On $\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\frac{n}{6^6}\,x\right)\,dx$ Reshetnikov gave the remarkable evaluation,
\begin{align}
I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\,dx \\
&=\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha^5-\alpha^4+1\right)\arctan\frac1\alpha-6\ln\left(1+\alpha^2\right)\right)\\
&=0.7857194\dots
\end{align}
where $\alpha$ is a quartic root. However, it seems this can be simplified a bit and generalized.

I. $p=5$: Given, 

$$I(n)=\int_0^1\arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{n}{4^4}\,x\right)\,\mathrm dx$$
Is it true that, in general,
$$\frac{12n}{5^5}\, I(n) = -12\Big(1\color{blue}-\frac{n}{5^5}\Big)\arctan\frac1\alpha+6\ln\Big(\frac{2}{1+\alpha^2}\Big)+3\pi+P(\alpha)$$
where $P(\alpha)=(1-\alpha)^3(5+3\alpha)$ and $\alpha$ is the largest positive root of the quintic,
$$\alpha^5-\alpha^4+\frac{n}{5^5}=0$$
provided real number $n<4^4\,$? (Note: By sheer coincidence, the choice of $n=4$ in the other post caused the quintic to factor.)

II. $p=7$: Given, 

$$J(n)=\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\,\frac{n}{6^6}\,x\right)\,\mathrm dx$$
is it true that,
$$\frac{60n}{7^7}\, J(n) = 60\Big(1\color{blue}+\frac{n}{7^7}\Big)\arctan\frac1\alpha-30\ln\Big(\frac{2}{1+\alpha^2}\Big)-15\pi-P(\beta)$$
where $P(\beta) = (1-\beta)^2(27-6\beta-9\beta^2+8\beta^3+10\beta^4)$ and $\beta$ is the largest positive root of,
$$\beta^7-\beta^6+\frac{n}{7^7}=0$$
provided real $n<6^6\,$?

Questions: 



*

*How do we prove the two conjectured equalities?

*What's the formula for $p=3$? (Mathematica takes too long to evaluate the integral that I couldn't use an integer relations subroutine on it.)

 A: I think you should start by the following general form 

$${}_{k}F_{k-1}\left(\frac{1}{k+1} ,\cdots ,\frac{k}{k+1};\frac{2}{k}
 \cdots ,\frac{k-1}{k},\frac{k+1}{k};\left(
 \frac{m(1-m^k)}{f_k}\right)^k \right) = \frac{1}{1-m^k}$$
Where 
$$f_k \equiv \frac{k}{(1+k)^{1+1/k}}$$

Put $k=4$
$${}_{4}F_{3}\left(\frac{1}{5} ,\cdots ,\frac{4}{5};\frac{2}{4}
 \cdots ,\frac{3}{4},\frac{5}{4};\left(
 \frac{m(1-m^4)}{f_4}\right)^4 \right) = \frac{1}{1-m^4}$$
The argument simplifies to 
$$ \left(\frac{m(1-m^4)}{\frac{4}{5^{1+1/4}}}\right)^4  = \frac{5^5 \,m^4(1-m^4)^4}{4^4}$$
Hence we have 

$${}_{4}F_{3}\left(\frac{1}{5} ,\cdots ,\frac{4}{5};\frac{2}{4} 
 \cdots ,\frac{3}{4},\frac{5}{4};\frac{5^5 (1-m)m^4}{4^4} \right) =
 \frac{1}{m}$$

Now suppose we want to find 
$$\int \arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{n}{4^4}\,x\right)\,\mathrm dx$$
Use $nx = 5^5m^4(1-m)$
$$\frac{5^5}{n}\int (4m^3(1-m)-m^ 4)\arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{5^5 (1-m)m^4}{4^4}\right)\,\mathrm dm$$
This simplifies to 
$$\frac{5^5}{n}\int (4m^3(1-m)-m^ 4)\arctan\,\frac{1}{m}\,\mathrm dm$$
The anti-derivative of this is elementary. 
Note that when we use substittution we will need the roots of 
$$m^5-m^4+\frac{n}{5^5}= 0$$
As conjectured by the OP.
Using the value $n=4$ we can verify Reshetnikov result

\begin{align}\int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\,dx
 &= \frac{5^5}{4}\int^{\alpha}_1 (4x^3(1-x)-x^ 4)\arctan
 \left(\frac{1}{x} \right)\,\mathrm dx\\&=0.7857194\dots \end{align}

Where $\alpha$ is the largest positive root of 
$$m^5-m^4+\frac{4}{5^5}= 0$$
