The function type/taylor returns true if expr is a series which is a polynomial in the variable in which it was expanded.  An order term may be present.

$\mathrm{series}\left(\sin \left(x\right)\,x\,5\right)$

$x-\frac{1}{6}{x}^3+O\left(x^5\right)$

$\mathrm{type}\left(\,'\mathrm{taylor}'\right)$

$\mathrm{true}$

$\mathrm{series}\left(\frac{1}{\sin \left(x\right)}\,x\,5\right)$

$x^{\mathrm{-1}}+\frac{1}{6}x+O\left(x^3\right)$

$\mathrm{type}\left(\,'\mathrm{taylor}'\right)$

$\mathrm{false}$

$\mathrm{type}\left(x^3\,'\mathrm{taylor}'\right)$

$\mathrm{false}$

$\mathrm{series}\left(\frac{x}{y}\,x\right)$

$\frac{1}{y}x$

$\mathrm{type}\left(\,'\mathrm{taylor}'\right)$

$\mathrm{true}$

$\mathrm{series}\left(\frac{x}{y}\,y\right)$

$x{y}^{\mathrm{-1}}$

$\mathrm{type}\left(\,'\mathrm{taylor}'\right)$

$\mathrm{false}$