Uncaught SystemException returned to top level calling integrate in V 14.1

Question

Report to WRI [CASE:5169570]

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May be someone with better debugging skills than me in Mathematica internals could figure what causes this. This integral worked OK in V 14.0 but now in V 14.1 it gives

 SystemException[MemoryAllocationFailure]

With message in console saying

Throw::sysexc: Uncaught SystemException returned to top level. Can be caught with Catch[\[Ellipsis], _SystemException].

Clearly this is some kind of regression ( I will report to WRI). BUt may be someone here can figure what causes it or if there is something a user can do to fix it using some option setting?

Does it happen on Linux or the mac? I am using windows 10. I have 128 GB RAM. So memory should not an issue.

ClearAll["Global`*"];
integrand = (a + a*Sec[e + f*x])^m*(d*Sec[e + f*x])^(1/2);
Integrate[integrand, x]

Screen shots

To answer comment about if V 14.0 result was correct or not. I could not verify it. So it looks not correct.

(* V 14 on windows *)
ClearAll["Global`*"];
integrand = (a + a*Sec[e + f*x])^m*(d*Sec[e + f*x])^(1/2)
anti = Integrate[integrand, x];
PossibleZeroQ[RootReduce[Cancel[Together[D[anti,x]-integrand]]]]

(* False *)

Tried few other attempts, but can't verify it.

Update

I found second one

integrand = Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2)^p;
Integrate[integrand, x]

(*SystemException["MemoryAllocationFailure"]*)

But thanks to suggestion below by Carl, changing "IntegrateOptions" now the internal error is gone:

Answer 1

The execution takes exceptional long time. This is the case since vs 11 generally, if special complex functions with a tail of conditions on the domains are implied.

Instead of the earlier versions intention do deliver a formal algebraic expression in minimal time, now any internal call of a special function checks for cases of domains or known values in order to yield an exact equation.

As a poisoned gift to the unexperienced user, the new integration packages are buggy, too, stepwise repair left to complaints of users.

Version 6.0

   integrand = (a + a*Sec[e + f*x])^m*(d*Sec[e + f*x])^(1/2);
     Integrate[integrand, x]^2 // FullSimplify

$$\frac{4 d}{f^2}\ \tan ^2 \left(\frac{1}{2} (e+f x)\right) \left(1-\tan ^2\left(\frac{1}{2} (e+f x)\right)\right)^{2 m} (a (\sec (e+f x)+1))^{2 m} F_1\left(\frac{1}{2};m+\frac{1}{2},\frac{1}{2};\frac{3}{2};\tan ^2\left(\frac{1}{2} (e+f x)\right),-\tan ^2\left(\frac{1}{2} (e+f x)\right)\right){}^2$$

in zero sec.

We have a standard substitution case

 (t^2 -> Tan[y/2]^2,  Cos[y]->(1-t^2)/(1+t^2), 2 dt -> (1+t^2) dy} 

 integrand = (a + a*Sec[e + f*x])^m  (d*Sec[e + f*x])^(1/2) dx

 Assuming[m \[Element] PositiveIntegers && 0 < t < 1, 
  FullSimplify[
        (integrand //. 
           {e + f x -> y,  dx -> dy/f, 
              Sec[y] -> (1 + t^2)/(1 - t^2), 
               dy -> (2 dt)/(1 + t^2)})^2]]

$$-\frac{d 4^{m+1} \left(\frac{a}{t^2-1}\right)^{2 m}}{f^2 \left(t^4-1\right)} \ \text{dt}^2$$

Rearranging and concentrating to the integral

  \[Integral]1  /( Sqrt[1 + t^2] (1 - t^2)^(m + 1/2)) \[DifferentialD]t //
   FullSimplify

$$t \ F_1\left(\frac{1}{2};m+\frac{1}{2},\frac{1}{2};\frac{3}{2};t^2,-t^2\right) /. \{t \to \tan\left(\frac{1}{2}(e + f*x)\right\}$$

Caveat: I didn't check anything.