How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$? How to evaluate
$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$
The integral arises in the computation of
$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$
as
$$ \scriptsize{\frac{\pi \ln(2)}{4\sqrt{2}} + \frac{\ln(2) \ln(3-2\sqrt{2})}{4\sqrt{2}} -3\ln(2) + \frac{\pi}{2}= \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx + \int_{0}^{1} \frac{x^{1/4}}{2(1+x)}\left(\tan^{-1}(x^{1/4}) - \tanh^{-1}(x^{1/4}) \right)}dx $$
A similar Integral
$$ \int_{0}^{1}\left( \frac{x^2 \ln(2)}{x^4-1} - \frac{x^2 \ln(1+x^4)}{x^4-1}\right)dx = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} \tag{2} $$
Where $ C $ = Catalan Constant
Unfortunately the same techniques I used to evaluate $(2)$ have not worked for $(1)$.
I know only for integration - By parts, U-Sub, and  using Taylor Series as well as Mathematica.
Q = Is there a closed form for Integral $(1)$?
EDIT
$$ (1) = \frac{-\pi^2}{8\sqrt{2}} + \frac{\pi \ln(8)}{4\sqrt{2}} +\frac{\ln(8) \ln(3-2\sqrt{2})}{4\sqrt{2}} -\frac{\pi \ln(3-2\sqrt{2})}{8\sqrt{2}} + 4\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} $$
$$\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} = \frac{1}{64}\left(\psi^{(1)}\left(\frac{7}{8}\right)-\psi^{(1)}\left(\frac{3}{8}\right)\right) + W $$
Where $W$ is some value.
 A: By expanding the denominator and integrating term-by-term we obtain:
$$
-\sum _{n=0}^{\infty } \frac{(-1)^n }{4 n+3}H_{n+\frac{3}{4}},
$$
where $H_n$ is the $n$th harmonic number, in the formula above--the harmonic number of fractional order.
Not sure if it is useful, but the sum can be performed with the help of, e.g., Mathematica yielding a relatively simple expression:
$$
\tfrac{1}{3} \Gamma \big(\tfrac{7}{4}\big)\, {}_2F_1^{(0,0,1,0)}\!\left(\tfrac{3}{4},1,\tfrac{7}{4},-1\right)-\frac{\gamma}{4 \sqrt{2}}\left[\pi -2\log(\sqrt{2}-1)\right].
$$
The superscript means that derivative is taken with respect to the parameter $c$ of the regularized hypergeometric function ${}_2F_1(a,b,c,z)$.
A: This may be counterproductive but we can obtain another form from @yarchik's solution.
$$
\begin{aligned}
S%
&=-\sum _{n=0}^{\infty } \frac{(-1)^n }{4n+3}H_{n+\frac{3}{4}},\\
&=-\frac{1}{4}\sum _{n=0}^{\infty } \frac{(-1)^n }{n+3/4}H_{n+\frac{3}{4}},\\
&=-\frac{1}{4}\int_0^1x^{-1/4}\sum _{n=0}^{\infty } (-x)^nH_{n+\frac{3}{4}}\,\mathrm dx,\\
\end{aligned}
$$
The remaining series can be solved exactly yielding
$$
\begin{aligned}
S%
&=\frac{1}{7}\int_0^1\frac{x^{3/4}}{1+x}F\left({1,7/4\atop 11/4};-x\right)\,\mathrm dx-\frac{H_{3/4}}{4}\int_0^1\frac{x^{-1/4}}{1+x}\,\mathrm dx,\\
&=\frac{1}{7}\int_0^1\frac{x^{3/4}}{1+x}F\left({1,7/4\atop 11/4};-x\right)\,\mathrm dx-\frac{H_{3/4}(-1)^{1/4}}{2}(\operatorname{arctan}((-1)^{1/4})- \operatorname{arctanh}((-1)^{1/4})),
\end{aligned}
$$
The hypergeometric function in the integral is zero-balanced and may reduce further.
