There must be a shortcut!

Let $\sqrt{2x+1} = a, \sqrt{2x-1} = b$
$S = \displaystyle \sum_{x=1}^{40} \dfrac{a^{2} +b^{2} + ab}{a+b} = \sum_{x=1}^{40} \dfrac{a^{3}-b^{3}}{a^{2}-b^{2}}$
$\therefore S = \displaystyle \sum_{x=1}^{40} \dfrac{(2x+1)^{\frac{3}{2}} - (2x-1)^{\frac{3}{2}}}{2x+1-(2x-1)} = \dfrac{(81)^{\frac{3}{2}}-1}{2} = 364$

After some calculations.

Just multiply numerator and denominator by $(\sqrt{2x+1}-\sqrt{2x-1})$ .

$\Rightarrow \dfrac {4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}=\dfrac{\sqrt{2x+1}(2x+1)+\sqrt{2x-1}(1-2x)}{2}$

Now, putting $x=1$ , $x=2$ , $x=3$ and $x=40$ , we observe that all terms cancel leaving $-\dfrac{1}{2}$ and $\dfrac{81\sqrt{81}}{2}$ .

$\therefore \dfrac{81\sqrt{81}-1}{2}=\dfrac{728}{2}=\boxed{364}$