Show that $$\int_{0}^{\infty} \frac{\tanh \left(\frac{x}{2}\right)+\tanh (2 x)}{x}\left(e^{\frac{3 x}{2}}-1\right)^{2} e^{-4 x} \mathrm{~d}x=\boxed{2\log\left(\frac{768\sqrt2\pi^4}{25\Gamma^3\left(\frac14\right)\Gamma^4\left(\frac5{16}\right)}\right)}$$
what I've done so far. For $\alpha > 0$, let $$I(\alpha) = \int_0^{\infty}\frac{e^{-\alpha x}\tanh x}x\;dx$$ Then we have $$I'(\alpha) = -\int_0^{\infty}e^{-\alpha x}\tanh x\;dx$$ Writing $$\tanh x = \frac{1-e^{-2x}}{1+e^{-2x}} = 1 + 2\sum_{n=1}^{\infty}(-1)^ne^{-2nx}$$ Thus, integrating term-by-term gives $$I'(\alpha) = -\frac1{\alpha} - 2\sum_{n=1}^{\infty}\frac{(-1)^n}{\alpha+2n}$$ Adding and subtracting an extra term, we write $$I'(\alpha) = \frac1{\alpha} + \sum_{n=0}^{\infty}(-1)^n\left(\frac1{2\left\lceil\frac{n+1}2\right\rceil}-\frac1{n+\frac{\alpha}2}\right)$$
Since this series now converges absolutely, we can split it into even and odd parts to get $$I'(\alpha) = \frac1{\alpha} + \sum_{n=0}^{\infty}\left(\frac1{2n+2}-\frac1{2n+\frac{\alpha}2}\right) - \sum_{n=0}^{\infty}\left(\frac1{2n+2}-\frac1{2n+\frac{\alpha}2+1}\right)$$
We can rewrite your $I'(\alpha)$ using the digamma function. Factoring out a $\frac12$ from each infinite sum, we get $$I'(\alpha)=\frac1{\alpha}+\frac12\left(\psi\left(\frac\alpha4\right)-\psi\left(\frac\alpha4+\frac12\right)\right)$$Integrating this, we get that $I(\alpha)=\ln(\alpha)+2\log\Gamma\left(\frac{\alpha}4\right)-2\log\Gamma\left(\frac{\alpha}4+\frac12\right)$. So the answer is just a combination of a ton of these $I(\alpha)$'s, giving $I(-6)+2I(0)-2I(5)+I(-3)-2I(\frac52)$.
Credit to @SangchulLee for finding this first.