I've encountered the following product, \begin{equation} \, _2F_1\left(3 d+2,3 d+2;6 d+4;1-\frac{1}{t^2}\right) \, _3F_2\left(-\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;t^2\right) \end{equation} of rather highly-structured hypergeometric functions and speculate that it might have some still further simplified form. This product is a factor in an integrand, \begin{equation} w(t,d)= t^{-3 (d+1)} \left(t^2-1\right)^d \, _2F_1\left(3 d+2,3 d+2;6 d+4;1-\frac{1}{t^2}\right) \, _3F_2\left(-\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;t^2\right), \end{equation} I am trying to integrate over $t \in [0,1]$, with $d$ being a free parameter, the positive integer values of which are the ones of (quantum-information-theoretic, https://arxiv.org/abs/1610.01410) interest.

I investigated using the transformation 15.8.3 in http://dlmf.nist.gov/15.8#E1 , which would convert the 2F1 argument $1-\frac{1}{t^2}$ to $t^2$, as in the 3F2 function, and which then might lead to some simplification, but this seemed to produce singularities due to the equality between the first two entries of the Gauss hypergeometric function. I seem to be able to transform the argument to $1-t^2$, but not $t^2$ itself.

After changing variables from $t$ to $t=\sqrt{T}$, for $d=2$, if one performs (in Mathematica) an indefinite integration of \begin{equation} \tilde{w}(T,d) =\frac{1}{2} (T-1)^d T^{-\frac{3 d}{2}-2} \, _2\tilde{F}_1\left(3 d+2,3 d+2;6 d+4;\frac{T-1}{T}\right) \, _3\tilde{F}_2\left(-\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;T\right), \end{equation} over $T$ for $d=2$, one obtains

(37 - 444 T + 2442 T^2 - 8140 T^3 + 7950 T^4 - 233982 T^5 - 572866 T^6 + 19458 T^7 + 729165 T^8 + 127450 T^9 - 62280 T^10 - 8790 T^11 + 2100 T^4 Log[1/T] + 78540 T^5 Log[1/T] + 508620 T^6 Log[1/T] + 901740 T^7 Log[1/T] + 401100 T^8 Log[1/T] - 52500 T^9 Log[1/T] - 35700 T^10 Log[1/T] - 2100 T^11 Log[1/T])/(2560481280000 (-1 + T)^12)

But for $d=4,6,8,\ldots$ (odd $d$'s are intractable), one gets sums of ($1 +\frac{3d}{2}$) MeijerG functions divided by $T^{1 +\frac{3d}{2}}$, descending from \begin{equation} T^{-\frac{3 d}{2}-1} G_{3,3}^{2,3}\left(\frac{1}{T}| \begin{array}{c} -3 d-1,-3 d-1,-\frac{3 d}{2} \\ 0,0,-\frac{3 d}{2}-1 \\ \end{array} \right) \end{equation} to \begin{equation} G_{3,3}^{2,3}\left(\frac{1}{T}| \begin{array}{c} 1,-3 d,-3 d \\ 1,1,0 \\ \end{array} \right), \end{equation} such as, for $d=4$,

```
-(MeijerG[{{-13, -13, -6}, {}}, {{0, 0}, {-7}}, 1/T]/(
```

4330257806605605116960008147304448000000000 T^7)) + MeijerG[{{-13, -13, -5}, {}}, {{0, 0}, {-6}}, 1/T]/( 909354139387177074561601710933934080000000 T^6) - ( 31 MeijerG[{{-13, -13, -4}, {}}, {{0, 0}, {-5}}, 1/T])/( 14549666230194833192985627374942945280000000 T^5) + MeijerG[{{-13, -13, -3}, {}}, {{0, 0}, {-4}}, 1/T]/( 466335456095988243364923954325094400000000 T^4) - ( 43 MeijerG[{{-13, -13, -2}, {}}, {{0, 0}, {-3}}, 1/T])/( 36374165575487082982464068437357363200000000 T^3) + ( 31 MeijerG[{{-13, -13, -1}, {}}, {{0, 0}, {-2}}, 1/T])/( 90935413938717707456160171093393408000000000 T^2) - MeijerG[{{-12, -12, 1}, {}}, {{1, 1}, {0}}, 1/ T]/24249443716991388654976045624904908800000000

I am studying the structure of such results ($d=4, 6, 8,\ldots,150$) and have found (using the Mathematica FindSequenceFunction command) that the coefficients of the highest power ($3 d-5$) of $T$ occurring in the denominators of the MeijerG terms follow the rule \begin{equation} \frac{2^{8 d+1} 3^{\frac{9 d}{2}+3} \Gamma \left(\frac{d}{2}+\frac{5}{6}\right)^3 \Gamma \left(\frac{d}{2}+\frac{7}{6}\right)^3 \Gamma \left(\frac{d+1}{2}\right)^2 \Gamma (d+1) \Gamma \left(\frac{3 d}{2}+1\right)^4 \Gamma (3 d+2)}{\pi ^4}, \end{equation} and the coefficients of the second-highest power ($3 d-6$) of $T$, the rule \begin{equation} \frac{3^{9 d+\frac{13}{2}} \Gamma \left(\frac{d}{2}+2\right) \Gamma \left(\frac{3 d}{2}\right) \Gamma \left(d+\frac{2}{3}\right)^2 \Gamma (d+1)^3 \Gamma \left(d+\frac{4}{3}\right)^3 \Gamma \left(d+\frac{5}{3}\right) \Gamma (3 d+2)}{16 \pi ^3 (d+1)^2}. \end{equation} The ratio of this expression to the previous one is \begin{equation} \frac{2^{-d-1} 3^{\frac{3 d}{2}+\frac{1}{2}} \Gamma \left(\frac{d}{2}+\frac{2}{3}\right) \Gamma \left(\frac{d}{2}+\frac{4}{3}\right) \Gamma (d+3)}{\sqrt{\pi } d^2 (d+1)^3 \Gamma \left(\frac{3 d}{2}\right) \Gamma \left(\frac{d+1}{2}\right)}. \end{equation}

In fact, the numerators of the coefficients of the highest ($3 d-5$) power are all -1, and of the second highest ($3 d-6$) power of $T$ occurring in the denominator are +1. The alternation in sign continues as the coefficients of the third, fourth,...highest powers are examined, but the numerators cease to be just -1 or 1.

The reciprocal of the coefficients of the third highest ($3 d-7$) power of $T$ occurring in the denominator follow the rule \begin{equation} -\frac{2^{8 d} 27^{3 \left(\frac{d}{2}-1\right)+4} \left(6 \left(\frac{d}{2}-1\right)+7\right) \left(\frac{d}{2}+2\right) \Gamma \left(2 \left(\frac{d}{2}-1\right)+3\right) \Gamma \left(3 \left(\frac{d}{2}-1\right)+4\right)^3 \Gamma \left(3 \left(\frac{d}{2}-1\right)+7\right) \Gamma \left(6 \left(\frac{d}{2}-1\right)+6\right) \Gamma \left(\frac{d}{2}+\frac{1}{2}\right)^2 \Gamma \left(\frac{d}{2}+\frac{5}{6}\right)^3 \Gamma \left(\frac{d}{2}+\frac{7}{6}\right)^3}{\pi ^4 \left(2 \left(\frac{d}{2}-1\right)+3\right)^2 \left(2 \left(\left(\frac{d}{2}-1\right) \left(\frac{d}{2}+4\right)+7\right) \left(\frac{d}{2}-1\right)+5\right)}. \end{equation} And similarly, for the fourth highest ($3 d-8$) power, we have, more succinctly, \begin{equation} \frac{144 \Gamma \left(\frac{d}{2}+4\right) \Gamma \left(\frac{3 d}{2}+3\right) \Gamma (3 d+2)^4}{d (d+1)^2 (d (d (d (8 d (d+8)+49)-322)-96)+288)}. \end{equation} We are examining the coefficients of still further lower powers.

As noted for even $d$ (including for $d=2$, as we have found), one gets sums of ($1 +\frac{3d}{2}$) MeijerG functions divided by $T^{1 +\frac{3d}{2}}$. We have observed that for each of $d=2,4,6,...222$, the sum of the coefficients of these ($1 +\frac{3d}{2}$) terms is zero.