Integrate of HypergeometricPFQ gives the wrong result

Question

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]

Thank you for your help.

Answer 1

As n is integer you can try

Table[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}], {n, 1, 10}]

for n up to 10 or higher.

The result -10Pi^2/3 for n=1 is just the integral over the term with the first Hypergeometric.