Yeah, it's a tricky Emden-Fowler equation. The only thing I can think of is to multiply by x' and integrate to get
[tex]\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{x} + C[/tex]
Now, swap for x from the original equation, i.e.,
[tex]x = \frac{E}{m x^{\prime \prime}}[/tex]
such that
[tex]\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{\frac{E}{m x^{\prime \prime}}} + C[/tex]
rearrange to get
[tex]x^{\prime \prime} = A e^{-\frac{m x^{\prime 2}}{2 E}}[/tex]
where A is a constant. Now make the change of variable
[tex]x^{\prime} = i \sqrt{\frac{2 E}{m}} z[/tex]
to get
[tex]i \sqrt{\frac{2E}{m}} z^{\prime} e^{-z^2} = A[/tex]
but of course
[tex]\frac{ d }{dt} erf{(z)} = \sqrt{\frac{2}{\pi}} z^{\prime} e^{-z^2}[/tex]
so integrating gives you
[tex]erf{(z)} = \alpha t + \beta[/tex]
where alpha and beta are constants that may or may not be complex
So there's a solution of the form
[tex]x(t) = i\sqrt{\frac{2 E}{m}} \int{erf^{-1}{(\alpha t + \beta)} dt}[/tex]
Note: the inverse erf function is integrable -- see
this Wolfram page.