Integral $\int_0^1\frac{x^{42}}{\sqrt{x^4-x^2+1}}\operatorname d \!x$ Could you please help me with this integral?
$$\int_0^1\frac{x^{42}}{\sqrt{x^4-x^2+1}} \operatorname d \!x$$

Update: user153012 posted a result given by a computer that contains scary Appel function, and Cleo gave much simpler closed forms for powers $n=42,\,43$. I am looking for a way to prove those forms. I also would like to find a more general result that would work for arbitrary integer powers, not just $42$.

Odd case: The change of variables $x^2=t$ transforms the integral into
$$\mathcal{I}_{2n+1}=\int_0^1\frac{x^{2n+1}dx}{\sqrt{x^4-x^2+1}}=\frac12\int_0^1\frac{t^ndt}{\sqrt{t^2-t+1}}$$
Further change of variables $t=\frac12+\frac{\sqrt3}{4}\left(s-\frac1s\right)$ allows to write $t^2-t+1=\frac3{16}\left(s+\frac1s\right)^2$ and therefore gives an integral of a simple rational function of $s$:
$$\mathcal{I}_{2n+1}=\frac12\int_{1/\sqrt3}^{\sqrt3}\left[\frac12+\frac{\sqrt3}{4}\left(s-\frac1s\right)\right]^n\frac{ds}{s}.$$
Even case: To demystify the result of Cleo, let us introduce
$$\mathcal{K}_n=\mathcal{I}_{2n}=\int_0^1\frac{x^{2n}dx}{\sqrt{x^4-x^2+1}}=\frac12\int_0^1\frac{t^{n-\frac12}dt}{\sqrt{t^2-t+1}}.$$
Note that 
$$\mathcal{K}_{n+1}-\frac12\mathcal{K}_n=\frac12\int_0^1 t^{n-\frac12}d\left(\sqrt{t^2-t+1}\,\right)=\frac12-\left(n-\frac12\right)\left(\mathcal{K}_{n+1}-\mathcal{K}_{n}+\mathcal{K}_{n-1}\right),$$
where the second equality is obtained by integration by parts. This gives a recursion relation
$$\left(n+\frac12\right)\mathcal{K}_{n+1}=n\mathcal{K}_{n}-\left(n-\frac12\right)\mathcal{K}_{n-1}+\frac12,\qquad n\geq1.$$
It now suffices to show that 
\begin{align*}
\mathcal{K}_0&=\int_0^1\frac{dx}{\sqrt{x^4-x^2+1}}=\frac12\mathbf{K}\left(\frac{\sqrt3}{2}\right),\\
\mathcal{K}_1&=\int_0^1\frac{x^2dx}{\sqrt{x^4-x^2+1}}=
\frac12\mathbf{K}\left(\frac{\sqrt3}{2}\right)-\mathbf{E}\left(\frac{\sqrt3}{2}\right)+\frac12.
\end{align*}
