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I do not know if this is related to this issue Why adding TimeConstrained to Integrate causes TerminatedEvaluation["RecursionLimit"]? or not. If you think it is the same, will append this to the above question. But these integrals return TerminatedEvaluation["RecursionLimit" as part of the anti-derivative itself which is different from the above question.

I am finding in V 13.2.1 many many integrals now return result but with TerminatedEvaluation["RecursionLimit"] as part of the anti-derivative!

Here are few examples of dozens found so far. This is on windows 10 pro.

enter image description here

Any one can find a workaround this? Is it related to the same problem in the above linked to question? This seems to be new in V 13.2.1 as in V 13.1 this used to work OK

enter image description here

Could someone with V 13.3 development version verify if this is fixed?

Also send email to WRI support now in case this is new issue. ID [CASE:5023624]

Code

Integrate[x^2*(a + b*Log[c*(d + e/Sqrt[x])^n])^2, x]
Integrate[(e*Sin[c + d*x])^(3/2)/(a + b*Cos[c + d*x])^2, x]
Integrate[(d*Sec[e + f*x])^(5/2)/Sqrt[b*Tan[e + f*x]], x]
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    $\begingroup$ Probably the same issue. They all appear to behave in 13.3. $\endgroup$ Commented Apr 23, 2023 at 16:26
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    $\begingroup$ 13.3.0 is live in the Cloud and it works now. $\endgroup$ Commented Jun 26, 2023 at 1:21
  • $\begingroup$ @ВалерийЗаподовников Thanks. Verified. $\endgroup$
    – Nasser
    Commented Jun 26, 2023 at 2:13

1 Answer 1

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$Version

(* "13.2.1 for Mac OS X ARM (64-bit) (January 27, 2023)" *)

Clear["Global`*"]

expr = x^2*(a + b*Log[c*(d + e/Sqrt[x])^n])^2;

Integrate[expr, x]

(* 0 *)

Assuming[x != 0, Integrate[expr, x]]

(* 0 *)

Assuming[x > 0, Integrate[expr, x]]

(* 0 *)

int = Assuming[x < 0, Integrate[expr, x]]

(* (a^2 x^3)/3 - (1/(180 d^6))
 b (d e n (b e n (-154 e^3 Sqrt[x] - 47 d e^2 x - 18 d^2 e x^(3/2) - 
          6 d^3 x^2) + 
       2 a (60 e^4 Sqrt[x] + 30 d e^3 x + 20 d^2 e^2 x^(3/2) + 15 d^3 e x^2 + 
          12 d^4 x^(5/2))) + 
    2 b d e n (60 e^4 Sqrt[x] + 30 d e^3 x + 20 d^2 e^2 x^(3/2) + 
       15 d^3 e x^2 + 12 d^4 x^(5/2)) Log[c (d - e/Sqrt[x])^n] - 
    60 b d^6 x^3 Log[c (d - e/Sqrt[x])^n]^2 - 
    60 b e^6 n^2 Log[-e + d Sqrt[x]]^2 + 
    2 e^6 n Log[-e + d Sqrt[x]] (60 a - 137 b n + 
       60 b Log[c (d - e/Sqrt[x])^n] + 60 b n Log[(d Sqrt[x])/e]) - 
    120 a d^6 x^3 Log[c (-e + d Sqrt[x])^n x^(-n/2)] + 
    120 b e^6 n^2 PolyLog[2, 1 - (d Sqrt[x])/e]) *)

However, int does not appear to be a valid anti-derivative of expr

expr == D[int, x] // FullSimplify[#, x < 0] &

(* x^2 (a + b Log[c (d + e/Sqrt[x])^n])^2 == 
 x^2 (b^2 Log[c (d - e/Sqrt[x])^n]^2 + 
    a (a + 2 b Log[c (-e + d Sqrt[x])^n x^(-n/2)])) *)
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    $\begingroup$ It is fixed in V 13.3, yes verified in the cloud. $\endgroup$
    – Nasser
    Commented Jun 26, 2023 at 2:13

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