Can $\Gamma(1/5)$ be written in this form? tldr
Feel free to skip to the integrals at the end.

I was curious as to whether or not for $n\in\mathbb N$, $\Gamma(1/n)$ satisfies the form
$$\Gamma(1/n)=a\pi^{b/n}\prod_k\operatorname{agm}(1,c_k)^{d_k}$$
Where $a,c_k,d_k$ are algebraic and $b$ is natural and $\operatorname{agm}$ is the arithmetic-geometric mean given by
$$\operatorname{agm}(x,y)=\lim_{n\to\infty}a_n\\a_0=x,~g_0=y\\a_{n+1}=\frac{a_n+g_n}2,\quad g_{n+1}=\sqrt{a_ng_n}$$
The first few values can be derived from this PDF.
$$\Gamma(1/1)=1$$
$$\Gamma(1/2)=\pi^{1/2}$$
$$\Gamma(1/3)=\frac{\pi^{2/3}2^{2/3}}{3^{1/12}\operatorname{agm}(1,3^{1/4}2^{-1/2})}$$
$$\Gamma(1/4)=\frac{2^{1/2}\pi^{3/4}}{\operatorname{agm}(1,2^{-1/2})}$$

However, it does not appear $\Gamma(1/5)$ can be written this way.  The PDF gives
$$\Gamma(1/5)=\pi^{1/5}2^{19/50}5^{1/2}\phi^{1/10}H_1^{2/5}H_2^{1/5}$$
$$\phi=5+\sqrt5\\H_1=\int_0^1\frac{dx}{\sqrt{1-x^5}}\\H_2=\int_0^1\frac{x~dx}{\sqrt{1-x^5}}$$
And I'm uncertain about whether or not $H_1$ or $H_2$ can be written in the form of
$$\frac\pi{2\operatorname{agm}(1,\sqrt{1-x})}=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-xt^2)}},~x>0$$


Can $\Gamma(1/5)$ be written in terms of this arithmetic-geometric mean?

 A: This is not an answer. I still decided to post, because this is too long for a comment. For what it's worth I would bet on a negative answer, i.e. we can't represent $H_1$ and $H_2$ in terms of $\text{agm}$.
I tried to link the integrals from the OP to Carlson Elliptic Integrals which are more general than the classic elliptic integrals and have a very nice computation algorithm (which seems to be what the OP wants).
However, the integrals are much more complicated than that.
With simple enough transformations, I was able to express $H_2$ in the following form:
$$H_2=4 \int_0^\infty \frac{dt}{\sqrt{t(t+2)(t^4+3t^3+4t^2+2t+1)(t^4+5t^3+10t^2+10t+5)}}$$
$$H_2=4 \int_0^\infty \frac{t^{-1/2}dt}{\sqrt{(t+z_1)(t+z_2)\dots(t+z_9)}}$$
Which translates to the general function $R_{-a}(\mathbf{b},\mathbf{z})$ - see here.

$$H_2=\frac{2^7}{35} R_{-4} \left(\mathbf{\frac{1}{2}},\mathbf{z} \right) $$

Where $\mathbf{\frac{1}{2}},\mathbf{z}$ are vectors of dimension $9$.
Since the most complicated of Carlson integrals only has $5$ factors under the root, there doesn't seem to be a way to represent $H_2$ in terms of Carlson integrals, or indeed, classic elliptic integrals.
Still, this representation might be useful if one wishes to study the properties of $R_{-a}(\mathbf{b},\mathbf{z})$ in general, which are rather nice.

Explicitly $z_k$ can be written the following way:
$$z_1=2$$
Now introducing:
$$a_+=\sqrt{5}+i \sqrt{2(5-\sqrt{5})}$$
$$a_-=\sqrt{5}-i \sqrt{2(5-\sqrt{5})}$$
$$b_+=\sqrt{5}+i \sqrt{2(5+\sqrt{5})}$$
$$b_-=\sqrt{5}-i \sqrt{2(5+\sqrt{5})}$$
We can write the rest as:
$$z_2=\frac{1}{4}(3-a_-)$$
$$z_3=\frac{1}{4}(3-a_+)$$
$$z_4=\frac{1}{4}(3+b_-)$$
$$z_5=\frac{1}{4}(3+b_+)$$
$$ $$
$$z_6=\frac{1}{4}(5+a_-)$$
$$z_7=\frac{1}{4}(5+a_+)$$
$$z_8=\frac{1}{4}(5-b_-)$$
$$z_9=\frac{1}{4}(5-b_+)$$

Of course, $H_2$ can also be expressed in a much more simple form as a Hypergeometric function:
$$H_2=\frac{1}{2} {_2F_1} \left(\frac{1}{2},\frac{2}{5}; \frac{7}{5};1 \right)$$

Meanwhile, $H_1$ can be represented as:
$$H_1=\int_0^\infty \frac{(t+1)dt}{\sqrt{t(t+1)(t^4+5t^3+10t^2+10t+5)}}$$

$$H_1=\frac{2}{3} R_{-1} \left(\mathbf{\frac{1}{2}},\mathbf{z} \right)+\frac{4}{3} R_{-2} \left(\mathbf{\frac{1}{2}},\mathbf{z} \right) $$

Where $$z_1=1$$
$$z_2=\frac{1}{4}(5+a_-)$$
$$z_3=\frac{1}{4}(5+a_+)$$
$$z_4=\frac{1}{4}(5-b_-)$$
$$z_5=\frac{1}{4}(5-b_+)$$
Even though it's simpler than $H_2$, it still doesn't reduce to Carlson elliptic integrals.

We can derive alternative representations for $H_1$ and $H_2$ in terms of $R$ functions, using the relation:
$${_2F_1}(a,b;c;z)=R_{-a}(b,c-b;1-z,1)$$
However, even these, shorter representations do not reduce to Carlson integrals, but rather to their values in terms of Gamma functions.
