We can't use the antiderivative returned by Integrate because it contains multi-valued functions with default branch-cuts and in order to use a multi-valued antiderivative, the integration path needs to be in an analytic domain of the function. In order to accomplish this, replace the ArcTan function in the RootSum antiderivative with an analytically-continuous version. No need to replace the Log expression as the integration path does not cross a default branch cut.
First integrate a rational version of the integrand:
integrand =
1/(-1509.04 + 1/(-1.64387 - 1.05604 Cos[0.88 \[Theta]]) +
1/(0.000622346 + 0.0000998034 Cos[0.88 \[Theta]]) +
397.918 Cos[0.88 \[Theta]]);
ClearAll[currentArcVal, theCurrentRoot, currentZ];
newIntegrand = Rationalize[integrand, 10^-16];
antiD = Integrate[newIntegrand, \[Theta]]
6250/11 RootSum[
189275993388171865138 + 1514214159714562294323 #1 +
4605782875704094238834 #1^2 + 6617810287079231882646 #1^3 +
4605782875704094238834 #1^4 + 1514214159714562294323 #1^5 +
189275993388171865138 #1^6 &, (1902663296339164 ArcTan[
Sin[(22 \[Theta])/25]/(Cos[(22 \[Theta])/25] - #1)] -
951331648169582 I Log[1 - 2 Cos[(22 \[Theta])/25] #1 + #1^2] +
29652457743237434 ArcTan[Sin[(22 \[Theta])/25]/(
Cos[(22 \[Theta])/25] - #1)] #1 -
14826228871618717 I Log[
1 - 2 Cos[(22 \[Theta])/25] #1 + #1^2] #1 +
77680004276476128 ArcTan[Sin[(22 \[Theta])/25]/(
Cos[(22 \[Theta])/25] - #1)] #1^2 -
38840002138238064 I Log[
1 - 2 Cos[(22 \[Theta])/25] #1 + #1^2] #1^2 +
29652457743237434 ArcTan[Sin[(22 \[Theta])/25]/(
Cos[(22 \[Theta])/25] - #1)] #1^3 -
14826228871618717 I Log[
1 - 2 Cos[(22 \[Theta])/25] #1 + #1^2] #1^3 +
1902663296339164 ArcTan[Sin[(22 \[Theta])/25]/(
Cos[(22 \[Theta])/25] - #1)] #1^4 -
951331648169582 I Log[
1 - 2 Cos[(22 \[Theta])/
25] #1 + #1^2] #1^4)/(1514214159714562294323 +
9211565751408188477668 #1 + 19853430861237695647938 #1^2 +
18423131502816376955336 #1^3 + 7571070798572811471615 #1^4 +
1135655960329031190828 #1^5) &]
It's messy for sure but the important part is the RootSum and the expression:
ArcTan[Sin[(22 [Theta])/25]/(Cos[(22 [Theta])/25] - #1)]
which is multivalued with the integration path crossing a default Mathematica branch-cut. First compute the six roots:
poly[x_] =
189275993388171865138 + 1514214159714562294323 x +
4605782875704094238834 x^2 + 6617810287079231882646 x^3 +
4605782875704094238834 x^4 + 1514214159714562294323 x^5 +
189275993388171865138 x^6;
theRoots = x /. NSolve[poly[x] == 0, x, WorkingPrecision -> 30]
In order to supply an analytically-continuous version of this function to the RootSum, we integrate six differential equations, one for each root of the RootSum across the integration path:
w0 = ArcTan[Sin[(22 z)/25]/(Cos[(22 z)/25] - theRoots[[4]])] /.
z -> myz[0]
wDerivTable = Table[
w'[z] == (D[
ArcTan[Sin[(22 z)/25]/(Cos[(22 z)/25] - theRoots[[i]])], z]),
{i, 1, Length@theRoots}
];
maxZ = 50 Pi;
arcSol = NDSolveValue[{#, w[0] == 0}, w, {z, 0, maxZ},
WorkingPrecision -> 25] & /@ wDerivTable
$\texttt{arcSol}$ now are the six analytically-continuous versions of the ArcTan expression in the RootSum.
In the RootSum, replace ArcTan with the six arcSol solutions, replace $\theta$ with z and # with currentRoot:
newRootSumF[currentArcVal_, theCurrentRoot_,
currentZ_] := (6250/
11 (1902663296339164 ArcTan[Sin[(22 \[Theta])/25]/(
Cos[(22 \[Theta])/25] - #1)] -
951331648169582 I Log[1 - 2 Cos[(22 \[Theta])/25] #1 + #1^2] +
29652457743237434 ArcTan[Sin[(22 \[Theta])/25]/(
Cos[(22 \[Theta])/25] - #1)] #1 -
14826228871618717 I Log[
1 - 2 Cos[(22 \[Theta])/25] #1 + #1^2] #1 +
77680004276476128 ArcTan[Sin[(22 \[Theta])/25]/(
Cos[(22 \[Theta])/25] - #1)] #1^2 -
38840002138238064 I Log[
1 - 2 Cos[(22 \[Theta])/25] #1 + #1^2] #1^2 +
29652457743237434 ArcTan[Sin[(22 \[Theta])/25]/(
Cos[(22 \[Theta])/25] - #1)] #1^3 -
14826228871618717 I Log[
1 - 2 Cos[(22 \[Theta])/25] #1 + #1^2] #1^3 +
1902663296339164 ArcTan[Sin[(22 \[Theta])/25]/(
Cos[(22 \[Theta])/25] - #1)] #1^4 -
951331648169582 I Log[
1 - 2 Cos[(22 \[Theta])/
25] #1 + #1^2] #1^4)/(1514214159714562294323 +
9211565751408188477668 #1 + 19853430861237695647938 #1^2 +
18423131502816376955336 #1^3 + 7571070798572811471615 #1^4 +
1135655960329031190828 #1^5)) /. {ArcTan[x__] ->
currentArcVal, #1 -> theCurrentRoot, \[Theta] -> currentZ}
Create the analytically-continuous antiderivative and plot it over the integration interval:
newAntiD[z_] := Module[{baseSum, endSum},
baseSum = 0;
For[i = 1, i <= 6, i++,
currentArcVal = arcSol[[i]][0];
theCurrentRoot = theRoots[[i]];
baseSum += newRootSumF[currentArcVal,theCurrentRoot,z];
];
endSum = 0;
For[i = 1, i <= 6, i++,
currentArcVal = arcSol[[i]][z];
theCurrentRoot = theRoots[[i]];
endSum += newRootSumF[currentArcVal,theCurrentRoot,z];
];
(endSum - baseSum)
];
Plot[Re@newAntiD[z], {z, 0, 125}]
Used Re since will have small imaginary residue from numerical integration.

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