Since $\frac1\phi=\phi-1$, we get
$$
\begin{align}
\int_0^\infty\frac{\sqrt[\phi]{x}\,\arctan(x)}{\left(x^\phi+1\right)^2}\mathrm{d}x
&=\int_0^\infty\frac{x^\phi\arctan(x)}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{1}\\
&=\int_0^\infty\frac{x^\phi(\frac\pi2-\arctan(x))}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{2}
\end{align}
$$
Average $(1)$ and $(2)$ to get
$$
\begin{align}
\int_0^\infty\frac{\sqrt[\phi]{x}\,\arctan(x)}{\left(x^\phi+1\right)^2}\mathrm{d}x
&=\frac\pi4\int_0^\infty\frac{x^\phi}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{3}\\
&=\frac\pi{4\phi}\int_0^\infty\frac{x}{\left(x+1\right)^2}\frac{\mathrm{d}x}{x}\tag{4}\\
&=\frac\pi{4\phi}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $\frac1\phi=\phi-1$
$(2)$: Substitute $x\mapsto\frac1x$
$(3)$: Average $(1)$ and $(2)$
$(4)$: Substitute $x\mapsto x^{1/\phi}$
$(5)$: $\int_0^\infty\frac{\mathrm{d}x}{(x+1)^2}=\left[-\frac1{x+1}\right]_0^\infty=1$
(As no one posted an answer, summarizing what I learned from the comments.)
Following up on the reference given by user @WhatsUp in the comments on the question, the Wikipedia article on “Particular values of the gamma function”, has in its section on products, as the last item under “Other rational relations include”, the relation
$$\frac{\Gamma\left(\frac{1}{5}\right)^2}{\Gamma\left(\frac{1}{10}\right)\Gamma\left(\frac{3}{10}\right)} = \frac{\sqrt{1+\sqrt{5}}}{2^{\tfrac{7}{10}}\sqrt[4]{5}}$$
which when squared would give
$$\frac{\Gamma(1/5)^4}{\Gamma(1/10)^2\Gamma(3/10)^2} = \frac{1+\sqrt{5}}{2^{7/5}\sqrt{5}}$$
which is what the question asks for.
The reference given on Wikipedia for this relation, namely the paper “Expressions for values of the gamma function” by Raimundas Vidūnas (seems to be published in Kyushu Journal of Mathematics, 2005, Vol 59, pp 267–283; DOI), does not seem to have this relation exactly.
(It's an interesting question how the relation got onto Wikipedia then. Looking at the edit history of the Wikipedia page, it appears that in February 2018 someone added it to the Wikipedia page, citing Knuth's 2017 lecture, as I did in the question here. The edit was immediately undone less than an hour later. Then a week later the user “WorstUsernameEver” restored it as a conjecture, and the next day found that it can be obtained by multiplying together two formulae from Vidūnas's paper. As I was typing below...)
However, the paper contains (section 2, page 3), among others, the expressions:
$$Γ(1/10) = \frac{\sqrt \phi}{\sqrt\pi 2^{7/10}} Γ(1/5)\,Γ(2/5) \tag 1$$
and
$$Γ(3/10) = \frac{\sqrt \pi \phi^⋆}{2^{3/5}\sqrt5} Γ(1/5)\,Γ(2/5)^{−1} \tag 2$$
where $\phi = 5 + \sqrt5$ and $\phi^⋆ = 5 - \sqrt{5}$.
Multiplying these two together gives
$$Γ(1/10)Γ(3/10) = \frac{\sqrt{5+\sqrt{5}}(5-\sqrt{5})}{2^{13/10}\sqrt5}Γ(1/5)^2$$
and so
$$\frac{Γ(1/5)^2}{Γ(1/10)Γ(3/10)} = \frac{2^{13/10}\sqrt5}{\sqrt{5+\sqrt{5}}(5-\sqrt{5})} \stackrel{?}{=} \frac{\sqrt{1+\sqrt{5}}}{2^{7/10}\sqrt[4]{5}}$$
and the algebra seems to work out.
(TODO: Understand the paper enough to prove $(1)$ and $(2)$.)
Best Answer
Ok, so my answer works but requires you to:
Edit (D S) :
I noticed that getting a better approximation for $\ln 10$ significantly reduces the work required for $\ln(\phi)$. So here it goes: $$\ln(10) = \ln(2)+\ln(5) = 3\ln(2) + \ln(5/4)$$ $$\begin{align}\ln(2) &= \ln((1+1/3)/(1-1/3)) \\ &= 2\sum_{k=0}^\infty \frac{(1/3)^{2k+1}}{2k+1}\\ &< 2\left(\sum_{k=0}^3 \frac{(1/3)^{2k+1}}{2k+1}\right)+ 2\left(\sum_{k=4}^\infty \frac{(1/3)^{2k+1}}{9}\right) \\ &= 2\left(\sum_{k=0}^3 \frac{(1/3)^{2k+1}}{2k+1}\right)+\frac{(1/3)^9}{4} \\ &= \frac{1910051}{2755620} < 0.693148 \end{align}$$and $$\begin{align}\ln(5/4) &= \ln((1+1/9)/(1-1/9)) \\ &= 2\sum_{k=0}^\infty \frac{(1/9)^{2k+1}}{2k+1}\\ &< 2\left(\sum_{k=0}^2 \frac{(1/9)^{2k+1}}{2k+1}\right)+ 2\left(\sum_{k=3}^\infty \frac{(1/9)^{2k+1}}{7}\right) \\ &= 2\left(\sum_{k=0}^2 \frac{(1/9)^{2k+1}}{2k+1}\right)+ \frac{(1/9)^5}{280} \\ &= \frac{3689393}{16533720}<0.223144 \end{align}$$ Hence $\ln10 < 3\times 0.693148+0.223144 = \color{blue}{2.302588}$.
Let $x$ be such that $$x^{x^2} = 1000\phi$$ or$$x^2\ln(x) = 3\ln(10)+\ln(\phi)$$ We are interested in showing $$3\ln(10)+\ln(\phi)<e^2 \iff 3\ln(10)+\ln(\phi)- e^2<0$$since $x^2\ln(x)$ is increasing in $x$.
First, we approximate $\ln(\phi)$. Note that the series expansion of $\ln(1+x)$ for $|x|>1$ is given by $$\ln(1+x) = \ln(x) - \sum_{k=1}^\infty(-1)^k\frac{x^{-k}}{k}$$So that $$2\ln(\phi) = \ln(\phi^2) = \ln(1+\phi) = \ln(\phi) + \sum_{k=1}^\infty(-1)^{k+1}\frac{\phi^{-k}}{k}$$ $$\iff \ln(\phi) = \sum_{k=1}^\infty(-1)^{k+1}\frac{\phi^{-k}}{k}< \sum_{k=1}^{17}(-1)^{k+1}\frac{\phi^{-k}}{k}$$since the magnitude of the terms in the sum are decreasing, and we have stopped at a positive term, so this approximation is greater.
Now, to calculate negative powers of $\phi$, you can just extend the Fibonacci sequence backwards, i.e, $$\ldots \color{blue}{-8,5,-3,2,-1,1},0,1,1,2,3 \ldots$$ For example, $\phi^{-1} = \color{blue}1\cdot \phi \color{blue}{- 1}$, $\phi^{-2} = \color{blue}{-1}\cdot \phi +\color{blue}2$, $\phi ^{-3} = \color{blue}2\cdot \phi \color{blue}{- 3}$ and so on.
In the end, you should get $$\sum_{k=1}^{17}(-1)^{k+1}\frac{\phi^{-k}}{k} = \frac{3433663565\sqrt 5 - 7666113149}{24504480}<\color{red}{0.48123}$$
Edit:
With our new approximation of $\ln(10)$, it is enough to take the sum till 13, i.e, $$\ln(\phi) < \sum_{k=1}^{13}(-1)^{k+1}\frac{\phi^{-k}}{k} = \frac{4109827\sqrt 5- 9120481}{144144} < \color{blue}{0.481266}$$
Next, we will approximate $e^2$. $$e^2 = \left(\sum_{k=0}^\infty\frac{1}{k!}\right)^2>\left(\sum_{k=0}^8\frac{1}{k!}\right)^2$$ (We don't use series expansion of $e^z$ as that requires more terms.) $$\left(\sum_{k=0}^8\frac{1}{k!}\right)^2 = \left(\frac{109601}{40320}\right)^2>\color{red}{7.38903}$$
Finally, $$\boxed{3\ln(10)+\ln(\phi)- e^2< 3\cdot2.3026 + 0.48123- 7.38903 = 0 \ \ }$$ Phew!
Edit:
Armed with our new bounds, we have $$\boxed{3\ln(10)+\ln(\phi) - e^2<3\cdot 2.302588+0.481266-7.38903 = 0}$$
Update DesmosTutu :
We have a trick to evaluate $\ln 10$
$$\operatorname{arcsinh}(10)=\ln(20)+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2n-1)!!}{2n(2n)!!(10)^{2n}},\operatorname{arcsinh}(10)\simeq 3$$
Now use expansion of $\operatorname{arcsinh(x)}$ and Engels expansion of $\ln 2$
Source: Wikipedia