About the integral $\int_{0}^{1}\frac{\log(x)}{\sqrt{1+x^{4}}}dx$ and elliptic functions

For a work we need to evaluate the following integral $$\int_{0}^{1}\frac{\log\left(x\right)}{\sqrt{1+x^{4}}}dx=\,-_{3}F_{2}\left(\frac{1}{4},\frac{1}{4},\frac{1}{2};\frac{5}{4},\frac{5}{4};-1\right).\tag{1}$$ Classical approaches seem to lead nowhere, but it is possible to translating the problem into the language of elliptic functions. Let $\text{sn}(u,k)$ be the Jacobi elliptic sine. We can prove that the evaluation of $(1)$ boils down to the evaluation of $$\int_{0}^{T/4}\log\left(-e^{-\pi i/4}\text{sn}\left(e^{3\pi i/4}z,-1\right)\right)dz$$ where $T=2K(1/2)$ and $K\equiv K(k)$ is the complete elliptic integral of the first kind with $k$ the elliptic modulus. I am not an expert in elliptic functions so I have difficulty to understand if this integral can be evaluated or not. However, I found this formula $$\log\left(\text{sn}\left(u,k\right)\right)=\log\left(\frac{2K}{\pi}\right)+\log\left(\sin\left(\frac{\pi u}{2K}\right)\right)-4\sum_{n\geq1}\frac{1}{n}\frac{q^{n}}{1+q^{n}}\sin^{2}\left(\frac{n\pi u}{2K}\right)$$ with $$q\equiv e^{-\pi\frac{K}{K^{\prime}}}=e^{\pi i\tau}$$ and $\left|\text{Im}\left(\frac{\pi u}{2K}\right)\right|<\frac{\pi}{2}\text{Im}\left(\tau\right)$. So, assuming that we can exchange the integral with the series, which I'm not sure about, the problem boils down to studying the following Lambert series $$\sum_{n\geq1}\frac{1}{n^{2}}\frac{q^{n}}{1+q^{n}}\sin\left(\frac{\pi nT}{4K}\right).\tag{2}$$ I have seen that similar series have been studied but this particular one has not (as far as I know). Clearly, there are a lot of heuristic passages and so I may have written nonsense.

Questions:

$1)$ Is it possible to find a closed form (in terms of special functions) of $(1)$?

$2)$ Assuming that the “elliptic approach” is correct, is there a closed form of $(2)$, maybe in terms of elliptic functions?

Thank you.

Update: The approach used by achille hui in a series of answers (see 1, 2) maybe can be helpful, even if I'm not sure about it.

Long Comment:

I found that you can approximate the original integral by the following process of repeated integration

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=-\frac{2 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}+2 \int_0^1 \frac{x^4 \log (x)}{\left(x^4+1\right)^{3/2}} \, dx$$

$$2 \int_0^1 \frac{x^4 \log (x)}{\left(x^4+1\right)^{3/2}} \, dx=\frac{1}{5 \sqrt{2}}-\frac{2 \Gamma \left(\frac{5}{4}\right)^2}{5 \sqrt{\pi }}+\frac{12}{5}\int _0^1\frac{ x^8 \log (x)}{\left(x^4+1\right)^{5/2}}$$

$$\frac{12}{5} \int_0^1 \frac{x^8 \log (x)}{\left(x^4+1\right)^{5/2}} \, dx=\frac{2}{15\sqrt{2}}-\frac{2 \Gamma \left(\frac{5}{4}\right)^2}{9 \sqrt{\pi }}+\frac{24}{9} \int_0^1 \frac{x^{12} \log (x)}{\left(x^4+1\right)^{7/2}} \, dx$$

$$\frac{24}{9} \int_0^1 \frac{x^{12} \log (x)}{\left(x^4+1\right)^{7/2}} \, dx=\frac{19}{195 \sqrt{2}}-\frac{2 \Gamma \left(\frac{5}{4}\right)^2}{13 \sqrt{\pi }}+\frac{112}{39} \int_0^1 \frac{x^{16} \log (x)}{\left(x^4+1\right)^{9/2}} \, dx$$

After four iterations above, it is found that

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx\approx \frac{14 \sqrt{2}}{65}-\frac{1624 \Gamma \left(\frac{5}{4}\right)^2}{585 \sqrt{\pi }}$$

with an error of less than $0.13$ %

Update 05/01/2022 - Playing around tonight I've found further conjectural simplifications to update comments above, but no closed form in terms of special functions:

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=\frac{1}{6} \left(-\, _3F_2\left(-\frac{3}{4},-\frac{3}{4},-\frac{1}{2};\frac{1}{4},\frac{1}{4};-1\right)+\sqrt{2}-\frac{4 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}\right)\tag{1}$$

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=\frac{1}{3} \left(\, _3F_2\left(-\frac{3}{4},-\frac{3}{4},\frac{1}{2};\frac{1}{4},\frac{1}{4};-1\right)-\sqrt{2}+\frac{4 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}\right)\tag{2}$$

$$\frac{4}{3} e^{i \pi/ 4}\, F\left(\left.i \sinh ^{-1}\left(e^{i \pi/ 4}\right)\right|-1\right)=-\frac{8 \Gamma \left(\frac{5}{4}\right)^2}{3 \sqrt{\pi }}$$

where $F(\phi |m)$ is elliptic integral of the first kind.

Assuming $(1)$ and $(2)$ are correct then subtracting one from the other gives: $$2 \, _3F_2\left(-\frac{3}{4},-\frac{3}{4},\frac{1}{2};\frac{1}{4},\frac{1}{4};-1\right)+\, _3F_2\left(-\frac{3}{4},-\frac{3}{4},-\frac{1}{2};\frac{1}{4},\frac{1}{4};-1\right)=3 \sqrt{2}-\frac{12 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}$$

Update 17/01/2022 - Another Mathematica derived series approximation that is more convenient to use in a CAS, yet related to my original observation is

$$\int_0^1 \frac{\log (x)}{\sqrt{x^4+1}} \, dx=-\underset{m\to \infty }{\text{lim}}\int_0^1 \left(\sum _{n=0}^m \frac{(2 n)!\, x^{4 n}}{n! \, \left(x^4+1\right)^{\frac{1}{2} (2 n+1)} \left(\prod _{k=0}^n (4 k+1)\right)}\right) \, dx$$

and $$ \sum _{n=0}^\infty \frac{(2 n)!\, x^{4 n}}{n! \, \left(x^4+1\right)^{\frac{1}{2} (2 n+1)} \left(\prod _{k=0}^n (4 k+1)\right)} \,=\frac{\, _2F_1\left(\frac{1}{2},1;\frac{5}{4};\frac{x^4}{x^4+1}\right)}{\sqrt{x^4+1}}$$

$$I=-\int\limits_0^\infty \dfrac{x e^{-x}}{\sqrt{1+e^{-4x}}}\,\text dx =-\int\limits_0^\infty \dfrac{x}{\sqrt{e^{2x}+e^{-2x}}}\,\text dx =-\dfrac1{\sqrt2}\int\limits_0^\infty \dfrac{x}{\sqrt{2\cosh^2x-1}}\,\text dx,$$ $$I=-\dfrac1{2}\int\limits_0^\infty \dfrac{x}{\cosh x\sqrt{1-\dfrac1{2\cosh^2x}}}\,\text dx =\sum\limits_{k=0}^\infty \dfrac1{(-2)^{k+1}}\dbinom{-^1\!/_4}kI_k,\tag1$$ where $$I_k=\int\limits_0^\infty \dfrac{x}{\cosh^{2k+1}x}\,\text dx.\tag2$$ Then \begin{cases} I_0=2C\approx1.831931184,\quad I_1=C-\dfrac12\approx0.415965594,\\[5pt] I_2=\dfrac{3}{4}C-\dfrac{11}{24}\approx 0.228640862,\quad I_3=\dfrac58C-\dfrac{299}{720} \approx 0.157200719,\\[6pt] I_4=\dfrac{35}{64}C-\dfrac{15371}{40320} \approx 0.119693486,\quad I_5=\dfrac{63}{128}C+\dfrac{142819}{403200} \approx 0.096613026\dots, \tag3 \end{cases} where $\;C\;$ is the Catalan constant.

At the same time, $$I_{k+1}=I_{k}-\int\limits_0^\infty \dfrac{x \tanh^2 x\,\text dx}{\cosh^{2k+1}x} =I_k+\int\limits_0^\infty \dfrac{x\sinh x}{\cosh^{2k+1}x}\,\text d\,\dfrac1{\cosh x}$$ $$=I_k-\int\limits_0^\infty \dfrac{-(2k+1)\sinh^2 x+\cosh^2 x}{\cosh^{2k+3}x}\,x\,\text dx-\dfrac1{2k+1}\,\dfrac1{\cosh^{2k+1}x}\bigg|_0^\infty$$ $$=I_k-(2k+1)I_{k+1}+2kI_k+\dfrac1{2k+1},$$ $$I_{k+1}=\dfrac{2k+1}{2k+2} I_k-\dfrac{1}{(2k+1)(2k+2)}.\tag4$$

Obtained series can converge faster than the closed form from OP.