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This was also reported to WRI. [CASE:4903465]

I'd like to know why these integrals no longer evaluate in V 13 and timeout after 3 minutes while using the same PC and OS they all evaluate quickly in time using V 12.3.

Is there is a workaround in V 13?

This is on windows 10. 64 GB RAM. These integrals come from Rubi's input test files.

The timeout is there to prevent the script from hanging forever.

Is this a bug in integrate?

I also include at bottom a link to the notebook in V 13 and V 12.3 with the code and output to download.

Clear["Global`*]
timeOut = 3*60;
TimeConstrained[Integrate[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^(1/3), x], timeOut]
TimeConstrained[Integrate[Cot[e + f*x]^4*(a + a*Sin[e + f*x])^(1/3), x], timeOut]
TimeConstrained[Integrate[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^(2/3), x], timeOut]
TimeConstrained[Integrate[Csc[c + d*x]*(a + a*Sin[c + d*x])^(4/3), x], timeOut]
TimeConstrained[Integrate[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^(4/3), x], timeOut]
TimeConstrained[Integrate[Csc[c + d*x]^2/(a + a*Sin[c + d*x])^(1/3), x], timeOut]
TimeConstrained[Integrate[Csc[c + d*x]^2/(a + a*Sin[c + d*x])^(4/3), x], timeOut]
TimeConstrained[Integrate[(d*Sin[e + f*x])^(5/2)/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])), x], timeOut]
TimeConstrained[Integrate[1/((d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x])),x], timeOut]
TimeConstrained[Integrate[1/((d*Sec[e + f*x])^(1/3)*(a + b*Tan[e + f*x])^2),x], timeOut]
TimeConstrained[Integrate[Tan[c + d*x]^(2/3)/Sqrt[a + b*Tan[c + d*x]], x], timeOut]
TimeConstrained[Integrate[1/(Tan[c + d*x]^(4/3)*Sqrt[a + b*Tan[c + d*x]]),x], timeOut]
TimeConstrained[Integrate[(a + a*Sec[c + d*x])^2/Sqrt[e*Sin[c + d*x]], x], timeOut]
TimeConstrained[Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^(3/2), x], timeOut]

Screen shot

Mathematica graphics

Link to the notebooks is here

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    $\begingroup$ Whenever you use assumptions, Mathematica V13 computes these integrals. $\endgroup$ Feb 8, 2022 at 1:32
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    $\begingroup$ What answer are you expecting here that you would not get from Technical Services? I'm thinking this falls into the category of "The question is out of scope for this site. The answer to this question requires either advice from Wolfram support..." $\endgroup$ Feb 8, 2022 at 14:44
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    $\begingroup$ @E.Chan-López That observation should be made into an answer as that does provide a fix. $\endgroup$
    – JimB
    Feb 8, 2022 at 16:34
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    $\begingroup$ Reminds me of this, where you posted that MMA lost the ability to compute a set of integrals in going from 11.2 to 11.3. As you know, that was subsequently fixed in 12.0: mathematica.stackexchange.com/questions/169209/… I gather that you typically evaluate new releases with Rubi's test suite, to see if there are issues, which is useful. Given that Wolfram is aware of what happened with 11.3, I would think that, by now, Wolfram should know to include this suite as part of their own Q&A SOP's. $\endgroup$
    – theorist
    Feb 8, 2022 at 19:55

2 Answers 2

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I was trying different settings, I found a solution to this problem.

Now they all evaluate in V 13 as they did in V 12.3.

You need to add this setting at the top (this is new in V 13.0)

SetSystemOptions["IntegrateOptions" -> "UseIntegrateAlgebraic" -> False];

The default was True

As an example of one of the above


Mathematica graphics


IntegrateAlgebraic is very good in solving many integrals that could not be solved easily before and it does it fast, but it looks like there is a small clitch somewhere integrating IntegrateAlgebraic into V 13 Integrate command.

I also noticed I needed to start with Clean kernel when changing the this setting from True to False or vis verse for the new settings to take effect. Not sure why.

The V13 notebook showing they all now evaluate can be downloaded from here.

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Whenever you use assumptions, Mathematica V13 computes these integrals.

For example:

enter image description here

I think it's convenient to use assumptions, since the integral has several parameters that could be interpreted as complex by Mathematica.

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    $\begingroup$ Note that the \[Element] Reals at the end of your assumptions is (I think) unnecessary; telling Mathematica that {a > 0, c > 0, d > 0, e > 0} automatically includes the assumptions that they're real-valued. $\endgroup$ Feb 8, 2022 at 20:08
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    $\begingroup$ @Michael Seifert You're right, I appreciate the comment. $\endgroup$ Feb 8, 2022 at 21:23

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