Chilly Cardano Curve

Using polar coordinates, we have $x = r \cos t , y = r \sin t$ . Substitute these in, you get,

$r^3 ( 6 \cos^3 t + 11 \cos^2 t \sin t + 6 \cos t \sin^2 t + \sin^3 t ) = r \cos t$

so that

$r^2 = \dfrac{ \cos t }{6 \cos^3 t + 11 \cos^2 t \sin t + 6 \cos t \sin^2 t + \sin^3 t}$

The area is given by

$A = \displaystyle 2 (\frac{1}{2} ) \int_0^{\frac{\pi}{2} } r^2 dt = \int_0^{\frac{\pi}{2}} \dfrac{ \cos t }{6 \cos^3 t + 11 \cos^2 t \sin t + 6 \cos t \sin^2 t + \sin^3 t} dt$

Multiply top and bottom by $\sec^3 t$

$A = \displaystyle \int_0^{\frac{\pi}{2}} \dfrac{ \sec^2 t }{6 + 11 \tan t + 6 \tan^2 t + \tan^3 t} dt$

Substitute $u = \tan t$ , then $du = \sec^2 t dt$ and the integral becomes

$A = \displaystyle \int_0^{\infty} \dfrac{du}{u^3 + 6 u^2 + 11 u + 6 }$

By partial fraction expansion we have $\dfrac{1}{u^3 + 6 u^2 + 11 u + 6 } = \dfrac{1}{2} \left( \dfrac{1}{u+1} - \dfrac{2}{u+2} + \dfrac{1}{u+3} \right)$

The indefinite integral of which is $\dfrac{1}{2} \ln \dfrac{ (u+1)(u+3) }{(u + 2)^2}$

Finally evaluating this between the limits yields

$A = 0 - \dfrac{1}{2} \ln \dfrac{3}{4} = \dfrac{1}{2} \ln \dfrac{4}{3}$

Therefore $A = 1 , B = 2 , C = 4 , D = 3$ and $ABCD = \boxed{24}$