Average area bound by tangent lines

We start by finding a formula for the tangent line, say $l$ , to the curve through $P$ . Using our knowledge about tangent lines, we find that $l: y = -\frac{x}{p^2}+\frac{2}{p}.$ By symmetry in each quadrant (rotating the first quadrant about 90 degrees will give you the second quadrant for example), we see that tangent lines to the curve to the rotated points will be either parallel, or perpendicular to the tangent line. If we were to search for the tangent line, say $k$ , to the point $P$ rotated around the origin by $\pi/2$ radians, then we will end up with $k: y = p^2\cdot x + 2p.$ We now want to find the intersection between those two tangent lines. Solving the system of equations containing the lines $l$ and $k$ results in the point $Q\left(-2\cdot\frac{p^3-p}{p^4+1},\ 2\cdot\frac{p^3+p}{p^4+1}\right).$ What we need to know is the length of the linesegment from O to Q (O here is the origin). We find that $d(O, Q) = \sqrt{4\cdot\frac{(p^3-p)^2+(p^3+p)^2}{(p^4+1)^2}} = \frac{2\sqrt{2(p^6-p^2)}}{p^4+1} = \frac{2p\sqrt{2}}{\sqrt{p^4+1}}.$ If we look at the right picture, we will see that $A(p) = 2(d(O,Q))^2 = \frac{16p^2}{p^4+1}.$ We want to find $\overline{A} = \frac{1}{1-0} \int_0^1 A(p) dp = 16 \cdot \int_0^1 \frac{p^2}{p^4+1} dp.$ Solving this integral requires some work. We can factor $p^4+1$ by $p^2+p\sqrt{2}+1$ and $p^2-p\sqrt{2}+1$ , after which we apply partial fractions decomposition. $I = \int_0^1 \frac{p^2}{p^4+1} dp = \int_0^1 \frac{p^2}{(p^2+p\sqrt{2}+1)(p^2-p\sqrt{2}+1)} dp = \int_0^1 \frac{p}{2\sqrt{2}(p^2-p\sqrt{2}+1)}-\frac{p}{2\sqrt{2}(p^2+p\sqrt{2}+1)} dp \implies$ $I = \int_0^1 \frac{2p-\sqrt{2}+\sqrt{2}}{4\sqrt{2}(p^2-p\sqrt{2}+1)}-\frac{2p+\sqrt{2}-\sqrt{2}}{4\sqrt{2}(p^2+p\sqrt{2}+1)} dp = \left.\frac{1}{4\sqrt{2}}\ln\left(\frac{p^2-p\sqrt{2}+1}{p^2+p\sqrt{2}+1}\right)\right|_0^1 + \int_0^1 \frac{\sqrt{2}}{4\sqrt{2}(p^2-p\sqrt{2}+1)} + \frac{\sqrt{2}}{4\sqrt{2}(p^2+p\sqrt{2}+1)} dp \implies$ $I = \frac{1}{4\sqrt{2}}\ln\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right) +\frac{1}{4}\int_0^1 \frac{1}{p^2-p\sqrt{2}+1} + \frac{1}{p^2+p\sqrt{2}+1} dp = \frac{1}{4\sqrt{2}}\ln\left(\frac{6-4\sqrt{2}}{2}\right) + \frac{1}{4}\int_0^1 \frac{1}{(p-\frac{1}{\sqrt{2}})^2+\frac{1}{2}}dp + \frac{1}{4}\int_0^1 \frac{1}{(p+\frac{1}{\sqrt{2}})^2+\frac{1}{2}} dp.$ Substitute $p = \frac{u+1}{\sqrt{2}}$ in the left integrand and $p = \frac{v-1}{\sqrt{2}}$ in the right integrand, then $dp = du/\sqrt{2}$ and $dp = dv/\sqrt{2}$ . So $I = \frac{1}{4\sqrt{2}}\ln(3-2\sqrt{2}) + \frac{1}{4}\int_{u(0)}^{u(1)} \frac{\sqrt{2}}{u^2+1} du + \frac{1}{4}\int_{v(0)}^{v(1)}\frac{\sqrt{2}}{v^2+1} dv = \frac{1}{4\sqrt{2}}\ln(3-2\sqrt{2}) + \left.\frac{1}{2\sqrt{2}}\arctan(p\sqrt{2}-1)\right|_0^1 + \left.\frac{1}{2\sqrt{2}}\arctan(p\sqrt{2}+1)\right|_0^1 \implies$ $I= \frac{1}{4\sqrt{2}}\ln(3-2\sqrt{2}) + \frac{1}{2\sqrt{2}}(\arctan(\sqrt{2}-1)-\arctan(-1)+\arctan(1+\sqrt{2})-\arctan(1)).$ Notice that $\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$ , so $\arctan(\sqrt{2}+1)+\arctan(\sqrt{2}-1)=\pi/2$ . Furthermore, $\arctan(\cdot)$ is an odd function, so we see that $I = \frac{1}{4\sqrt{2}}\ln(3-2\sqrt{2}) + \frac{\pi}{4\sqrt{2}}.$ We can now conclude that $\overline{A} = 16 \cdot \frac{1}{4\sqrt{2}}\ln(3-2\sqrt{2}) + \frac{\pi}{4\sqrt{2}} = 2\sqrt{2}(\pi + \ln(3-2\sqrt{2}))$ , and thus $a+b+c+d+e=2+2+3-2+2=\boxed{7\,}.$