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.
 A: 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.
A: $x=\sqrt{2}\tan{u},dx=\sqrt{2}\sec^2{u}du,\sqrt{1+x\sqrt{x^2+2}}dx=\sqrt{2+4\tan{u}|\sec{u}|}\sec^2{u}du=\dfrac{\sqrt{2\cos^2{u} \pm 4\sin{u}}}{|\cos{u}|\cos^2{u}}du=\pm\dfrac{\sqrt{2\cos^2{u} \pm 4\sin{u}}}{cos^4{u}}d\sin{u}=\pm\dfrac{\sqrt{2(1-v^2) \pm 4v}}{(1-v^2)^2}dv ,v=\sin{u}$
consider case "+",$2(1-v^2)+4v=2(2-(v-1)^2)=2(2-y^2),y=v-1$ 
$\dfrac{\sqrt{2(1-v^2) +4v}}{(1-v^2)^2}dv=\dfrac{\sqrt{4-2y^2}}{(y(y+2))^2}dy$
$\dfrac{1}{y^2(y+2)^2}=\dfrac{1}{4(y+1)}\left(\dfrac{1}{y^2}-\dfrac{1}{(y+2)^2}\right)=\dfrac{1}{4y}\left(\dfrac{1}{y}-\dfrac{1}{(y+1)}\right)-\dfrac{1}{4(y+2)}\left(\dfrac{1}{y+1}-\dfrac{1}{(y+2)}\right)$
note :
$\dfrac{\sqrt{4-2y^2}}{y^2},\dfrac{\sqrt{4-2y^2}}{y}...$can be solved,so the problem can be solved.
A: Substitute $x=\frac{2-t}{2\sqrt t}$. Then, $dx=-\frac {2+t}{4t\sqrt t}\>dt$ and
\begin{align}
&\int\sqrt{1+x\sqrt{x^2+2}}\>dx \\
= & -\frac18 \int \frac{t+2}{t^2} \sqrt{ 4+4t-t^2}\>dt
= -\frac18 \int \frac{\sqrt{ 4+4t-t^2}}{t-2}d\left( \frac{(t-2)^2}t\right)\\
=& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}- \int \frac1{t\sqrt{ 4+4t-t^2}}dt\\
=& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}+\frac12 \tanh^{-1} \frac{t+2}{\sqrt{ 4+4t-t^2}}+C
\end{align}
A: 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}} &= \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 &= \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}
