Limit calculated incorrect

[osmanfb]

This one is related to taking the definite integral of Sqrt[(1+x)/(1-x)] from -1 to 1. Mathilda does not evaluate this integral. But the indefinite integral is calculated correctly:
In[1]:= Integrate[Sqrt[(1+x)/(1-x)],x]
Out[1]= -2 Sqrt[(1 + x)/(1 - x)]/(1 + (1 + x)/(1 - x)) + 2 ArcTan[Sqrt[(1 + x)/(1 - x)]]
The first term at x=-1 and x=1 is calculated by Limit[]. But the second term has problems at +1
In[10]:= Limit[2 ArcTan[Sqrt[(1 + x)/(1 - x)]], x -> 1]
Out[10]= -Pi
Which should be Pi. Wolframscript, Maxima, Reduce. Axiom/Fricas all calculate the definite integral correctly.

In[12]:= Limit[2 ArcTan[Sqrt[(1 + x)/(1 - x)]],x -> 0.999]
Out[12]= 3.09687

In[13]:= Limit[2 ArcTan[Sqrt[(1 + x)/(1 - x)]],x -> 0.9999]
Out[13]= -3.14159
I don't know why but somehow switches sign as it gets closer to +1.
But without the Limit , ArcTan seems to be working ok:
In[14]:= 2 ArcTan[Sqrt[(1 + 0.999)/(1 - 0.999)]]
Out[14]= 3.09687

In[15]:= 2 ArcTan[Sqrt[(1 + 0.9999)/(1 - 0.9999)]]
Out[15]= 3.12745

[stblake]

That's a good find. Fixed with the following two commits:

Commit 1 — a9f9dd2 (the original bug): ArcTan at infinities + Limit
compose-at-directed-infinity + constant-factor linearity → Limit[2
ArcTan[Sqrt[(1+x)/(1-x)]], x→1] = Pi.

Commit 2 — 988fc80 (the gap you asked to close):

• Inverse trig/hyperbolic at every infinity flavor. The reciprocal functions
  (f[z]=g[1/z], 1/z→0) return their value at 0 for
  Infinity/-Infinity/ComplexInfinity/DirectedInfinity: ArcCot→0, ArcSec→Pi/2, ArcCsc→0,
  ArcCoth→0, ArcCsch→0, ArcSech→I Pi/2. Primary functions map a directionless infinity to
  ComplexInfinity: ArcSin/ArcCos/ArcSinh/ArcCosh. Plus ArcTanh[±Infinity] = ∓I Pi/2.
• Limit sums a convergent Plus at finite points.

$ ./Mathilda

Mathilda 0.015 - A small, open source computer algebra system.

This program is free, open source software and comes with ABSOLUTELY NO WARRANTY.

End a line with '\' to enter a multiline expression. Press Return to evaluate.
Exit by evaluating Quit[] or CONTROL-C.

In[1]:= Limit[2 ArcTan[Sqrt[(1 + x)/(1 - x)]], x -> 1]
Out[1]= Pi

In[2]:= Integrate[Sqrt[(1+x)/(1-x)], {x, -1, 1}]
Out[2]= Pi

[stblake]

The reported integral can now be computed in two different ways in Mathilda:

$ ./Mathilda

Mathilda 0.016 - A small, open source computer algebra system.

This program is free, open source software and comes with ABSOLUTELY NO WARRANTY.

End a line with '' to enter a multiline expression. Press Return to evaluate.
Exit by evaluating Quit[] or CONTROL-C.

In[1]:= Integrate[1/(1 + x^4), {x, -Infinity, Infinity}, Method -> "Residue"] // Timing
Out[1]= {0.002656, Pi/Sqrt[2]}

In[2]:= Integrate[1/(1 + x^4), {x, -Infinity, Infinity}, Method -> "NewtonLeibniz"] // Timing
Out[2]= {0.655872, Pi/Sqrt[2]}