$\int_0^{\frac{\pi}{2}}\left(\sec^4\frac{x}{2}\right)dv$

$\left(2a+4b^3\right)$

$\left(x^2+3x+2\right)^2$

$5+9a-9a$

$\left(4x^3\right)^5$

$-10a+8a+a-8a$

$\left(\frac{\left(a^{-2}b^{-3}\right)}{\left(x-y^2\right)}\right)\left(\frac{\left(x^2b\right)}{\left(a^{\left(-\frac{2}{3}\right)}y^{-3}\right)}\right)$