Infinite series with harmonic numbers related to elliptic integrals
Let me share some results for the integral form of the series:
$$S(p)=\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{H_n}{2n+1} p^n$$
Where $p$ can be $ \pm 1$ or any other number provided the series converges.
First, we will use the integral expression for $H_n$:
$$H_n=\int_0^1 \frac{1-x^n}{1-x} dx$$
Now we exchange the summation and the integration, keeping in mind that the terms inside the integral diverge separately and we are only allowed to consider them as such under the integral sign.
$$S(p)=\int_0^1 \frac{dx}{1-x} \left( \sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n (2n+1)} p^n-\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n (2n+1)} (px)^n \right)$$
In other words, we need to find the closed form for:
$$S_1(p)=\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n (2n+1)} p^n$$
To get this into the hypergeometric form we transform the general term:
$$\frac{(\tfrac34)_n}{(\tfrac54)_n (2n+1)}=\frac{(\tfrac34)_n (1)_n}{2(\tfrac54)_n (n+1/2)n!}=\frac{1}{2} \frac{\Gamma(1/2)}{\Gamma(3/2)} \frac{(\tfrac34)_n (1)_n (\tfrac12)_n}{(\tfrac54)_n (\tfrac32)_n n!}$$
In other words:
$$S_1(p)={_3 F_2} \left( 1, \frac{1}{2}, \frac{3}{4}; \frac{3}{2}, \frac{5}{4}; p \right)-1$$
Which means:
$$S(p)=\int_0^1 \frac{dx}{1-x} \left( {_3 F_2} \left( 1, \frac{1}{2}, \frac{3}{4}; \frac{3}{2}, \frac{5}{4}; p \right)-{_3 F_2} \left( 1, \frac{1}{2}, \frac{3}{4}; \frac{3}{2}, \frac{5}{4}; px \right) \right)$$
To simplify this further and get rid of special functions, let's use the reduction formula (see Wikipedia):
$${_3 F_2} \left( a_1, a_2, a_3; b_1, b_2; p \right)=\frac{\Gamma(b_2)}{\Gamma(a_3) \Gamma(b_2-a_3)} \int_0^1 t^{a_3-1} (1-t)^{b_2-a_3-1} {_2 F_1} \left( a_1, a_2; b_1; p t \right) dt$$
We are free to choose which parameters to reduce, so let's get rid of the most complicated ones:
$${_3 F_2} \left( 1, \frac{1}{2}, \frac{3}{4}; \frac{3}{2}, \frac{5}{4}; p \right)=\frac{\Gamma(5/4)}{\Gamma(3/4) \Gamma(1/2)} \int_0^1 t^{-1/4} (1-t)^{-1/2} {_2 F_1} \left( 1, \frac{1}{2}; \frac{3}{2}; p t \right) dt$$
And finally, the wonderful thing is:
$${_2 F_1} \left( 1, \frac{1}{2}; \frac{3}{2}; p t \right)=\frac{\tanh^{-1} \sqrt{p t} }{\sqrt{p t}}$$
So:
$$S(p)=\frac{\Gamma(5/4)}{\Gamma(3/4) \sqrt{ \pi}} \int_0^1 \int_0^1 \frac{dx dt}{(1-x)t^{1/4} \sqrt{1-t}} \left( \frac{\tanh^{-1} \sqrt{p t} }{\sqrt{p t}} - \frac{\tanh^{-1} \sqrt{p x t} }{\sqrt{p x t}} \right) $$
We obtain the integral form valid for all the sums (5) and (6).
In addition, the arctangents also have a simple integral form, which makes this function expressible as a triple integral of algebraic functions.
$$\frac{\tanh^{-1} \sqrt{p t} }{\sqrt{p t}}=\int_0^1 \frac{dy}{1-pty^2}$$
After some simple transformation we get the following triple integral formula over a unit cube:
$$S(p)=\frac{\Gamma(5/4)}{\Gamma(3/4) \sqrt{ \pi}} p \int_0^1 \int_0^1 \int_0^1 \frac{t^{3/4} y^2 ~ dx dt dy}{\sqrt{1-t}(1-p t y^2)(1-p x t y^2)} $$
This result checks out numerically, for example:
Integrating w.r.t. $x$ we get another double integral representation:
$$S(p)=-\frac{\Gamma(5/4)}{\Gamma(3/4) \sqrt{ \pi}} \int_0^1 \int_0^1 \frac{ \log (1-p t y^2) dt dy}{t^{1/4} \sqrt{1-t}(1-p t y^2)} $$
The Clausen's product formula is (see [1] pg.188) $$ \left({}_2F_1\left(a,b;a+b+1/2;z\right)\right)^2={}_3F_2(2a,a+b,2b;a+b+1/2,2a+2b;z)\tag 1 $$ Set in $(1)$ $a=1/2$, $b=1/4$, then $$ {}_3F_2\left(1,\frac{1}{2},\frac{3}{4};\frac{5}{4},\frac{3}{2};z\right)=\left({}_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};z\right)\right)^2\tag 2 $$ Now from [2] pg.87 Entry 30, we have the next theorem of Ramanujan:
Theorem (Ramanujan).
Let $\alpha+\beta+1=\gamma+\delta$, $c=\Gamma(\alpha)\Gamma(\delta)/\{\Gamma(\gamma)\Gamma(\delta)\}$, and $$ y=c\frac{{}_2F_1\left(\alpha,\beta;\delta;1-x\right)}{{}_2F_1\left(\alpha,\beta;\gamma;x\right)}. $$ Then $$ y'=-\frac{x^{-\gamma}(1-x)^{-\delta}}{{}_2F_1^2(\alpha,\beta;\gamma;x)}. $$
Using the theorem of Ramanujan with $\alpha=1/4$, $\beta=1/2$, $\delta=5/4$, $\gamma=1/2$, we get $$ c=\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} $$ and because ${}_2F_1\left(\alpha,\beta;\gamma;x\right)=\frac{1}{\sqrt[4]{1-x}}$, we get $$ y'=-\frac{1}{(1-x)^{3/4}\sqrt{x}}. $$ Also $$ y=\frac{(1-x)^{1/4}\Gamma\left(\frac{1}{4}\right){}_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};1-x\right)}{\Gamma\left(\frac{5}{4}\right)}=-\int^{x}_{c_0}\frac{1}{(1-t)^{3/4}\sqrt{t}}dt $$ Hence $$ {}_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};1-x\right)=-\frac{\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}(1-x)^{-1/4}\int^{x}_{c_0}\frac{1}{(1-t)^{3/4}\sqrt{t}}dt. $$ Hence $$ {}_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};x\right)=\frac{1}{4x^{1/4}}\int^{x}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\tag 3 $$ Hence $$ {}_3F_2\left(1,\frac{1}{2},\frac{3}{4};\frac{5}{4},\frac{3}{2};x\right)=\left(\frac{1}{4x^{1/4}}\int^{x}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\right)^2\tag 4 $$ Setting this into $S(p)$ in the answer of Yuriy S, we get $$ S(p)= $$ $$ \int^{1}_{0}\left(\frac{1}{1-x}\left(\frac{1}{4p^{1/4}}\int^{p}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\right)^2-\frac{1}{1-x}\left(\frac{1}{4(xp)^{1/4}}\int^{xp}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\right)^2\right)dx= $$ $$ =\frac{1}{16\sqrt{p}}\int^{1}_{0}\frac{1}{1-x}\left[\left(\int^{p}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\right)^2-\left(\int^{p}_{0}\frac{dt}{t^{3/4}\sqrt{1-tx}}\right)^2\right]dx\tag 5 $$ Hence $$ \frac{d}{dp}(\sqrt{p}S(p))=\int^{1}_{0}\frac{1}{1-x}\frac{{}_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};p\right) \sqrt{1-p x}-\sqrt{1-p}\cdot {}_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};p x\right)}{2 \sqrt{(1-p) p} \sqrt{1-p x}}dx= $$ $$ =\frac{1}{2\sqrt{p}}\int^{1}_{0}\frac{1}{1-x}\left(\frac{{}_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};p\right)}{ \sqrt{1-p}}-\frac{{}_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};p x\right)}{\sqrt{1-p x}}\right)dx.\tag 6 $$
References
[1]: J.M. Borwein and P.B. Borwein. ''Pi and the AGM''. 1987 ed., New York: John Wiley and Sons Inc., (1987).
[2]: B.C. Berndt. ''Ramanujan`s Notebooks Part II''. Springer-Verlag, New York. (1989).