Who's up to the challenge? 6

Let $I$ be the integral, then we have:

$\begin{aligned} I & = \int_0^\infty e^{-x^3} \sin x^3 \space dx \quad \quad \small \color{#3D99F6}{\text{Let } u = x^3, \space du = 3 x^2 \space dx} \\ & = \int_0^\infty \frac{e^{-u} \sin u}{3 u^{\frac{2}{3}}} du \\ & =\Im \int_0^\infty \frac{1}{3} \left(u^{-\frac{2}{3}} e^{-u} e^{ui}\right) du \\ & = \frac{1}{3} \Im \int_0^\infty u^{-\frac{2}{3}} e^{-(1-i)u} du \quad \quad \small \color{#3D99F6}{\text{Let } z = (1-i)u, \space dz = (1-i) \space du} \\ & = \frac{1}{3} \Im \left((1-i)^{-\frac{1}{3}} \int_0^\infty z^{\color{#3D99F6}{\frac{1}{3}}-1} e^{-z} dz \right) \\ & = \frac{1}{3} \Im \left( \left( \sqrt{2} e^{-\frac{\pi}{4}i} \right)^{-\frac{1}{3}} \Gamma \left(\frac{1}{3}\right) \right) \\ & = \frac{1}{3} \Im \left( \frac{e^{\frac{\pi}{12}i} \Gamma \left(\frac{1}{3}\right)}{\sqrt [6]{2}} \right) \\ & = \frac{\sin \frac{\pi}{12} \Gamma \left(\frac{4}{3}\right)}{\sqrt [6]{2}} = \frac{(\sqrt{3}-1) \Gamma \left(\frac{4}{3}\right)}{2 \sqrt{2} \sqrt [6]{2}} = \frac{(\sqrt{3}-1) \Gamma \left(\frac{4}{3}\right)}{\sqrt [3]{2^5}} \end{aligned}$

$\Rightarrow A+B+C+D+E = 3+4+3+2+5 = \boxed{17}$

I used differentiation under the integral sign...