$\tan\left(\frac{1}{\cos\left(\frac{x}{\sqrt{x^2+\:16}}\right)}\right)$
$10x+2\left(5x-4\right)$
$\frac{dy}{dx}\left(-x^{2}+6x-4\right)$
$\frac{3x^3-2x^4-1-x^2}{x-12}$
$\frac{1}{4}m^2+\frac{3}{5}m^2-n^3+5n^3$
$5x^4-3x^2+2x^2+6x$
$\int x^3\left(x^4-1\right)^9dx$
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