A calculus problem by Mateus Gomes

Relevant wiki: $u$ -Substitution

$\begin{aligned} I & = \int_0^3 \sqrt{9+x^2} \ dx \\ & = \int_0^3 3\sqrt{1+\left(\color{#3D99F6}{\frac x3}\right)^2} \ dx & \small \color{#3D99F6}{\text{Let }\frac x3 = \tan \theta, \ dx = 3 \sec^2 \theta \ d \theta} \\ & = \int_0^\frac \pi 4 9 \sec^3 \theta \ d \theta & \small \color{#3D99F6}{\text{By integration by part }} \\ & = 9\tan \theta \sec \theta \ \bigg|_0^\frac \pi 4 - 9 \int_0^\frac \pi 4 \tan^2 \theta \sec \theta \ d \theta \\ & = 9\sqrt 2 - 9 \int_0^\frac \pi 4 (\sec^2 \theta -1) \sec \theta \ d \theta \\ & = 9\sqrt 2 - I + 9 \int_0^\frac \pi 4 \sec \theta \ d \theta \\ & = \frac {9\sqrt 2}2 + \frac 92 \int_0^\frac \pi 4 \sec \theta \ d \theta \\ & = \frac {9\sqrt 2}2 + \frac 92 \ln (\tan \theta + \sec \theta) \ \bigg|_0^\frac \pi 4 \\ & = \frac {9\sqrt 2}2 + \frac 92 \ln (1 + \sqrt 2) \end{aligned}$

$\implies A+B+C+D+E+F = 9+2+2+9+2+1+2 = \boxed{27}$