Four semi-circles are shown with $AB:BC:CD = 1:2:3$. What is the ratio of the shaded area to the unshaded area in the semi circle with diameter $AD$? Express your answer as a common fraction. [asy]
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));
filldraw(arc((6,0),6,0,180)--cycle);
filldraw(arc((3,0),3,0,180)--cycle,fillpen=white); filldraw(arc((8,0),2,0,180)--cycle,fillpen=white); filldraw(arc((11,0),1,0,180)--cycle,fillpen=white);
label("$A$",(12,0),S); label("$B$",(10,0),S); label("$C$",(6,0),S); label("$D$",(0,0),S);
[/asy]
Let the radius of the large semicircle be $6x$.  The diameter of the smallest semicircle is $\frac{1}{1+2+3} = \frac16$ of the diameter of the largest semicircle, so the radius of the smallest semicircle is $x$.  Similarly, the radius of the next smallest semicircle is $2x$, and the radius of the next semicircle is $3x$.  The unshaded area then is the sum of the areas of the three smallest semicircles: \[\frac12(x)^2\pi + \frac12 (2x)^2 \pi + \frac12(3x)^2\pi = \frac12(x^2 + 4x^2 + 9x^2)\pi = (7x^2)\pi.\] The largest semicircle has area $\frac12(6x)^2\pi = 18x^2\pi$, so the shaded area is \[18x^2\pi - 7x^2 \pi = 11x^2\pi.\] Therefore, the desired ratio is \[\frac{11x^2\pi}{7x^2\pi} = \frac{11}{7}.\]