The hasfun command searches an expression e for the function f, or, if f is a list or set of names (of functions), then hasfun searches e for any one of these functions. It returns true if it finds one, false otherwise.

If the third optional argument x is specified, the hasfun function tests if the expression e has function(s) f of x, in the sense that x occurs in one or more arguments of the call to f. A typical application is in integration where you might want to test if the integrand has a special function of a variable x.

For a positive match, x has to occur as an exact subexpression as defined by Maple's op function. See has for examples of what this means.

If the subexpression x that needs to occur in the arguments to f is a list or set, it needs to be enclosed in a list or set itself, otherwise hasfun will search for the members of x only.

$e≔\sin \left(x\right)+\exp \left(3y\right)+1$

$e≔\sin \left(x\right)+{\ⅇ}^{3y}+1$

$\mathrm{hasfun}\left(e\,\exp \right)$

$\mathrm{true}$

$\mathrm{hasfun}\left(e\,\cos \right)$

$\mathrm{false}$

$\mathrm{hasfun}\left(e\,\exp \,y\right)$

$\mathrm{true}$

$\mathrm{hasfun}\left(e\,\exp \,x\right)$

$\mathrm{false}$

$\mathrm{hasfun}\left(e\,\exp \,\left[x\,y\right]\right)$

$\mathrm{true}$

$\mathrm{hasfun}\left(e\,\left\{\cos \,\sin \right\}\,x\right)$

$\mathrm{true}$

$e≔\mathrm{D}\left(x\right)\left(t\right)-x\left(t\right)$

$e≔\mathrm{D}\left(x\right)\left(t\right)-x\left(t\right)$

$\mathrm{hasfun}\left(e\,\mathrm{D}\left(x\right)\,t\right)$

$\mathrm{true}$

$e≔g\left(2f\left(x+y+z\right)-2\right)$

$e≔g\left(2f\left(x+y+z\right)-2\right)$

$\mathrm{hasfun}\left(e\,f\,x+y\right)$

$\mathrm{false}$

$e≔1+\mathrm{Hypergeom}\left(\left[m\,n\right]\,\left[\right]\,z\right)$

$e≔1+\mathrm{Hypergeom}\left(\left[m\,n\right]\,\left[\right]\,z\right)$

$\mathrm{hasfun}\left(e\,\mathrm{Hypergeom}\,\left[k\,m\right]\right)$

$\mathrm{true}$

$\mathrm{hasfun}\left(e\,\mathrm{Hypergeom}\,\left[\left[k\,m\right]\right]\right)$

$\mathrm{false}$

$\mathrm{hasfun}\left(e\,\mathrm{Hypergeom}\,\left[\left[m\,n\right]\right]\right)$

$\mathrm{true}$