How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve
$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$
I've chosen the substitution variables
$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$
However, I am completly stuck at
$$\int\sqrt{1+xu} dx$$
Which let me believe I've chosen wrong substitution variables.
I've then tried letting $u=x^2+2$ or simply $u=x$, but it does not help me at all solving it.
Would someone please give me an hint on this ?
Thanks.

I thought about Wolfram and Maple too, needs a little help. But I was wrong.
$$\int {\sqrt {1 + x\sqrt {{x^2} + 2} } } dx = \int {\sqrt {1 + \frac{1}{2}\sqrt {{x^2} + 2} } } 2xdx$$
\begin{gathered}
  y = {x^2} + 2 \\ 
  dy = 2xdx \\ 
\end{gathered} 
\begin{gathered}
  \int {\sqrt {1 + \frac{1}{2}\sqrt y } } dy = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt y  + 2} \right)^{3/2}}\left( {3\sqrt y  - 4} \right) \\ 
   = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt {{x^2} + 2}  + 2} \right)^{3/2}}\left( {3\sqrt {{x^2} + 2}  - 4} \right) \\ 
\end{gathered} 
Now Wolfram: This is an identity:
$$\int {\sqrt {1 + \frac{1}{2}\sqrt y } } dy = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt y  + 2} \right)^{3/2}}\left( {3\sqrt y  - 4} \right)$$
Resubstituting gives me:
$$\frac{2}{{15}}\sqrt 2 {\left( {\sqrt y  + 2} \right)^{3/2}}\left( {3\sqrt y  - 4} \right) = \frac{2}{{15}}\sqrt 2 {\left( {\sqrt {{x^2} + 2}  + 2} \right)^{3/2}}\left( {3\sqrt {{x^2} + 2}  - 4} \right)$$
Differentiation:
$$\frac{d}{{dx}}\left( {\frac{2}{{15}}\sqrt 2 {{\left( {\sqrt {{x^2} + 2}  + 2} \right)}^{3/2}}\left( {3\sqrt {{x^2} + 2}  - 4} \right)} \right) = \sqrt 2 x\sqrt {2 + \sqrt {{x^2} + 2} } $$`
And:
$$\sqrt 2 x\sqrt {2 + \sqrt {{x^2} + 2} }  \ne \sqrt {1 + x\sqrt {{x^2} + 2} }$$
Very poor! For me. Made a mistake by myself.
