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

$9y+4y$

$\left(y^2+1\right)\left(y^2-13\right)$

$-\left|-8+\left(4-7\right)+\left(2-5-3\right)\right|+\left|\left(6-3\right)-\left(2-5-6\right)-12\right|$

$\frac{1}{x^3+5x}$

$\cos\left(x\right).\left(\sin\left(x\right)-1\right)=0$

$\int\left(\cos\left(\arccos\left(2x\right)\right)\right)dx$