Let me add two remarks

1) the exact value of the jump position seems to be xj = 2^(2/3) - 1

N[2^(2/3) - 1, 20]
(* 0.58740105196819947475 *)

2) an antiderivative without a jump can be calculated as follows

ad2F1[x_] = Integrate[1/Sqrt[1 + 2 x^3], {x, 0, t}, Assumptions -> t > 0] /. t -> 1 + x

(* Out[116]= (1 + x) Hypergeometric2F1[1/3, 1/2, 4/3, -2 (1 + x)^3] *)

Hence we find a hypergeometric function instead of the "bizarre" elliptic integral.

Proof

D[ad2F1[x], x] // Simplify

(* Out[109]= 1/Sqrt[1 + 2 (1 + x)^3] *)

Yes, for the indefinite integral you're right; it is still in Version 10.0.0 the case. The difference of valued indefinite integrals does not convert into the Hypergeometric2F1[] form and the bug reappears

In[11]:= N[(ReplaceAll[#, y -> 2] - ReplaceAll[#, y -> 1]) &[Integrate[1/Sqrt[2 y^3 + 1], y]]]
Out[11]= -2.96267 - 1.92762 I

In[15]:= N[FullSimplify[(ReplaceAll[#, y -> 2] - ReplaceAll[#, y -> 1])] &[Integrate[1/Sqrt[2 y^3 + 1], y]]]
Out[15]= -2.96267 - 1.92762 I

In[17]:= (Integrate[1/Sqrt[2 y^3 + 1], y] /. y -> x + 1) === Integrate[1/Sqrt[2 (1 + x)^3 + 1], x]
Out[17]= True

that means this is a bug not about a path singularity, because the indefinite integral does not go along a path. It's something about turning back to the right leaf. For example, look at this, the derivative does not turn back to the original integrand, but presents an equation of six-th degree instead:

Dear Udo, thanks for your comments.

But if you try (I am using Mathematica 9)

In[1]:= sol = Integrate[1/Sqrt[2 y^3 + 1], y]

Out[1]= (1/(3^(1/4) Sqrt[1 + 2 y^3]))(-1)^(1/6) 2^( 2/3) Sqrt[(-1)^(5/6) (-1 + (-2)^(1/3) y)] Sqrt[ 1 + (-2)^(1/3) y + (-2)^(2/3) y^2] EllipticF[ArcSin[Sqrt[-(-1)^(5/6) - I (-2)^(1/3) y]/3^(1/4)], (-1)^( 1/3)]

In[2]:= (sol /. y -> 2) - (sol /. y -> 1) // N

Out[2]= -2.96267 - 1.92762 I

You still have a wrong answer. So, in my opnion, there is a bug in Integrate function.

Regarding the crash associated with plotting a long expression, I wonder if the bug is related to this post: http://community.wolfram.com/groups/-/m/t/314527?p_p_auth=3uJCatzr

Postscript: I tried the plot in http://community.wolfram.com/groups/-/m/t/314527?p_p_auth=3uJCatzr, but using a shorter expression for the first argument to Plot. The plot in that post still crashes (randomly, it seems). So, perhaps the bugs are not relatives.

I filed a bug report.

Plotting theDerivative using

Plot[theDerivative,{x,0,1}] 

does not crash my Kernel, but plotting the full expression for theDerivative as written in my post does crash it reproducibly.

A side observation...

If one computes

theIntegral = Integrate[1/Sqrt[2 (1 + x)^3 + 1], x]

and then computes

theDerivative = D[theIntegral, x] // FullSimplify

which yields

-(((-1)^(5/6) Sqrt[3] Sqrt[
     1 + 2 (1 + x)^3])/(Sqrt[-(-1)^(5/6) (1 + 2^(1/3) + 2^(1/3) x)]
      Sqrt[(-1)^(5/6) (-1 + (-2)^(1/3) (1 + x))] Sqrt[
     1 + (-2)^(1/3) (1 + x) + (-2)^(2/3) (1 + x)^2] Sqrt[
     3 - (-1)^(1/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 1]] Sqrt[
     3 + (-1)^(5/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 2]])) 

and then attempts to Plot this using

Plot[-(((-1)^(5/6) Sqrt[3] Sqrt[
      1 + 2 (1 + x)^3])/(Sqrt[-(-1)^(5/6) (1 + 2^(1/3) + 2^(1/3) x)]
       Sqrt[(-1)^(5/6) (-1 + (-2)^(1/3) (1 + x))] Sqrt[
      1 + (-2)^(1/3) (1 + x) + (-2)^(2/3) (1 + x)^2] Sqrt[
      3 - (-1)^(1/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 1]] Sqrt[
      3 + (-1)^(5/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 2]])) , {x, 0, 1}]

the result is that it crashes the Mathematica Kernel. (I will report this.)

In[1]:= SystemInformation["Small"]

Out[1]= {"Kernel" -> {"SystemID" -> "MacOSX-x86-64", 
   "ReleaseID" -> "10.0.0.0 (5098698, 5098537)", 
   "CreationDate" -> 
    DateObject[{2014, 6, 29}, TimeObject[{20, 38, 32}]]}, 
 "FrontEnd" -> {"OperatingSystem" -> "MacOSX", 
   "ReleaseID" -> "10.0.0.0 (5098698, 2014062704)", 
   "CreationDate" -> 
    DateObject[{2014, 6, 27}, TimeObject[{23, 17, 40.}]]}}

Interesting. I've reported the crash and it's in the bug system. And I just added your example to the report.

@otsocol, Yes, I was looking at that antiderivative.

As to whether the antiderivative itself is correct, About all I can say is it gives a numerical result consistent with NIntegrate if one only integrates from 0 to 1/2 (which excludes that singularity point). This might or might not answer your question.

I see what you mean

sln = DSolve[{y'[x] == 1/Sqrt[2 (1 + x)^3 + 1], y[0] == 0}, y]

Plot [Im[y[x] /. sln], {x, 0, 1}, PlotRange -> All ] shows that y[x] develops an imaginary part at 0.59

Frank: In brief: Mathematica is using the Newton-Leibniz method to do the definite integral. It computes the antiderivative. It looks for points on the path of integration where the antiderivative might have discontinuities. It evaluates by taking limits at endpoints and all these points.

The antiderivative has a jump discontinuity near x=0.59, which of course is between 0 and 1. The code that looks for discontinuities is failing to find it. Not entirely a surprise (finding them involves some hard-core equation solving, often transcendental). But it's worth further investigation.

There is a path singularity around .59 that is not found. Will investigate.