Rationalisation?

Consider the given integral:

$I = \int_{0}^{1} \frac{dx}{\sqrt{1+x}+\sqrt{1-x}+2}$

Let:

$x = \cos\left(2\theta\right)$ $\implies dx = -2\sin\left(2\theta\right) \ d\theta$

Moreover, when $x=0$ , $\theta = \pi/4$ and when $x=1$ , $\theta = 0$ . Applying this chage of variable transforms the integral to:

$I = \int_{\pi/4}^{0} \frac{-2\sin\left(2\theta\right) \ d\theta}{\sqrt{2}\cos{\theta} + \sqrt{2}\sin{\theta} + 2}$

Note that:

$\sqrt{2}\cos{\theta} + \sqrt{2}\sin{\theta} = 2\sin\left(\theta+\frac{\pi}{4}\right)$

$I = \int_{0}^{\pi/4} \frac{2\sin\left(2\theta\right) \ d\theta}{2\sin\left(\theta+\frac{\pi}{4}\right) + 2}$

$I = \int_{0}^{\pi/4} \frac{2\cos{\theta}\sin{\theta} \ d\theta}{\sin\left(\theta+\frac{\pi}{4}\right) + 1}$

Let $z = \theta+\frac{\pi}{4}$ . This change of variables transforms the integral to:

$I = \int_{\pi/4}^{\pi/2} \frac{2\cos\left(z-\frac{\pi}{4}\right)\sin\left(z-\frac{\pi}{4}\right) \ dz}{\sin{z} + 1}$

$I = \int_{\pi/4}^{\pi/2} \frac{\sin\left(2z-\frac{\pi}{2}\right) \ dz}{\sin{z} + 1}$

$I = \int_{\pi/4}^{\pi/2} \frac{-\sin\left(\frac{\pi}{2}-2z\right) \ dz}{\sin{z} + 1}$

$I = \int_{\pi/4}^{\pi/2} \frac{-\cos\left(2z\right) \ dz}{\sin{z} + 1}$

$I = -\int_{\pi/4}^{\pi/2} \frac{\left(1 - 2\sin^2{z}\right)\left(1 -\sin{z}\right) \ dz}{\cos^2{z}}$

Using trigonometric relations, this can be further simplified to:

$I = -\int_{\pi/4}^{\pi/2} \left(2 - \sec^2{z} - \sec{z}\tan{z} + \frac{2\sin^3{z}}{\cos^2{z}}\right)dz$

$I = \lim_{b \to \pi/2} \left(-\int_{\pi/4}^{b} \left(2 - \sec^2{z} - \sec{z}\tan{z} + \frac{2\sin^3{z}}{\cos^2{z}}\right)dz\right)$

I am leaving out intermediate steps from here and presenting the final answer.

$I = 2\sqrt{2}-1 -\frac{\pi}{2}$

Therefore:

$a = 2$ $b = 2$ $c = 2$ $d = -1$

Putting $u^2 = 1-x$ and then $u = \sqrt{2}\sin\theta$ makes the integral equal to $\begin{aligned} I & = \; \int_0^1 \frac{2u}{\sqrt{2-u^2} + u + 2}\,du \; = \; 2\int_0^1 \frac{u\big(u+2 - \sqrt{2-u^2}\big)}{(u+2)^2 - (2-u^2)}\,du \\ & = \; \int_0^1 \frac{u^2 + 2u - u\sqrt{2-u^2}}{(u+1)^2}\,du \; = \; \int_0^1 \left\{ 1 - \frac{1}{(u+1)^2} - \frac{u\sqrt{2-u^2}}{(u+1)^2}\right\}\,du \\ &= \; \frac12 - \int_0^1 \frac{u\sqrt{2-u^2}}{(u+1)^2}\,du \; = \;\frac12 - 2\sqrt{2}\int_0^{\frac14\pi} \frac{\sin \theta \cos^2\theta}{(\sqrt{2}\sin\theta + 1)^2}\,d\theta \end{aligned}$ Integrating by parts this last integral becomes $\begin{aligned} I & = \; \frac12 - 2\sqrt{2}\left[-\frac{\sin\theta \cos\theta}{\sqrt{2}(\sqrt{2}\sin\theta + 1)}\right]_0^{\frac14\pi} - 2\sqrt{2}\int_0^{\frac14\pi} \frac{1 - 2\sin^2\theta}{\sqrt{2}(\sqrt{2}\sin\theta + 1)}\,d\theta \\ &= \; 1 + 2\int_0^{\frac14\pi} \big(\sqrt{2}\sin\theta - 1\big)\,d\theta \; = \; -\tfrac12\pi + 2\sqrt{2} - 1 \end{aligned}$ making the answer $2+ 2 + 2 - 1 = \boxed{5}$ .