An expression expr is of type radalgfun if it is an expression in the variable(s) V over the domain K extended by (polynomial) RootOfs or radicals.

The domain specification K must be a type name, such as rational or anything.  If K is omitted, then it defaults to type constant.

The optional argument V is an indeterminate or a list or set of indeterminates.  If V is not specified, then all the indeterminates of expr, which are names, are used.  That is, expr must be an algebraic function in all of its variables.

$\mathrm{type}\left(\frac{x}{1-x}\,\mathrm{radalgfun}\left(\mathrm{rational}\,x\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(\frac{\mathrm{sqrt}\left(x\right)}{1-\mathrm{RootOf}\left(z^3+z+x\,z\right)}\,\mathrm{radalgfun}\left(\mathrm{rational}\,x\right)\right)$

$\mathrm{true}$

$f≔1+2\mathrm{RootOf}\left(x^3-2+\mathrm{sqrt}\left(y\right)\,x\right)+yz$

$f≔1+2\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2+\sqrt{y}\right)+yz$

$\mathrm{type}\left(f\,\mathrm{radalgfun}\left(\mathrm{anything}\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(f\,\mathrm{radalgfun}\left(\mathrm{rational}\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(f\,\mathrm{radalgfun}\left(\mathrm{rational}\,y\right)\right)$

$\mathrm{false}$

$\mathrm{type}\left(f\,\mathrm{radalgfun}\left(\mathrm{rational}\,\left[y\,z\right]\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(f\,\mathrm{algfun}\left(\mathrm{rational}\,\left[y\,z\right]\right)\right)$

$\mathrm{false}$

$g≔{\left(2+\mathrm{RootOf}\left(x^5+x+y\,x\right)\right)}^{-\frac{2}{3}}$

$g≔\frac{1}{{\left(2+\mathrm{RootOf}\left({\mathrm{\_Z}}^5+\mathrm{\_Z}+y\right)\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

$\mathrm{type}\left(g\,\mathrm{radalgfun}\left(\mathrm{rational}\,y\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(g\,\mathrm{algfun}\left(\mathrm{rational}\,y\right)\right)$

$\mathrm{false}$

$\mathrm{type}\left(g\,\mathrm{radfun}\left(\mathrm{rational}\,y\right)\right)$

$\mathrm{false}$