Primitive of a continuous function over $\Bbb R$

Question

A continuous function over $\Bbb R$ has a primitive that is also continuous over $\Bbb R$.

However, it often happens with Mathematica and other CAS that the result is not continuous, especially with trigonometric integrals computed with the Weierstrass substitution. The result can't be correct because from a periodic integrand this substitution leads to a periodic primitive, while the primitive over $\Bbb R$ is not periodic in general.

For instance:

In[1]:= Integrate[3/(5 - 4 Cos[x]), x]
Out[1]= 2 ArcTan[3 Tan[x/2]]

My question: Is there a way in Mathematica (an option, a package...) to obtain a continuous primitive in this case?

A correct one could be:

$$x + 2 \arctan\left(\frac{\sin x}{2 - \cos x}\right)$$

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There are publications about this:

• D. J. Jeffrey, "The Importance of Being Continuous"
• David J. Jeffrey, Albert D. Rich, "The evaluation of trigonometric integrals avoiding spurious discontinuities"

See also this related question on Math.SE.

I wonder if Mathematica has implementations of the methods mentioned in those articles.

Answer 1

Yes, it is possible. First compute in a standard way

Integrate[3/(5-4 Cos[x]),x]
(*2 ArcTan[3 Tan[x/2]]*)
Plot[2 ArcTan[3 Tan[x/2]],{x,-10π,10π}]

Now try the Rubi package

Get["Rubi`"]
Int[3/(5-4 Cos[x]),x]
(*x+2 ArcTan[Sin[x]/(2-Cos[x])]*)
Plot[x+2 ArcTan[Sin[x]/(2-Cos[x])],{x,-10π,10π}]

Answer 2

Integrate[f[x], x] gives an antiderivative whose derivative is generically equal to f[x]. From the docs:

For indefinite integrals, Integrate tries to find results that are correct for almost all values of parameters.

A primitive function of a continuous function $f$ such as the OP seeks is given by $$F(x) = \int_a^x f(t) \; dt \,.$$

prim = Integrate[3/(5 - 4 Cos[t]), {t, 0, x}, 
  Assumptions -> -Infinity < x < Infinity]

Plot[prim, {x, -30, 30}]