A logarithmic integral, generalization of a result of Shalev As many of you are already aware, I and Marco Cantarini are currently working on the applications of fractional operators to hypergeometric series, extending the class of $\phantom{}_{p+1} F_p$s whose closed form is provided by FL-expansions (like the ones appearing here, soon in Bollettino UMI). That said, any insightful contribution here might easily result in a collaboration, or at least a citation, in our next work. Fractional operators allow to state that a closed form for any of the following integrals
$$ \int_{0}^{1}\frac{-\log u}{\sqrt{1+6u+u^2}}\,du \tag{A} $$
$$ \int_{0}^{1}\frac{\operatorname{arctanh}(u)}{\sqrt{(1-u^2)(2-u^2)}}\,du\tag{B} $$
$$ \int_{0}^{+\infty}\frac{z}{\sqrt{3+\cosh z}}\,dz\tag{C} $$
$$ \int_{0}^{1}\frac{K(x)}{\sqrt{x}(2-x)}\,dx\tag{D} $$
(where $K(x)=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-x\sin^2\theta}}$) result in a closed form for many $\phantom{}_3 F_2$s with quarter-integer parameters. However, we have not been able to find a closed form for any of the previous integrals. It is relevant to point out that
$$\int_{0}^{1}\frac{-\log x}{\sqrt{x}\sqrt{1-x\sin^2\theta}}\,dx = \frac{4}{\sin\theta}\left[\theta\log(2\sin\theta)+\frac{1}{2}\operatorname{Im}\operatorname{Li}_2(e^{2i\theta})\right]$$
for any $\theta\in\left(0,\frac{\pi}{2}\right)$ thanks to Shalev/nospoon, and the coefficients of the Maclaurin series of $\frac{1}{\sqrt{1+6u+u^2}}$ are given by central Delannoy numbers, i.e. Legendre polynomials evaluated at $3$. I guess this is enough context, so:

How can we express $(A),(B),(C)$ or $(D)$ in terms of standard mathematical constants and values of the $\Gamma$ function and polylogarithms?

Small update/context expansion: if we attack $(D)$ through Taylor series, the problem boils down to finding the twisted hypergeometric series
$$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{\mathscr{H}_n}{2^n},\qquad \mathscr{H}_n=\sum_{k=0}^{n}\frac{1}{2k+1} \tag{E} $$
while in order to tackle $(C)$ through $\int_{0}^{+\infty}\frac{x\,dx}{(\cosh x)^m}$ it might be worth to exploit integral representations for the Riemann $\zeta$ function and Dirichlet L-function $L(\chi_4,s)$, like
$$ \int_{0}^{+\infty}\frac{x^s}{\cosh^2 x}\,dx = \zeta(s)\frac{2(2^s-2)\Gamma(s+1)}{4^s}, $$
$$ \int_{0}^{+\infty}\frac{x^s}{\cosh x}\,dx = 2\,\Gamma(s+1)\,L(\chi_4,s+1)$$
and integration by parts.
March 17th 2019 Update: I realize there was a typo in the original question. $(A)$ should have been
$$ \int_{0}^{1}\frac{-\log u}{\sqrt{\color{red}{u}(1+6u+u^2)}}\,du \tag{A} $$
but I am confident that Shalev's substitution $u=\frac{(1-t)}{t(1+t)}$ simplifies the structure of the integral in this case, too.

I'll use the ingenious method of user @FDP from 
the IntegralsAndSeries forum.
The idea is to make the following chain of substitutions:
$$ x = \frac{ 1 + \sqrt{2} \cos y }{ 1 - \sqrt{2} \cos y} $$
$$ z = \frac{y}{2} - \frac{\pi}{8} $$
$$ t = \tan z $$
Together with the following observations:
$$ \frac{ 1 + \sqrt{2} \cos y }{ 1 - \sqrt{2} \cos y} = \cot \left( \frac{y}{2}+\frac{\pi}{8}\right)\cot \left( \frac{y}{2}-\frac{\pi}{8}\right) \tag1 $$
$$ 1  + \sqrt{2} \cos y = 2 \sqrt{2} \cos \left( \frac{y}{2}+\frac{\pi}{8}\right)\cos \left( \frac{y}{2}-\frac{\pi}{8}\right) \tag2$$
$$ \sqrt{2} \cos z \cos \left( z + \frac{\pi}{4} \right) = \frac{ 1 - \tan z}{ 1 + \tan^2 z} \tag3$$
We find that 
$$\begin{align}
I=\int_0^1 \frac{- \ln x}{\sqrt{1+ 6 x +x^2}} \mathrm{d}x
\\&= \int_1^\infty \frac{\ln x}{x\sqrt{1+ 6 x +x^2}} \mathrm{d}x
\\&= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{\ln \left( \dfrac{ 1 + \sqrt{2} \cos y }{ 1 - \sqrt{2} \cos y}\right)}{1+ \sqrt{2} \cos y} \mathrm{d}y
\\&=  \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\ln \left( \cot \left( \frac{y}{2}+\frac{\pi}{8}\right)\cot \left( \frac{y}{2}-\frac{\pi}{8}\right)\right)}{2 \sqrt{2} \cos \left( \frac{y}{2}+\frac{\pi}{8}\right)\cos \left( \frac{y}{2}-\frac{\pi}{8}\right)}\mathrm{d}y
\\&= \int_0^{\frac{\pi}{8}} \frac{\ln  \left( \cot z \cot \left( z + \frac{\pi}{4} \right) \right)}{ \sqrt{2} \cos z \cos \left( z + \frac{\pi}{4} \right)} \mathrm{d}z
\\&= \int_0^{\frac{\pi}{8}} \frac{\ln  \left( \frac{1}{\tan z} \frac{1-\tan z}{1+\tan z} \right)}{1- \tan z} (1+ \tan^2 z) \mathrm{d}z
\\&= \int_0^{\sqrt{2}-1} \frac{ \ln \left( \frac{1}{t} \frac{1-t}{1+t} \right)}{1-t} \mathrm{d}t
\\&= \frac{\pi^2}{12} +\frac12 \ln(2+\sqrt{2}) \ln(2-\sqrt{2}) + \operatorname{Li}_2\left(\frac{1}{\sqrt{2}}\right)-\operatorname{Li}_2 \left( 2-\sqrt{2}\right).
\end{align}$$
Where we just finished off with basic polylog stuff.

Edit.
Now that I think of it, we could have just made the simple Euler substitution $$\sqrt{ x^2 + 6 x +1} = x +t$$ to get 
$$ I = \int_1^{2 \sqrt{2}-1} \frac{ \ln \left( \frac{1-t^2}{2 (t-3)} \right)}{t-3} \mathrm{d}t $$
And from here the integral can be reduced to polylogs with little difficulty.
