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When answering this related question I proved that if we define $B(\lambda)$ as:

$$\begin{eqnarray*} B(\lambda)&=&\frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{\lambda^2+1-2\lambda\cos\theta}\,d\theta=\sum_{n\geq 0}\left(\frac{\binom{2n}{n}}{(2n-1)\,4^n}\right)^2\lambda^{2n}\\&=&\phantom{}_2 F_1\left(-\frac{1}{2},-\frac{1}{2};1;\lambda^2\right)\end{eqnarray*}$$ we have: $$ \int_{0}^{1} \lambda\,B(\lambda)\,d\lambda = \frac{16}{9\pi} $$ since $B(\lambda)$ satisfies the ODE: $$ \lambda\, B = (\lambda^2+1)B'+(\lambda-\lambda^3)B''.$$ With the same technique it is not difficult to prove that, if we define $A_k$ as:

$$ A_k = \int_{0}^{1}x^{2k+1}\,B(x)\,dx $$

we have $A_0=\frac{16}{9\pi}$ and:

$$ (2k+3)^2\,A_k = \frac{16}{\pi}+(2k)^2\,A_{k-1}. \tag{0}$$

Now the question. Is it possible to give a nice closed form to $A_k$, given $(0)$?

$A_k$ appears to be related with a partial sum of the series converging to the Catalan constant, namely: $$\frac{1}{2\pi}\sum_{j=0}^{k+1}\frac{(-1)^j}{(2j+1)^2}.$$

From the recurrence relation it follows that: $$ A_k = \frac{16}{\pi(2k+3)^2}+\frac{16}{\pi}\left(\frac{(2k)!!}{(2k+3)!!}\right)^2\sum_{j=1}^{k}\left(\frac{(2k+3-2j)!!}{(2k-2j)!!}\right)^2$$ or: $$ A_k = \frac{16}{\pi(2k+3)^2}+\frac{16}{\pi}\left(\frac{(2k)!!}{(2k+3)!!}\right)^2\sum_{j=1}^{k}\left(\frac{(2j+1)!}{2^{2j-1}(j-1)! j!}\right)^2.$$ Update: By robjohn's argument shown here, we also have: $$ A_k = \frac{1}{\pi}\int_{0}^{\pi/2}\int_{-\cos\theta}^{\cos\theta}(\rho+\cos\theta)^2(\rho^2+\sin^2\theta)^k\,d\rho\,d\theta$$ or: $$ A_k = \frac{2}{\pi}\int_{0}^{\pi/2}\int_{0}^{\cos\theta}(\rho^2+\cos^2\theta)(\rho^2+\sin^2\theta)^k\,d\rho\,d\theta.$$

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