The expression f is converted to factored normal form. This is the form numerator/denominator, where the numerator and denominator are relatively prime polynomials with integer coefficients.

The normal command provides basic simplification and recognizes expressions equal to zero in the domain of rational functions. This includes any expression constructed from sums, products and integer powers of integers and variables.

If f is a list, set, range, series, equation, relation, or function, normal is applied recursively to the components of f. In the case of a series, for example, this means that the coefficients of the series are normalized.

If f contains subexpressions not in the domain of rational functions, such as square roots, powers, and functions,  normal is first applied recursively to these objects. They are then frozen (see the frontend function) to unique names so that the form of f can be simplified.  For such cases, normal may not recognize when an expression is equal to zero.

If normal is called with the second argument, expanded, the numerator and denominator will be expanded polynomials.

$\mathrm{normal}\left(x^2-\left(x+1\right)\left(x-1\right)-1\right)$

$0$

$\mathrm{normal}\left(\frac{x^2-y^2}{{\left(x-y\right)}^3}\right)$

$\frac{x+y}{{\left(x-y\right)}^2}$

$\mathrm{normal}\left(\frac{{f\left(x\right)}^2-1}{f\left(x\right)-1}\right)$

$f\left(x\right)+1$

$\mathrm{normal}\left(\left\{\frac{2}{x}+\frac{y}{3}=0\,\frac{1}{x}-\frac{5}{x^2}=1\right\}\right)$

$\left\{\frac{x-5}{x^2}=1\,\frac{yx+6}{3x}=0\right\}$

$\mathrm{normal}\left(\sin \left(x\left(x+1\right)-x\right)\right)$

$\sin \left(x^2\right)$

$\mathrm{normal}\left(\frac{1}{x}-\frac{1}{x^2}<5\right)$

$\frac{x-1}{x^2}<5$

$g≔\left[\mathrm{seq}\left(\frac{7ix}{i^{2}x+1}+\frac{1}{x}\&comma;i=1..4\right)\right]$

$g≔\left[\frac{7x}{x+1}+\frac{1}{x}\&comma;\frac{14x}{4x+1}+\frac{1}{x}\&comma;\frac{21x}{9x+1}+\frac{1}{x}\&comma;\frac{28x}{16x+1}+\frac{1}{x}\right]$

$\mathrm{normal}\left(g\right)$

$\left[\frac{7x^2+x+1}{x\left(x+1\right)}\&comma;\frac{14x^2+4x+1}{\left(4x+1\right)x}\&comma;\frac{21x^2+9x+1}{\left(9x+1\right)x}\&comma;\frac{28x^2+16x+1}{\left(16x+1\right)x}\right]$

$f≔\mathrm{series}\left(\frac{xy}{x+y}+\frac{1}{x}\&comma;x=1\&comma;3\right)$

$f≔\frac{y}{1+y}+1+\left(\frac{y-\frac{y}{1+y}}{1+y}-1\right)\left(x-1\right)+\left(-\frac{y^2}{{\left(1+y\right)}^3}+1\right){\left(x-1\right)}^2+O\left({\left(x-1\right)}^3\right)$

$\mathrm{normal}\left(f\right)$

$\frac{2y+1}{1+y}-\frac{2y+1}{{\left(1+y\right)}^2}\left(x-1\right)+\frac{y^3+2y^2+3y+1}{{\left(1+y\right)}^3}{\left(x-1\right)}^2+O\left({\left(x-1\right)}^3\right)$

$\mathrm{normal}\left(\frac{1}{x}+\frac{x}{x+1}\right)$

$\frac{x^2+x+1}{x\left(x+1\right)}$

$\mathrm{normal}\left(\frac{1}{x}+\frac{x}{x+1}\&comma;\mathrm{expanded}\right)$

$\frac{x^2+x+1}{x^2+x}$