The combine/radical function is used to combine products of radicals of the same power that appear in an expression.

The main transformation made by combine/radical is

where $x$ and $y$ are both positive, (i.e. $\mathrm{signum}\left(x\right)=\mathrm{signum}\left(y\right)$ $=1$) and $0<m,n<d$ where $m,n,d$ are integers.

For example: $2^{\frac{1}{2}}{3}^{\frac{1}{2}}\to 6^{\frac{1}{2}}$ 

If the condition $0<m<d$ above is not satisfied, then let $m=qd+r$ where $0<r<d$. Then expand $x^{\frac{m}{d}}=x^{\frac{qd+r}{d}}$ $=x^{q+\frac{r}{d}}$ as $x^q$ $x^{\frac{r}{d}}$ so that the condition is now satisfied.  If $x$ is an integer, Maple applies this transformation automatically.  

For example: $2^{\frac{3}{2}}\to 2\cdot 2^{\frac{1}{2}}$.

Suppose the sign of $x$ is known to be negative, i.e. $\mathrm{signum}\left(x\right)=\mathrm{-1}$. Then expand $x^{\frac{m}{d}}$ as ${\left(\mathrm{-1}\right)}^{\frac{m}{d}}$ ${\left(-x\right)}^{\frac{m}{d}}$ so that the positive part ${\left(-x\right)}^{\frac{m}{d}}$ can now be combined.  

For example: ${\left(-2\right)}^{\frac{1}{2}}\cdot 3^{\frac{1}{2}}\to \mathrm{I}\cdot 2^{\frac{1}{2}}\cdot 3^{\frac{1}{2}}\to \mathrm{I}\cdot 6^{\frac{1}{2}}$ 

The new radical ${\left(x^m{y}^n\right)}^{\frac{1}{d}}$ needs to be simplified. The new radicand $x^m{y}^n$ is simplified by applying normal(x^m*y^n, expanded). For example:$\sqrt{10+\sqrt{2}}\sqrt{10-\sqrt{2}}\to \sqrt{\left(10+\sqrt{2}\right)\left(10-\sqrt{2}\right)}\stackrel{\mathrm{normal}/\mathrm{expanded}}{\to }\sqrt{98}$ which after simplification yields $7\sqrt{2}$

If the sign of $x$ and $y$ is not known, then combine will not combine the radicals because that is not correct for all $x$ and $y$ in general. For example $\sqrt{x}\sqrt{y}\ne \sqrt{xy}$ for negative $x$ and $y$.  The user has two possibilities to force Maple to combine radicals of unknown sign. The first is to use assume to tell Maple the sign of the radicands. The second is to specify the optional argument symbolic which will assume all radicands of unknown sign are real and positive.

If the sign of one radicand is unknown -- for example, suppose that $\mathrm{signum}\left(x\right)=1$ and $\mathrm{signum}\left(y\right)$ is unknown -- then combine will still combine the radicands $x$ and $y$ because $x$ is known to be positive.

$\mathrm{f1}≔\sqrt{2}\sqrt{6}\&plus;\sqrt{2}\sqrt{x\&plus;1}$

$\mathrm{f1}≔\sqrt{2}\sqrt{6}+\sqrt{2}\sqrt{x+1}$

$\mathrm{combine}\left(\mathrm{f1}\&comma;\mathrm{radical}\right)$

$2\sqrt{3}+\sqrt{2x+2}$

$\mathrm{f2}≔\sqrt{4-\sqrt{3}}\sqrt{4\&plus;\sqrt{3}}$

$\mathrm{f2}≔\sqrt{4-\sqrt{3}}\sqrt{4+\sqrt{3}}$

$\mathrm{combine}\left(\mathrm{f2}\&comma;\mathrm{radical}\right)$

$\sqrt{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}$

$\mathrm{normal}\left(\&comma;\mathrm{expanded}\right)$

$\sqrt{13}$

$\mathrm{f3}≔\sqrt{x}\sqrt{y}\&plus;\sqrt{2}{\left(\sqrt{x\&plus;1}\right)}^3\sqrt{y}$

$\mathrm{f3}≔\sqrt{x}\sqrt{y}+\sqrt{2}{\left(x+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{y}$

$\mathrm{combine}\left(\mathrm{f3}\&comma;\mathrm{radical}\right)$

$\sqrt{x}\sqrt{y}+\left(x+1\right)\sqrt{y}\sqrt{2x+2}$

$\mathrm{combine}\left(\mathrm{f3}\&comma;\mathrm{radical}\&comma;\mathrm{symbolic}\right)$

$\sqrt{xy}+\left(x+1\right)\sqrt{2}\sqrt{\left(x+1\right)y}$

$\mathrm{assume}\left(x<0\right)$

$\mathrm{combine}\left(\mathrm{f3}\&comma;\mathrm{radical}\right)$

$\mathrm{I}\sqrt{-\mathrm{x~}y}+\left(\mathrm{x~}+1\right)\sqrt{y}\sqrt{2\mathrm{x~}+2}$