<s>[INST] Tricky Integral -- $\int_0^1 \sqrt{x^2-4x+3} \arcsin(x)~dx$ TL;DR: I can't get a closed form for the integral below.
$$ \int_0^1 \sqrt{x^2-4x+3} \arcsin(x)~dx $$

I got an interesting question from a coworker a while ago:

Question:
The quantities $a$, $b$, and $c$ are chosen uniformly and independently from $[0, 1]$.
a) What is the probability a triangle can be constructed with $a$, $b$, and $c$ as side lengths?
b) Given we can form such a triangle, what is its expected area?

I can do a) pretty easily -- each constraint like $a < b + c$ cuts off a corner of the cube with area $1/6$, and the cut-off bits are disjoint, so the remaining area is $1/2$.
Part b) is where things get hairy. I can reduce the problem down to a single integral. I feel like it should be expressible in terms of known constants, though I admit I have no good reason to believe this.
$$
\frac{3}{40} \int_0^1 x \sqrt{3-4x+x^2} \left( \sqrt{1 - x^2} + \frac{\arcsin{x}}{x} \right)~dx
$$
That can be split into two parts:
$$
\frac{3}{40} \int_0^1 x \sqrt{(3-4x+x^2)(1 - x^2)}~dx + \frac{3}{40} \int_0^1 \sqrt{3-4x+x^2} \arcsin(x)~dx
$$
The first part can be solved exactly.
$$
\begin{align*}
\int_0^1 x \sqrt{(3-4x+x^2)(1 - x^2)}~dx &= \int_0^1 x \sqrt{(3-x)(1-x)(1-x)(1+x)}~dx \\
&= \int_0^1 x(1-x) \sqrt{(3-x)(1+x)}~dx \\
&= \frac{1}{12} (32 - 9 \sqrt{3} - 4\pi) \textrm{ by Mathematica}
\end{align*}
$$
The second part is still pretty stubborn.
Mathematica tells me the integral (without the 3/40 constant) is approximately 0.452854, but doesn't given an exact form. Does anyone have any ideas how to evaluate this further?
 [/INST] This is not an answer.
We could use
$$\sqrt{x^2-4x+3}=\sum_{n=0}^\infty a_n\, x^n$$ with
$$a_n=\frac{2(2 n-3)\, a_{n-1}-(n-3)\, a_{n-2}}{3 n} \qquad \text{where}\qquad a_0=\sqrt{3}\qquad a_1=-\frac{2}{\sqrt{3}}$$ and
$$\int_0^1 x^n\arcsin(x)\,dx=\frac{\pi }{2( n+1)}-\frac{\sqrt{\pi }\,\,\Gamma \left(\frac{n}{2}+1\right)}{(n+1)^2
  \,\, \Gamma \left(\frac{n+1}{2}\right)}$$ but the convergence is very slow.
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