This is going to be a long answer.
To start, you can use the polygamma function of order $0$ (also known as the digamma function) to shorten the expressions a little bit. This is tricky though - the series definition
$$\psi^{[n]}(z)=(-1)^{n+1}n!\sum_{k=0}^\infty \frac{1}{(z+k)^{n+1}}$$
is only valid for $n>0$, whereas for $n=0$ one must use the derivative definition
$$\psi^{[0]}(z)=\frac{\mathrm{d}}{\mathrm{d}z}\ln(\Gamma(z))=\frac{\Gamma'(z)}{\Gamma(z)}$$
Because for $n=0$ the series definition doesn't converge.
Despite this however, you can "abuse" the series definition for finite sums and write
$$\sum_{k=1}^n \frac{1}{ak+b}=\frac{1}{a}\left(\psi^{[0]}\left(\frac{b}{a}+n+1\right)-\psi^{[0]}\left(\frac{a+b}{a}\right)\right)$$
This is difficult, but not impossible to prove. We can then write our sum as
$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sum_{k=1}^{n}\frac{1}{4k-1}=\sum_{n=1}^{\infty} \left[\frac{(-1)^n}{n}\frac{1}{4}\left(\psi^{[0]}\left(n+\frac{3}{4}\right)-\psi^{[0]}\left(\frac{3}{4}\right)\right)\right]$$
In order to calculate $\psi^{[0]}(3/4)$ we can use the two formulas
$$\psi^{[0]}(1-z)-\psi^{[0]}(z)=\pi\cot(\pi z)$$
$$\psi^{[0]}(2z)=\frac{1}{2}\psi^{[0]}(z)+\frac{1}{2}\psi^{[0]}\left(z+\frac{1}{2}\right)+\ln 2$$
And plug in $z=1/4$ to obtain a linear system
$$\begin{bmatrix}
1 & -1\\
1/2 & 1/2
\end{bmatrix}\begin{bmatrix}
\psi ^{[ 0]}( 3/4)\\
\psi ^{[ 0]}( 1/4)
\end{bmatrix} =\begin{bmatrix}
\pi \cot( \pi /4)\\
\psi ^{[ 0]}( 1/2) -\ln 2
\end{bmatrix}$$
If we use the well known identity $\psi^{[0]}(1/2)=-\gamma-2\ln 2$ ($\gamma$ being the Euler-Mascheroni constant) we can solve the system to obtain
$$\psi^{[0]}(1/4)=-\frac{\pi}{2}-\gamma-\ln(8)$$
$$\psi^{[0]}(3/4)=\frac{\pi}{2}-\gamma-\ln(8)$$
Now,
$$\mathcal{S}=\frac{1}{4}\sum_{n=1}^\infty\frac{(-1)^n\psi^{[0]}\left(n+\frac{3}{4}\right)}{n}+\frac{1}{4}\left(\gamma+\ln(8)-\frac{\pi}{2}\right)\sum_{n=1}^\infty\frac{(-1)^n}{n}$$
The latter is a well known sum and is equal to $-\ln(2)$. As for the former, perhaps you can use the asymptotic expansion of the digamma function
$$\psi^{[0]}(z)\asymp \ln(z)-\frac{1}{2z}-\sum_{n=1}^\infty\frac{B_{2n}}{2nz^{2n}}$$
$B_k$ being the $k$th Bernoulli number. The first few terms are
$$\psi^{[0]}(z)\approx \ln(z)-\frac{1}{2z}-\frac{1}{12z^2}+\frac{1}{120z^4}-\frac{1}{252z^6}+\frac{1}{240z^8}+...$$
Hopefully $\psi^{[0]}(z)\approx \ln(z)-\frac{1}{2z}-\frac{1}{12z^2}$ will already give us a reasonable approximation. So now,
$$\mathcal{S}\approx \frac{1}{4}\underbrace{\sum_{n=1}^{\infty}\frac{(-1)^n\ln\left(n+\frac{3}{4}\right)}{n}}_{S_1}-\frac{1}{8}\underbrace{\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+3/4)}}_{S_2}-\frac{1}{48}\underbrace{\sum_{n=1}^\infty\frac{(-1)^n}{n(n+3/4)^2}}_{S_3}-\frac{\ln(2)}{4}\left(\gamma+\ln(8)-\frac{\pi}{2}\right)$$
With some work, the second sum can be decomposed into partial fractions to obtain
$$S_2=\sum_{n=1}^\infty\frac{(-1)^n}{n(n+3/4)}=4\left(\frac{4}{9}-\frac{\pi}{3\sqrt{2}}-\frac{\ln(2)}{3}+\frac{\sqrt{2}}{3}\ln(1+\sqrt{2})\right)$$
Or perhaps one could use the properties of the Lerch transcendent:
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^2+an}=\frac{\Phi(-1,1,a+1)-\ln(2)}{a}$$
To compute the Lerch transcendent one can use an integral identity
$$\Phi(z,s,\alpha)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}e^{-ax}}{1-ze^{-x}}\mathrm{d}x ~~~|~~~\operatorname{Re}(s),\operatorname{Re}(\alpha)>0~;~z\in\mathbb{C}~\backslash~ [1,\infty)$$
I'm not the greatest at integration, but Mathematica produces
$$\Phi\left(-1,1,\frac{7}{4}\right)=\int_0^\infty\frac{e^{-7x/4}}{1+e^{-x}}$$
$$=\frac{4}{3}-2(-1)^{1/4}\arctan((-1)^{1/4})+2(-1)^{1/4}\operatorname{arctanh}((-1)^{1/4})$$
Which, by using the complex definitions of arctan, arctanh and ln, one can arrive at the form we got before.
The first sum, $S_1$, is the most bothersome. Not only does it not have any reasonable closed form representations, but it converges quite slowly. So, I'm going to use an Euler Transform to accelerate the convergence of the series. For an alternating series we can use the transform
$$\sum_{n=0}^\infty (-1)^n a_n=\sum_{n=0}^\infty (-1)^n\frac{\Delta^n a_0}{2^{n+1}}$$
Using the forward difference operator:
$$\Delta^n a_0=\sum_{k=0}^n(-1)^k~{}_{n}\mathrm{C}_k ~a_{n-k}$$
First, an index shift:
$$S_1=\sum_{n=1}^\infty\frac{(-1)^n\ln\left(n+\frac{3}{4}\right)}{n}=-\sum_{n=0}^\infty\frac{(-1)^n\ln\left(n+\frac{7}{4}\right)}{n+1}$$
Let $a_n=\ln(n+7/4)/(n+1).$ Euler's transform tells us that
$$S_1=-\lim_{N\to\infty}\sum_{n=0}^N\left[\frac{(-1)^n}{2^{n+1}}\left(\sum_{k=0}^n(-1)^k~{}_n\mathrm{C}_k\frac{\ln\left(n-k+\frac{7}{4}\right)}{n-k+1}\right)\right]$$
Which converges to 5 decimal precision with only $N=11$. With $N=35$ it converges to 12 decimal precision. See my implementation on Desmos. So approximately speaking
$$S_1\approx −0.288525102601$$
Now for the third sum. According to Mathematica, it does actually have a "closed form", but it's pretty horrific. I can't be bothered to typeset it all, so I'll just post a screenshot.
It makes use of the Hurwitz zeta function. Anyway, the numerical value is
$$S_3\approx -0.276850451954$$
So, finally,
$$\mathcal{S}\approx \frac{−0.288525102601}{4}+\frac{0.276850451954}{48}-\frac{1}{2}\left(\frac{4}{9}-\frac{\pi}{3\sqrt{2}}-\frac{\ln(2)}{3}+\frac{\sqrt{2}\ln(1+\sqrt{2})}{3}\right)-\frac{\ln(2)}{4}\left(\gamma+\ln(8)-\frac{\pi}{2}\right)\approx -0.198728103723$$
You might ask yourself: why did we do all this work? The answer: speed of convergence. If we look at the partial sums of the original:
$$\mathcal{S}_N=\sum_{n=1}^N\left[\frac{(-1)^n}{n}\sum_{k=1}^n\frac{1}{4k-1}\right]$$
It converges very poorly. See my implementation on Desmos. Even at $N=40$, with respect to the approximate sum that I found, the partial sums of the above jump around with a relative error of $\mathbf{15\%}$ (!) So yes, our work wasn't all pointless :)
Note
$$\frac1{x^4-1} =\frac1{2(x^2-1)} - \frac1{2(x^2+1)}\\
\frac1{x^2(x^4-1)} =\frac1{2(x^2-1)} + \frac1{2(x^2+1)}-\frac1{x^2}
$$
and rewrite the integral as
\begin{align}
I &= \int_{0}^{\infty} \left( \frac{\tan^{-1}x^2}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)} \right)dx\\
&=\frac{\pi^2}{16}+ \frac12\int_0^{\infty} \frac{\tan^{-1}x^2}{1+x^2}dx
- \int_0^{\infty} \frac{\tan^{-1}x^2}{x^2}dx
+ \frac12\int_0^{\infty} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx\tag1
\end{align}
where $\int_0^{\infty} \frac{\tan^{-1}x^2}{1+x^2}dx=\frac{\pi^2}8$, $
\int_0^{\infty} \frac{\tan^{-1}x^2}{x^2}dx=\frac{\pi}{\sqrt2}$ and
\begin{align}
&\frac12\int_0^{\infty} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx
=\int_0^{1} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx\\
\overset{IBP}=&-\int_0^1 \frac x{1+x^4} \ln \frac{1-x}{1+x}dx
\overset{x\to \frac{1-x}{1+x}}=-\int_0^\infty\frac {\ln x}{1+6x^2+ x^4}dx\\
=&-\frac1{4\sqrt2}\left( \int_0^\infty\frac{\ln x}{x^2+ (\sqrt2-1)^2}dx -\int_0^\infty\frac{\ln x}{x^2+ (\sqrt2+1)^2}dx \right)\\
=&-\frac1{8\sqrt2}\left(\frac{\ln(\sqrt2-1)}{\sqrt2-1} - \frac{\ln(\sqrt2+1)}{\sqrt2+1} \right)
\end{align}
Substitute above results into (1) to obtain
$$I= \frac{\pi^2}8-\frac\pi{\sqrt2}\left(1+\frac{\ln(\sqrt2-1)}{8(\sqrt2-1)} - \frac{\ln(\sqrt2+1)}{8(\sqrt2+1)} \right)
$$
Best Answer
$$I=\int_0^1 \frac{x^2\ln(1-x^4)}{1+x^4}dx=\frac{5\pi^2}{48\sqrt 2}+\frac{\pi\ln 2}{4\sqrt 2}+\frac{\pi\ln(\delta_S)}{2\sqrt 2}+\frac{\ln^2(\delta_S)}{2\sqrt 2}-\frac{\ln 2 \ln(\delta_S)}{2\sqrt 2 }$$ $$+\frac{3\operatorname{Li}_2\left(-(\delta_S)^2\right)}{8\sqrt 2}-\frac{3\operatorname{Li}_2\left(-1/(\delta_S)^2\right)}{8\sqrt 2}+\frac{\operatorname{Li}_2\left(1/(\delta_S)^4\right)}{8\sqrt 2}+\frac{\operatorname{Ti}_2\left(-\delta_S\right)}{\sqrt 2}-\frac{\operatorname{Ti}_2\left(1/\delta_S\right)}{\sqrt 2}$$ Where $\delta_S=1+\sqrt 2\, $ is the silver ratio, $\operatorname{Li}_2(x)$ is the dilogarithm and $\operatorname{Ti}_{2}(x)$ is the inverse tangent integral.
To obtain this closed form we'll start by moving the bounds from $(0,1)$ to $(0,\infty)$. That's simply because afterwards the plan is to differentiate under the integral sign and usually having the bounds as $(0,\infty)$ simplifies the final result a lot when dealing with rational functions. $$I=\frac12\int_{-1}^1\frac{x^2\ln(1-x^4)}{1+x^4}dx\overset{x\to\frac{1-x}{1+x}}=\frac12\int_0^\infty \frac{(1-x)^2}{1+6x^2+x^4}\ln\left(\frac{8x(1+x^2)}{(1+x)^4}\right)dx$$ $$=\frac{3\pi \ln 2}{4\sqrt 2}-\frac{3\ln 2\ln(1+\sqrt 2)}{2\sqrt 2}+\frac12X-2Y$$ $$X=\int_0^\infty \frac{(1-x)^2\ln\left(1+x^2\right)}{1+6x^2+x^4}dx;\quad Y=\int_0^\infty \frac{(1-x)^2\ln\left(1+x\right)}{1+6x^2+x^4}dx$$ Also one integral from above vanished via the substitution $x\to \frac{1}{x}$: $$\int_0^\infty \frac{(1-x)^2\ln x}{1+6x^2+x^4}dx=0$$ As mentioned above, to evaluate $X$ and $Y$ we'll start by differentiating under the integral sign and even though the rational functions from above are somewhat more complicated than if we kept the original ones, we expect to have an easier time when integrating back.
$$X(a)=\int_0^\infty\frac{(1-x)^2\ln\left(1+ax^2\right)}{1+6x^2+x^4}dx\Rightarrow X'(a)=\int_0^\infty \frac{(1-x)^2x^2}{(1+6x^2+x^4)(1+ax^2)}dx$$ $$=\frac{2}{1-6a+a^2}\int_0^\infty\left(\frac{ax}{1+ax^2}-\frac{ax+x^3}{1+6x^2+x^4}\right)dx$$ $$+\frac{1-a}{1-6a+a^2}\int_0^\infty\frac{1}{1+ax^2}dx-\frac{1}{1-6a+a^2}\int_0^\infty\frac{1-a-(a-5)x^2}{1+6x^2+x^4}dx$$ $$=\frac{\ln a}{1-6a+a^2}-\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{a-3}{1-6a+a^2}+\frac{\pi}{2\sqrt a}\frac{1-a}{1-6a+a^2}+\frac{\pi}{2\sqrt 2}\frac{a-3}{1-6a+a^2}$$ $$\Rightarrow X=\int_0^1\left(\frac{\ln a}{1-6a+a^2}-\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{a-3}{1-6a+a^2}\right)da+\frac{\pi\ln 2}{2\sqrt 2}+\frac{\pi\ln (1+\sqrt 2)}{2\sqrt 2}$$
$$Y(a)=\int_0^\infty\frac{(1-x)^2\ln\left(1+ax\right)}{1+6x^2+x^4}dx\Rightarrow Y'(a)=\int_0^\infty \frac{(1-x)^2x}{(1+6x^2+x^4)(1+ax)}dx$$ $$=\int_0^\infty\left(\frac{(1+5a^2-2a^3)x+(1+a)^2x^3}{(1+6a^2+a^4)(1+6x^2+x^4)}-\frac{a(1+a)^2}{(1+6a^2+a^4)(1+ax)}\right)dx$$ $$+\int_0^\infty \left(\frac{a(1+a)^2-(2-5a-a^3)x^2}{(1+6a^2+a^4)(1+6x^2+x^4)}\right)dx$$ $$=-\frac{(1+a)^2\ln a}{1+6a^2+a^4}-\frac{a(3-a+a^2)+1}{1+6a^2+a^4}\frac{\ln(1+\sqrt 2)}{\sqrt 2}+\frac{\pi}{2\sqrt 2}\frac{a(3+a+a^2)-1}{1+6a^2+a^4}$$ $$\Rightarrow Y=-\int_0^1\frac{(1+a)^2\ln a}{1+6a^2+a^4}da-\frac{\pi\ln(1+\sqrt 2)}{8\sqrt 2}-\frac{3\ln 2\ln(1+\sqrt 2)}{4\sqrt 2}-\frac{\pi^2}{16\sqrt 2}+\frac{3\pi\ln 2}{8\sqrt 2}$$
Let's reorder things a little, just to see what we have so far (combining with the $X$ and $Y$ integrals): $$I=\frac{\pi^2}{8\sqrt 2}+\frac{\pi\ln 2}{4\sqrt 2}+\frac{\pi\ln(1+\sqrt 2)}{2\sqrt 2}+2\int_0^1\frac{(1+x)^2\ln x}{1+6x^2+x^4}dx$$ $$+\frac12\int_0^1\left(\frac{\ln x}{1-6x+x^2}+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{3-x}{1-6x+x^2}\right)dx$$
So all that's left is to evaluate the remaining two integrals. For the first one we have:
$$\int_0^1\frac{(1+x)^2\ln x}{1+6x^2+x^4}dx=\int_0^1\left(\frac{1}{2\sqrt 2}\frac{(1+\sqrt 2)-x}{(1+\sqrt 2)^2+x^2}-\frac{1}{2\sqrt 2}\frac{(1-\sqrt 2)-x}{(1-\sqrt 2)^2+x^2}\right)\ln x \, dx$$ $$\small =\frac{1}{8\sqrt 2}\left(\operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)-\operatorname{Li}_2\left(-(1-\sqrt 2)^2\right)\right)+\frac{1}{2\sqrt 2}\left(\operatorname{Ti}_2\left(-(1+\sqrt 2)\right)-\operatorname{Ti}_2\left(-(1-\sqrt 2)\right)\right)$$ The result from above follows using: $$\int_0^1 \frac{x\ln x}{a^2+x^2}dx\overset{x^2\to x}=\frac14\int_0^1 \frac{\ln x}{a^2+x}dx=\frac14\operatorname{Li}_2\left(-\frac{1}{a^2}\right)$$ $$\int_0^1 \frac{a\ln x}{a^2+x^2}dx\overset{x\to ax}=\int_0^\frac1a\frac{\ln a +\ln x}{1+x^2}dx\overset{IBP}=-\operatorname{Ti}_2\left(\frac1a\right)$$
The second one is a little troublesome since we must keep the terms togheter to avoid divergence issues. So we'll do a couple substitutions first: $$\int_0^1\left(\frac{\ln x}{1-6x+x^2}+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{3-x}{1-6x+x^2}\right)dx$$ $$\overset{x\to\frac{1-x}{1+x}}=\int_0^1\left(\frac12\frac{1}{2x^2-1}\ln\left(\frac{1-x}{1+x}\right)+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\left(\frac{2x}{2x^2-1}-\frac{1}{x+1}\right)\right)dx$$ $$=-\frac{\ln 2\ln(1+\sqrt 2)}{\sqrt 2}+\int_0^1\left(\frac12\frac{1}{2x^2-1}\ln\left(\frac{1-x}{1+x}\right)+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{2x}{2x^2-1}\right)dx$$ $$\overset{\sqrt 2 x\to x}=-\frac{\ln 2\ln(1+\sqrt 2)}{\sqrt 2}-\frac{1}{\sqrt 2}\int_0^\sqrt 2\left(\frac12\frac{1}{1-x^2}\ln\left(\frac{\sqrt 2 -x}{\sqrt 2 + x}\right)+\frac{x}{1-x^2}\ln(1+\sqrt 2)\right)$$ $$\overset{x\to\frac{1-x}{1+x}}=\frac{\ln^2(1+\sqrt 2)}{\sqrt 2}-\frac{\ln 2 \ln(1+\sqrt 2)}{\sqrt 2 }-\frac{1}{4\sqrt 2}\int_{-(1-\sqrt 2)^2}^1\ln\left(\frac{1+(1+\sqrt 2)^2x}{1+\frac{x}{(1+\sqrt 2)^2}}\right)\frac{dx}{x}$$ $$\scriptsize{=\frac{\ln^2(1+\sqrt 2)}{\sqrt 2}-\frac{\ln 2 \ln(1+\sqrt 2)}{\sqrt 2 }-\frac{1}{4\sqrt 2}\left(\frac{\pi^2}{6}-\operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)+\operatorname{Li}_2\left(-(1-\sqrt 2)^2\right)-\operatorname{Li}_2\left((1-\sqrt 2)^4\right)\right)}$$ The last integral was split into two and the result follows by using the definition of the dilogarithm.
Finally putting everything togheter gives the announced closed form.