Points $A$, $B$, $C$ lie on a line in that order with $AB=BC=2$. Two semicircles with diameters $\overline{AB}$ and $\overline{BC}$ are drawn on the same side of the line, and a circle with center $B$ and radius $\sqrt3$ is drawn. The area of the region inside the circle but outside the two semicircles can be written in the form $\dfrac{a\pi+b\sqrt c}{d}$, where $a,b,c,d$ are positive integers, $c$ is squarefree, and $a,b,d$ have no common factor greater than $1$. What is $a+b+c+d$?

1. The circle has area $\pi(\sqrt3)^2=3\pi$. The lower half of the circle is untouched by the semicircles, so we only need the part of each semicircle that lies inside the circle. By symmetry, it is enough to handle the semicircle on $BC$ (center $O=(3,0)$, radius $1$) and double the result.
2. Put $B=(0,0)$ temporarily, so $O=(1,0)$. The circles $x^2+y^2=3$ and $(x-1)^2+y^2=1$ meet where $2x-1=2$, i.e. at $P=\left(\frac32,\frac{\sqrt3}2\right)$ in the upper half-plane. Then $\angle PBO=30^\circ$ (since $\tan\angle PBO=\frac{\sqrt3/2}{3/2}=\frac1{\sqrt3}$) and $OP=1$, $\angle BOP=120^\circ$.
3. The part of the semicircle inside the big circle is the sector of the big circle with center $B$, radius $\sqrt3$ and angle $30^\circ$, together with the circular segment of the small circle cut off by chord $BP$: $$\frac12\cdot3\cdot\frac\pi6+\left(\frac12\cdot1^2\cdot\frac{2\pi}3-\frac12\cdot1\cdot1\cdot\sin120^\circ\right)=\frac\pi4+\frac\pi3-\frac{\sqrt3}4=\frac{7\pi}{12}-\frac{\sqrt3}4.$$
4. Doubling for the two semicircles gives $\frac{7\pi}6-\frac{\sqrt3}2$, so the shaded area is $$3\pi-\frac{7\pi}6+\frac{\sqrt3}2=\frac{11\pi}6+\frac{\sqrt3}2=\frac{11\pi+3\sqrt3}{6}\approx6.63.$$
5. Thus $a=11,b=3,c=3,d=6$ and $a+b+c+d=\boxed{23}$.