Arctan and Tan inverse!

\begin{aligned} I & = \int_1^{16} \tan^{-1}\color{#3D99F6}{\sqrt{\sqrt{x}-1}} \ dx \quad \quad \small \color{#3D99F6}{\text{Let }u = \sqrt{\sqrt{x}-1}, \ x = (u^2+1)^2, \ dx = 4u(u^2+1) \ du} \\ & = \int_0^\sqrt{3} 4u(u^2+1) \tan^{-1} u \ du \quad \quad \small \color{#3D99F6}{\text{By integration by parts: }f' = 4u(u^2+1), \ g = \tan^{-1} u} \\ & = (u^4+2u^2)\tan^{-1} u \ \bigg|_0^\sqrt{3} - \int_0^\sqrt{3} \frac{u^4+2u^2}{u^2+1} \ du \\ & = 5\pi - \int_0^\sqrt{3} \left(u^2 + \color{#3D99F6}{\frac{u^2}{u^2+1}} \right) du \quad \quad \small \color{#3D99F6}{\text{Let }u = \tan \theta, \ du = \sec^2 \theta \ d\theta} \\ & = 5\pi - \frac{u^3}{3} \ \bigg|_0^\sqrt{3} - \int_0^\frac{\pi}{3} \tan^2 \theta \ d\theta \\ & = 5\pi - \sqrt{3} - \int_0^\frac{\pi}{3} (\sec^2 \theta - 1) \ d\theta \\ & = 5\pi - \sqrt{3} - (\tan \theta - \theta) \ \bigg|_0^\frac{\pi}{3} \\ & = 5\pi - \sqrt{3} - \sqrt{3} + \frac{\pi}{3} \\ & = \frac{16\pi}{3} - 2\sqrt{3} \end{aligned}

$\implies A+B+C+D+E = 16+1+3+2+3 = \boxed{25}$

Relevant wiki: Definite Integrals

Aha! Here's my approach! Hope you enjoy the ride... :D. Let $I = \large \int_1^{16} \arctan \sqrt{\sqrt x-1}\mathrm \ dx$

Substitute $\sqrt{\sqrt x-1} = \tan t\\\text{So, }\dfrac{dx}{4\sqrt x\times\sqrt{\sqrt x-1}} = \sec^2t \,dt\\\sqrt x = \sec^2 t\\\Rightarrow dx = 4\tan t \sec^4 t dt=4(\tan t+\tan^3 t)\sec^2 t\, dt$

Also note, $\begin{aligned}2\tan^2 t + \tan^4 t &= 2\tan^2t + \tan^2 t\left(\sec^2 t - 1\right)\\& = \tan^2 t + \tan^2 t \sec^2 t\\& = \sec^2 t +\tan^2 t \sec^2 t -1\end{aligned}$

Hence, using parts $\begin{aligned}I& = \int_0^{\pi/3} 4t(\tan t+\tan^3 t)\sec^2 t dt\\ &=t\left(2\tan^2 t+ \tan^4 t\right)_0^{\pi/3} - \int_0^{\pi/3} 2\tan^2 t+ \tan^4 t dt\\ &= 5\pi - \int_0^{\pi/3} \sec^2 t +\tan^2 t \sec^2 t -1 dt\\ &=\dfrac {15\pi}3 - \left(\tan t + \dfrac{\tan^3 t}3 - t\right)_0^{\pi/3}\\ &=\dfrac {15\pi}3 - \sqrt 3-\dfrac{3\sqrt 3}3 + \dfrac\pi3\\ &= \boxed{\dfrac{16\pi}3 - 2\sqrt 3}\end{aligned}$