Using some mathematical software like mathematica, one can find that
$$\int_0^1\frac{dx}{\sqrt{4x^5+1}}=\frac{3+\sqrt5}{2}\int_1^{+\infty}\frac{dx}{\sqrt{4x^5+1}}=0.84984011984360712818935\cdots. $$
The equality is accurate up to at least $100$ digits. This form looks like some kind of elliptic integral, but the polynomial inside square root is of degree $5$. So this should be some kind of period integral about the hyperelliptic algebraic plane curve $y^2=4x^5+1$, which has automorphism group $C_2\times C_5$ by the well known classification. But I have no idea why the point $x=1,y=\sqrt 5$ is special on this curve.
The "complete" integral $\int_0^\infty\frac{dx}{\sqrt{4x^5+1}}$ is easy to calculate using Gamma function. We get $\Gamma(3/10)\Gamma(6/5)/2^{2/5}\sqrt\pi$ by mathematica. And the "incomplete" part $\int_0^1\frac{dx}{\sqrt{4x^5+1}}$ equals to $_2F_1[1/5, 1/2, 6/5; -4]$. How to get a relation?
Similar questions about elliptic integrals:
integral $\int_0^a \frac{1}{\sqrt{1+x^6}} dx$ where $a=\frac{1}{\sqrt{\sqrt{3}-1}}$
Given algebraic $a$, find the closed form of $\int_0^a \dfrac{dx}{\sqrt{1-x^4}}$
Aside: Is there a general result for ratios of these incomplete Beta integrals?
