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.
Consider the integral
\begin{align}
I = \int \sqrt{1 + x \sqrt{x^{2} + 2}} \, dx
\end{align}
Make the substitution $x = \sqrt{2} \, csch(t)$ to obtain the integral
\begin{align}
I = - \sqrt{2} \, \int \sqrt{1 + 2 \, csch(t) \, coth(t)} \cdot csch(t) \, coth(t) \, dt.
\end{align} 
Now Wolfram can calculate the integral and provides the result
\begin{align}
- \frac{I}{\sqrt{2}} &amp;= \frac{1}{2} \sinh(t) \sqrt{2 coth(t) csch(t)+1} \left[\frac{g(t)}{\sqrt{4 cosh(t)+cosh(2 t)-1}} - csch^2(t) \right] 
\end{align}
where
\begin{align}
g(t) = \ln\left(\tanh^2\left(\frac{t}{2}\right)\right) - \ln\left(1 + \tanh^2\left(\frac{t}{2}\right) + \sqrt{-\tanh^4\left(\frac{t}{2}\right) + 2 \tanh^2\left( \frac{t}{2}\right) +1} \right)
\end{align}
By making a careful backward substitution the result becomes
\begin{align}
I &amp;= \frac{1}{\sqrt{2}} \left[ x \sqrt{1+x\sqrt{x^{2}+2}} + \ln\left( 1 + \frac{(x + \sqrt{x^2 + 2})^{2}}{2} + \sqrt{\frac{(x + \sqrt{x^2 + 2})^{4}}{4} + (x + \sqrt{x^2 + 2})^{2} - 1} \,  \right) \right]
\end{align}