Dtyjlui

Let $x_0,x in (0,1/10)$ and define:
begin{equation}
g(x):= F_{2,1}left[frac{1}{13},frac{1}{17},frac{1}{5}; 100 x^2 right]
end{equation}
Then the following identity holds true:
begin{eqnarray}
&&intlimits_{x_0}^x frac{left(2404500 xi ^3+13746115 xi ^2-47073 xi -161109right) }{(1-10 xi )^{5513/7140} xi ^{116/35} (10 xi +1)^{5683/7140}} g(xi )dxi =\
&&
left.frac{(1-10 x)^{1627/7140} (10 x+1)^{1457/7140} }{x^{116/35}} left(-97461 (10 x-1) (10 x+1) x^2 g'(x)-1785 left(840 x^2-13 x-39right) x g(x)right) right|_{x_0}^x
end{eqnarray}

In[815]:= x =.; x0 =.; xi =.; Clear[g]; Clear[F]; Clear[f]; Clear[f1];

g[x_] := Hypergeometric2F1[1/13, 1/17, 1/5, 100 x^2];

F[x_] := ((1 - 10 x)^(1627/7140) (1 + 10 x)^(1457/7140))/x^(

   116/35) (- 97461 x^2 (-1 + 10 x) (1 + 10 x) g'[x] - 

     1785 x (-39 - 13 x + 840 x^2) g[x]);

f[x_] := NIntegrate[ (1 - 10 xi)^(1627/7140 - 1) (1 + 10 xi)^(

    1457/7140 - 

     1) (-161109 - 47073 xi + 13746115 xi^2 + 2404500 xi^3)/( xi^(

      116/35) ) g[xi], {xi, x0, x}, WorkingPrecision -> 40];



{x, x0} = RandomReal[{0, 1/10}, 2, WorkingPrecision -> 40];

f1[x_] := F[x] - F[x0];

{f[x], f1[x]} // MatrixForm

This identity comes on one hand from applying the inverse gauge transformation (see https://www.math.fsu.edu/~eimamogl/hypergeometricsols/hypergeometricsols.mpl) to the differential operator:
begin{equation}
L:=frac{d^2}{d x^2} + frac{140500 x^2+663}{1105 x (10 x-1) (10 x+1)} frac{d}{d x}+ frac{400}{221 (10 x-1) (10 x+1)}
end{equation}
and to the gauge
begin{equation}
G:=-frac{39 left(1785 left(100 x^2-1right) x frac{d}{d x} -514500 x^2+425 x+5916right)}{x left(2404500 x^3+13746115 x^2-47073 x-161109right)}
end{equation}

On the other hand the identity in question comes from solving the in-homogeneous first order ODE
begin{equation}
G left[f(x) right] = g(x)
end{equation}
by the standard way of variation of constants.

Note 1: The function $g(x)$ satisfies the equation:
begin{equation}
Lleft[ g(x) right] =0
end{equation}

In[855]:= x =.; FullSimplify[(D[#, {x, 2}] + (663 + 140500 x^2)/(

      1105 x (-1 + 10 x) (1 + 10 x)) D[#, x] + 

     400/(221 (-1 + 10 x) (1 + 10 x)) #) & /@ {Hypergeometric2F1[1/13,

     1/17, 1/5, 100 x^2]}]



Out[855]= {0}

Note 2: The gauge $G$ corresponds to the first element of the global integral basis of the operator $L$, a basis which has been normalized at infinity (see https://arxiv.org/pdf/1606.01576.pdf for details).

Update: Here is actually another example of a similar integral identity:
Now define:
begin{equation}
g(x):= F_{2,1}left[frac{1}{11},frac{1}{3},frac{1}{2}; 100 x^2 right]
end{equation}
Then the following holds:
begin{eqnarray}
&&!!!!!!!!intlimits_{x_0}^x frac{sqrt[3]{1-10 xi } (10 xi +1)^{2/15} left(42500 xi ^3+170850 xi ^2-4620 xi -4389right)}{1155 xi ^{24/5}} g(xi) d xi = \
&&!!!!!!!!frac{(1-10 x)^{4/3} (10 x+1)^{17/15}}{x^{24/5}}left.left( -frac{5}{14} (10 x-1) (10 x+1) x^2 g'(x)-frac{1}{77} left(500 x^2-55 x-77right) x g(x)right)right|_{x_0}^x
end{eqnarray}

g[x_] := Hypergeometric2F1[1/11, 1/3, 1/2, 100 x^2];

F[x_] := ((1 - 10 x)^(4/3) (1 + 10 x)^(17/15))/x^(

   24/5) (-(5/14) x^2 (-1 + 10 x) (1 + 10 x) g'[x] - 

     1/77 x (-77 - 55 x + 500 x^2) g[x]);

f[x_] := (NIntegrate[( (1 - 10 xi)^(1/3) (1 + 10 xi)^(

      2/15) (-4389 - 4620 xi + 170850 xi^2 + 42500 xi^3))/(

     1155 xi^(24/5)) g[xi], {xi, x0, x}, WorkingPrecision -> 40]);

{x, x0} = RandomReal[{0, 1/10}, 2, WorkingPrecision -> 40];

{f[x], F[x] - F[x0]} // MatrixForm

The identity above has been derived in exactly the same way as the one before except that now we started from an abscissa transformed $x rightarrow 100 x^2$ Gaussian hypergeometric operator with parameters $(a,b,c)=(1/11,1/3,1/2)$ and then used the first element of the infinity-normalized global integral basis to construct the relevant inverse gauge transformation.

Update 1:
Here are two other examples which I generated this morning.

Firstly we define:
begin{equation}
g(x):=F_{2,1}left[-frac{1}{4},-frac{1}{6},frac{1}{2};9 x^2right]
end{equation}
Then the following holds true:
begin{eqnarray}
&&intlimits_{x_0}^x frac{sqrt[12]{1-9 xi ^2} e^{-4 sqrt{frac{2}{97}} tan ^{-1}left(sqrt{frac{2}{97}} (5 xi +8)right)}}{left(10 xi ^2+32 xi +45right)^{3/4}}
frac{45 left(2400 xi ^3+10547 xi ^2+8640 xi +5625right)}{left(90 xi ^4+288 xi ^3+395 xi ^2-32 xi -45right)}
g(xi) dxi=\
&&left.frac{sqrt[12]{1-9 x^2} e^{-4 sqrt{frac{2}{97}} tan ^{-1}left(sqrt{frac{2}{97}} (5 x+8)right)}}{x left(10 x^2+32 x+45right)^{3/4}}
left(450 x^2 g(x)-90 x left(10 x^2+32 x+45right) g'(x)right)right|_{x_0}^x
end{eqnarray}

g[x_] := Hypergeometric2F1[-1/4, -1/6, 1/2, 9 x^2];

F[x_] := (1 - 9 x^2)^(1/12)/(

   E^(4 Sqrt[2/97] ArcTan[Sqrt[2/97] (8 + 5 x)])

     x (45 + 32 x + 10 x^2)^(

    3/4)) (-90 x (45 + 32 x + 10 x^2) g'[x] + 450 x^2 g[x]);

{x0, x} = RandomReal[{0, 1/3}, 2, WorkingPrecision -> 40];

X1 = NIntegrate[ ((1 - 9 xi^2)^(1/12))/(

   E^(4 Sqrt[2/97]

      ArcTan[Sqrt[2/97] (8 + 5 xi)]) (45 + 32 xi + 10 xi^2)^(

    3/4)) (45  (5625 + 8640 xi + 10547 xi^2 + 2400 xi^3))/(-45 - 

      32 xi + 395 xi^2 + 288 xi^3 + 90 xi^4) g[xi], {xi, x0, x}, 

  WorkingPrecision -> 40]

X2 = F[x] - F[x0]

Secondly we define:
begin{equation}
g(x):=x^{1/3} F_{2,1}left[-frac{2}{15},-frac{1}{12},frac{2}{3};49 x^2right]
end{equation}
Then the following holds true:
begin{eqnarray}
&&intlimits_{x_0}^x left(-49 xi +2 sqrt{246}+2right)^{frac{sqrt{246}-1599}{1845}} left(49 xi +2 sqrt{246}-2right)^{-frac{13}{15}-frac{sqrt{frac{2}{123}}}{15}} cdot\
&&
left(1-49 xi ^2right)^{7/60}
frac{20 left(-1323 xi ^3-13798 xi ^2+210 xi +700right)}{2401 xi ^4-196 xi ^3-1029 xi ^2+4 xi +20} g(xi) dxi=\
&&
left.frac{left(-49 x+2 sqrt{246}+2right)^{frac{sqrt{frac{2}{123}}}{15}-frac{13}{15}} left(49 x+2 sqrt{246}-2right)^{frac{-1599-sqrt{246}}{1845}} left(1-49 x^2right)^{7/60}}{x}
left(frac{75}{98} left(441 x^2-20 x-100right) g(x)-frac{1125}{98} x left(49 x^2-4 x-20right) g'(x)right)right|_{x_0}^x
end{eqnarray}

g[x_] := x^(1/3) Hypergeometric2F1[-(2/15), -(1/12), 2/3, 49 x^2];

F[x_] := (1 - 49 x^2)^(

   7/60)/((2 + 2 Sqrt[246] - 49 x)^(13/15 - Sqrt[2/123]/15)

     x (-2 + 2 Sqrt[246] + 49 x)^((1599 + Sqrt[246])/

    1845)) (-(1125/98) x (-20 - 4 x + 49 x^2) g'[x] + 

     75/98 (-100 - 20 x + 441 x^2) g[x]);

{x0, x} = RandomReal[{0, 1/7}, 2, WorkingPrecision -> 40];

NIntegrate[(1 - 49 xi^2)^(

  7/60)/((2 + 2 Sqrt[246] - 49 xi)^((1599 - Sqrt[246])/

   1845) (-2 + 2 Sqrt[246] + 49 xi)^(13/15 + Sqrt[2/123]/15)) (

  20     (700 + 210 xi - 13798 xi^2 - 1323 xi^3))/(

  20 + 4 xi - 1029 xi^2 - 196 xi^3 + 2401 xi^4) g[xi], {xi, x0, x}, 

 WorkingPrecision -> 40]



F[x] - F[x0]

Update 2: In order to avoid confusion let me write down in a concise way how all those results were generated.
Let $L:=d^2/d x^2 + a_1(x) d/d x + a_0(x)$ and $G:=r_1(x) d/d x + r_0(x)$ be a pair of differential operators or order two and order one respectively with coefficients being rational functions in $x$.
Then let $Q(x):= int r_0(x)/r_1(x) dx$ and let $(R_0(x),R_1(x))$ be the inverse gauge of the operator $L$ and the gauge $G$. In other words the functions $(R_0(x),R_1(x))$ are such that:
begin{equation}
left( r_0(x) + r_1(x) frac{d}{d x}right)left(R_0(x) + R_1(x) frac{d}{d x}right) f(x) = L f(x) quad (i)
end{equation}
Note that the inverse gauge can always be found by making a polynomial ansatz for $R_0(x),R_1(x)$ inserting that ansatz to the equation above and then by matching the coefficients at respective derivatives solving for the unknown polynomial coefficients.

Now let the function $g(x)$ solve the ODE below:
begin{equation}
L g(x) = 0 quad (ii)
end{equation}

Then the following identity holds true:
begin{equation}
intlimits_{x_0}^x frac{exp(Q(xi)}{r_1(xi)} g(xi) dxi =
exp(Q(x)) left.left[ R_0(x) g(x) + R_1(x) g^{'}(x) right] right|_{x_0}^x
end{equation}

Note that the proof of this statement follows in a straightforward way from differentiating the right hand side and then using the properties $(i)$ and $(ii)$. As such this identity is not astounding . However the tricky parts is to find the inverse gauge and also to solve the ODE $(ii)$.

Having said all this my final question is now how do you prove such identities as above in a different way?