Rather weird integral

$$\int \frac{\log(y)}{(y+3)\sqrt{y}} dy$$ $t=\sqrt{y},\;\; y=t^2,\;\;dy=2t\,dt$ $$=4\int \frac{\log(t)}{t^2+3} dt$$ $u=\log(t),\;\;du=\frac 1t,\;\; v=\frac{1}{\sqrt{3}}\arctan\left(\frac{t}{\sqrt{3}}\right),\;\; dv = \frac{1}{t^2+3}$ $$=4\left(\frac{\log(t)}{\sqrt{3}}\arctan\left(\frac{t}{\sqrt{3}}\right)-\frac{1}{\sqrt{3}}\int\frac{\arctan\left(\frac{t}{\sqrt{3}}\right)}{t}dt\right)$$ Looking at the final integral, we get that $$\int\frac{\arctan\left(\frac{t}{\sqrt{3}}\right)}{t}dt$$ $$=\frac{i}{2}\left(\int\frac{\log\left(1-\frac{t}{\sqrt{3}}\right)}{t}dt-\int\frac{\log\left(1+\frac{t}{\sqrt{3}}\right)}{t}dt\right)$$ $$=\frac{i}{2}\left(I_1 - I_2\right)$$ $u=\frac{t}{\sqrt{3}},\;\; t=u\sqrt{3},\;\; dt=du\sqrt{3}$ $$I_1 = -\int\frac{-\log(1-u)}{u}du = -\operatorname{Li_2}(u) = -\operatorname{Li_2}\left(\frac{t}{\sqrt{3}}\right)$$ $u=\frac{-t}{\sqrt{3}},\;\; t=-u\sqrt{3},\;\; dt=-du\sqrt{3}$ $$I_2 = -\int\frac{-\log(1-u)}{u}du = -\operatorname{Li_2}(u) = -\operatorname{Li_2}\left(\frac{-t}{\sqrt{3}}\right)$$ Putting this all together, we get $$\frac{4\log(t)}{\sqrt{3}}\arctan\left(\frac{t}{\sqrt{3}}\right)+\frac{i}{2\sqrt{3}}\left[\operatorname{Li_2}\left(\frac{t}{\sqrt{3}}\right)+\operatorname{Li_2}\left(\frac{-t}{\sqrt{3}}\right)\right]$$ $$=\color{red}{\frac{4\log(t)}{\sqrt{3}}\arctan\left(\frac{t}{\sqrt{3}}\right)+\frac{i}{4\sqrt{3}}\operatorname{Li_2}\left(\frac{t^2}{3}\right)}$$ This uses the Polylogarithm. Here is a link to the Wikipedia page on the topic.

This is more of a comment since I'm not sure this is what you're looking for but if you use $t=\sqrt y$ then $dy=2t\ dt$. The integrand becomes

$$2t \frac{\ln t^2}{(t^3+3t)}=4\frac{\ln t}{t^2+3}$$

Then by parts, $u=4\ln t$ and $dv=\frac{1}{t^2+3}$. This gives $du=\frac{4}{t}$ and $dv$ is a function of arctan.

This leave you with a new integrand which looks like $\frac{\arctan t}{t}$ which is POSSIBLE with polylogarithmic functions but not with elementary functions.