Sin and squares?

Substituting $u=\sqrt{\lambda}x$ , this integral becomes

$\frac{1}{\sqrt{\lambda}}\int\limits_{0}^{\infty}\sin(\lambda(u^2+u^{-2}))du=\frac{1}{\sqrt{\lambda}}\int\limits_{0}^{\infty}\sin(\lambda((u-u^{-1})^2+2))du=\frac{1}{2\sqrt{\lambda}}\int\limits_{-\infty}^{\infty}\sin(\lambda((u-u^{-1})^2+2))du$

By Glasser's Master Theorem, this integral can be reduced to

$\frac{1}{2\sqrt{\lambda}}\int\limits_{-\infty}^{\infty}\sin(\lambda(u^2+2))du=\frac{1}{2\sqrt{\lambda}}\int\limits_{-\infty}^{\infty}\sin({\lambda}u^2+2\lambda))du=\frac{1}{\sqrt{\lambda}}\int\limits_{0}^{\infty}\sin({\lambda}u^2)\cos(2\lambda)+\cos({\lambda}u^2)\sin(2\lambda)du$

We then make a final substitution $t=\sqrt{\lambda}u$ to get

$\frac{1}{\lambda}\int\limits_{0}^{\infty}\sin(t^2)\cos(2\lambda)+\cos(t^2)\sin(2\lambda)du$

Using the Fresnel integral, which states that $\int\limits_{0}^{\infty}\sin(x^2)dx=\int\limits_{0}^{\infty}\cos(x^2)dx=\sqrt{\frac{\pi}{8}}$ , we can simplify this integral to

$\frac{\sqrt{\pi}}{2\lambda\sqrt{2}}(\sin(2\lambda)+\cos(2\lambda))=\boxed{\frac{\sqrt{\pi}}{2\lambda}\sin(2\lambda+\frac{\pi}{4})}$

Substitute $\displaystyle x$ by $\displaystyle \dfrac{1}{\lambda x}$ and let $\displaystyle I=\int_{0}^{\infty} \sin(\lambda^{2}x^{2}+\dfrac{1}{x^{2}}) dx$ we get,

$\displaystyle \lambda I=\int_{0}^{\infty} \sin(\lambda^{2}x^{2}+\dfrac{1}{x^{2}}) d\dfrac{-1}{x}$

So, by adding the 2 integrals above(after multiplying the first one by $\lambda$ ), we arrive at,

$\displaystyle 2\lambda I=\int_{0}^{\infty} \sin(\lambda^{2}x^{2}+\dfrac{1}{x^{2}}) d(\lambda x-\dfrac{1}{x})=\int_{0}^{\infty} \sin((\lambda x-\dfrac{1}{x})^{2}+2\lambda) d(\lambda x-\dfrac{1}{x})$

Let $\displaystyle x=\lambda x-\dfrac{1}{x}$ , we have

$\displaystyle 2 \lambda I=\int_{-\infty}^{\infty} \sin(x^{2}+2\lambda ) dx=\int_{-\infty}^{\infty} \sin(x^{2})\cos(2\lambda )+\cos(x^{2})\sin(2\lambda ) dx$

Now,

$\displaystyle \int_{-\infty}^{\infty} \sin(x^{2}) dx=\int_{-\infty}^{\infty} \cos(x^{2}) dx=\sqrt{\dfrac{\pi}{2}}$

$\displaystyle 2 \lambda I=\sqrt{\pi}\sin(2\lambda+\dfrac{\pi}{4})$

$\displaystyle I=\dfrac{\sqrt{\pi}}{2\lambda}\sin(2\lambda+\dfrac{\pi}{4})$

So, $\displaystyle A=\boxed{\dfrac{1}{2}}, B=C=\boxed{2}, D=\boxed{4}$

$\displaystyle A+B+C+D=0.5+2+2+4=\boxed{8.5}$