Simplify $7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$

We will use a well-known$^{[1]}$ formula for sum of arctangents: $$\arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) \pmod \pi\tag1$$ The exact equality holds for $uv<1$, for other values there is additional term — an integer multiple of $\pi$ (it is easy to determine in each case).

Using this formula, we can establish the following identities: $$\begin{align} \arctan\varphi\phantom{^1}&=\frac\pi2-\frac12\,\arctan2,\tag2\\ \arctan\varphi^3&=\frac\pi4+\frac12\,\arctan2,\tag3\\ \arctan\varphi^5&=\pi-\frac32\,\arctan2.\tag4 \end{align}$$ Plugging these into the original expression in the question and expanding parentheses, we can see that all $\arctan2$ terms cancel, and we get the result $$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5=\frac{7\pi^2}8.\tag5$$

The answer of @Vladimir Reshetnikov: is hard to improve on, so we'll try to answer some of the questions posed in the comments.

If $\phi=\frac{1+ \sqrt{5}}{2}$ is the positive root of $x^2 - x-1=0$ then \begin{eqnarray} \tan(2 \arctan(\phi^n)) = \frac{2\phi^n}{ 1- \phi^{2n}} = \begin{cases} -\frac{2}{L_{n}} \text{ if $n$ odd} \\ -\frac{2}{F_{n}\sqrt{5}} \text{ if $n$ even } \end{cases} \end{eqnarray}

where $(F_n)$ is the Fibonacci sequence and $(L_n)$ is the Lucas sequence.

Indeed, for every $v$ we have $$(1+ iv)^2 = 1- v^2 + 2 i v= -v\cdot ( v -\frac{1}{v} - 2 i)$$ Now consider $v = \phi^n$. We get $$(1+ i \phi^n)^2 = - \phi^n\cdot ( \phi^n - \phi^{-n} - 2 i)$$ Note that $\phi$, $-\frac{1}{\phi}$ are the roots of $x^2 - x -1=0$. Therefore, if $n$ odd then $\phi^n - \phi^{-n}= \phi^n + \left(-\frac{1}{\phi}\right)^n=L_n$, while if $n$ even then $\phi^n - \phi^{-n} = \phi^n - \left(-\frac{1}{\phi}\right)^n= F_n \sqrt{5}$ .

Assume $n$ is odd. In order to write $2\arctan(\phi^n)$ as a combination of several $\arctan$'s of rational numbers, we decompose the number $L_n - 2 i$ in the Gaussian integers. It appears that for $n >5$ the number $L_n - 2 i$ keeps involving some distinct Gaussian primes from the previous ones.

Assume now $n$ is even. We need to decompose the number $2 + F_{n} \sqrt{-5}$ into some products of elements. Note that in the ring $\mathbb{Z}[\sqrt{-5}]$ the decomposition into irreducibles is not unique, so we have some problems.

As a preliminary result,

$$\frac{(1+i\phi)^7(1+i\phi^3)^2}{(1+i\phi^5)}=(-5+10i)(7+3\sqrt{5})\tag{1}$$ gives, switching to arguments, $$ 7\arctan\phi+2\arctan\phi^3-\arctan\phi^5 = 3\pi-\arctan 2. \tag{2} $$ In the same way, $$ 7\arctan\phi^{-1}+2\arctan\phi^{-3}-\arctan\phi^{-5} = \pi+\arctan 2. \tag{3} $$

In order to deal with the squared arctangent with Parseval-like techniques, it may be useful to recall that the Fourier transform of the arctangent function is given by $i\sqrt{\frac{\pi}{2}}\frac{e^{-|s|}}{s}$. Keep working.