# Integrate of HypergeometricPFQ gives the wrong result

Good morning, I computed the following integral using Integrate in version 12.2 and it gives the wrong result. Can you help me understand what I am doing wrong? Here is the integral:

Integrate[(
n (1 + n) (1 + 2 n) (3 +
2 n) (-HypergeometricPFQ[{1, 1, 1 - 2 n, 4 + 2 n}, {2, 3, 3}, 1] +
t HypergeometricPFQ[{1, 1, 1 - 2 n, 4 + 2 n}, {2, 3, 3}, t]) Log[
t])/(-1 + t), {t, 0, 1}, {u, 0, 1},
Assumptions -> n \[Element] Integers && n > 0]


the result is

-(1/6) n (1 + n) (1 + 2 n) (3 + 2 n) \[Pi]^2 HypergeometricPFQ[{1, 1,
1 - 2 n, 4 + 2 n}, {2, 3, 3}, 1]


For instance, this result is $$-\frac{10 \pi ^2}{3}$$ for $$n=1$$. However if I set $$n=1$$ inside Integrate or use NIntegrate i get $$-17.5$$, which is the correct result. Why is there such a difference? Here is the code

NIntegrate[((
n (1 + n) (1 + 2 n) (3 +
2 n) (-HypergeometricPFQ[{1, 1, 1 - 2 n, 4 + 2 n}, {2, 3, 3},
1] + t HypergeometricPFQ[{1, 1, 1 - 2 n, 4 + 2 n}, {2, 3, 3},
t]) Log[t])/(-1 + t)) /. n -> 1, {t, 0, 1}, {u, 0, 1}]


and

Integrate[(
n (1 + n) (1 + 2 n) (3 +
2 n) (-HypergeometricPFQ[{1, 1, 1 - 2 n, 4 + 2 n}, {2, 3, 3},
1] + t HypergeometricPFQ[{1, 1, 1 - 2 n, 4 + 2 n}, {2, 3, 3},
t]) Log[t])/(-1 + t) /. n -> 1, {t, 0, 1}, {u, 0, 1},
Assumptions -> n \[Element] Integers && n > 0]


• A possible bug. In my version (12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)), the first integral returns unevaluated, and the second two return -17.5 and -35/2. What version are you using? Sep 23, 2021 at 15:45
• This returns unevaluated in 12.3.1 on windows 10. May be there was a bug and WRI fixed it? screen shot !Mathematica graphics it is always best to use latest version of software. Sep 23, 2021 at 15:45
• This also returns -35/2 in MM 12.3.1 for Mac OS (ARM). Sep 23, 2021 at 21:02
• Remains unevaluated for me in "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" as well as in V12.3.1. Sep 25, 2021 at 13:36

Table[Integrate[(n (1 + n) (1 + 2 n) (3 + 2 n)