$\sqrt{1+\left(4x^2\left(x+1\right)\right)}$
$x^2+\frac{5}{6}x$
$\left(3a^47b^5\right)^4$
$\int x\left(3x+4\right)^{\frac{1}{3}}dx$
$\left(\frac{\left(3e^x\right)\left(\sqrt{5x-1}\right)}{\left(x^2+1\right)^2}\right)$
$x^2-5x-5$
$x^2+4x\:=\:20$
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