After substitution $x^4\to x,$ we get a 10-degree polynomial that factors over the rationals extended with $\sqrt{19}$ and a root of a solvable quintic
$$\!\!\!\!\!\!\!\!\!\!\small 322687697779 \,z^5-67934252164 \,z^4+5974826887 \,z^3-229104318 \,z^2+3132036 \,z+216=0,\tag{$\small\spadesuit$}$$
so your 40-degree root can be represented in radicals. Let
$$\small\begin{align}
\alpha &=30 \sqrt{198616747730+65513019062 \sqrt{5}},\\
\beta &=30 \sqrt{198616747730-65513019062 \sqrt{5}},\\
\eta &=\sqrt[5]{11410567+2790935 \sqrt{5}+\alpha\;}+\sqrt[5]{11410567+2790935 \sqrt{5}-\alpha\;}\\
&+\sqrt[5]{11410567-2790935 \sqrt{5}+\beta\;}-\sqrt[5]{2790935 \sqrt{5}-11410567+\beta\;},\,\text{and}\\
\gamma& =8-\left(\frac{2}{19}\right)^{4/5} \eta,
\end{align}$$
then
$$\small\begin{align}&\!\!\!\frac{\eta(19i)}{\eta(i)}\\
&=\frac{\sqrt[4]{100680000 +7361892000{\tiny\text{ }}\gamma+76992000 \sqrt{19}\,\gamma -1888138300{\tiny\text{ }}\gamma ^2+145028140{\tiny\text{ }}\gamma ^3-4533799{\tiny\text{ }}\gamma ^4}}{20\cdot 2^{3/4} \cdot19^{3/8}\cdot \sqrt[4]{1203\,}}.
\end{align}$$
I found the quintic $\small(\spadesuit)$ using my computer program that uses a combination of brute force and heuristics to find an extension of rationals in which a given polynomial can be factored.
Mathematica expression for the root:
With[{α = 30 Sqrt[198616747730 + 65513019062 √5],
β = 30 Sqrt[198616747730 - 65513019062 √5]},
With[{η = (11410567 + 2790935 √5 + α)^(1/5)
+ (11410567 + 2790935 √5 - α)^(1/5)
+ (11410567 - 2790935 √5 + β)^(1/5)
- (2790935 √5 - 11410567 + β)^(1/5)},
With[{γ = 8 - (2/19)^(4/5) η},
(100680000 + 7361892000 γ + 76992000 √19 γ
- 1888138300 γ^2 + 145028140 γ^3 - 4533799 γ^4)^(1/4)
/(20 2^(3/4) 19^(3/8) 1203^(1/4))]]]