It is well known that $$\sqrt{4+3\sqrt{2}}=\sqrt[4]{2}(1+\sqrt{2})\tag{1}$$, and similarly, $$\sqrt{10+6\sqrt{5}}=\sqrt[4]{5}(1+\sqrt{5})\tag{2}$$$$\sqrt{6+4\sqrt{3}}=\sqrt[4]{3}(1+\sqrt{3})\tag{3}$$$$ \sqrt{30+28\sqrt{3}}=\sqrt[4]{3}(5+\sqrt{3})\tag{4}$$ etc.
I wonder if there is an underlying rule that can be applied here; something like how the form $\sqrt[4]{r}(a+\sqrt{r})$ can be simplified into the form $\sqrt{a+b\sqrt{r}}$.
Let $x = \sqrt[4]{r}(a+b\sqrt{r})$, so $$x^2 = (\sqrt[4]{r}(a+b\sqrt{r}))^2 = \sqrt{r}(a^2+2ab\sqrt{r}+b^2r) = \sqrt{r}(a^2+rb^2)+2abr$$ So $$x = \sqrt{2abr+(a^2+rb^2)\sqrt{r}}$$