\mathrm{false}$

$\mathrm{type}\left(\mathrm{sqrt}\left(x+1\right)\,\mathrm{radfun}\right)$

$\mathrm{true}$

$\mathrm{type}\left(\mathrm{sqrt}\left(x+\mathrm{sqrt}\left(2\right)\right)\,\mathrm{radfun}\left(\mathrm{rational}\,x\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(\mathrm{sqrt}\left(x+\mathrm{sqrt}\left(y\right)\right)\,\mathrm{radfun}\right)$

$\mathrm{true}$

$\mathrm{type}\left(\mathrm{sqrt}\left(x+\exp \left(x\right)\right)\,\mathrm{radfun}\right)$

$\mathrm{false}$