Nested Square Root Integral

If $y = \sqrt{x^2 + \sqrt{x^2 + \sqrt{x^2 + \sqrt{x^2...}}}}$ , then $y = \sqrt{x^2 + y}$ , which rearranges to $y^2 - y - x^2 = 0$ , which by the quadratic equation solves to $y = \frac{1 + \sqrt{1 + 4x^2}}{2}$ for $x > 0$ . Therefore, $y = \sqrt{x^2 + \sqrt{x^2 + \sqrt{x^2 + \sqrt{x^2...}}}} = \frac{1 + \sqrt{1 + 4x^2}}{2}$ and

$\int_{0}^{1} \sqrt{x^2 + \sqrt{x^2 + \sqrt{x^2 + \sqrt{x^2...}}}} dx$

$= \displaystyle \lim_{t\to 0^+} \int_{t}^{1} \frac{1 + \sqrt{1 + 4x^2}}{2} dx$

$= \frac{1}{2} \displaystyle \lim_{t\to 0^+} \int_{t}^{1} (\sqrt{4x^2 + 1} + 1) dx$

$= \frac{1}{2}[\frac{\text{arsinh}(2x)}{4} + \frac{x\sqrt{4x^2 + 1}}{2} + x]_{0^+}^1$

$= \frac{1}{2}[\frac{\text{arsinh}(2)}{4} + \frac{\sqrt{5}}{2} + 1]$

$= \frac{\text{arsinh}(2) + 2\sqrt{5} + 4}{8}$

Therefore, $a = 2$ , $b = 2$ , $c = 5$ , $d = 4$ , $e = 8$ , and $a \times b \times c \times d \times e = \boxed{640}$ .