More identities of the Ramanujan Double factorial type. Ramanujan discovered the following identity 
$$x=\sum_{n=0}^\infty (-1)^n \left[\dfrac{(2n-1)!!}{(2n)!!}\right]^3=\left[\dfrac{\Gamma\left(\frac98\right)}{\Gamma\left(\frac54\right)\Gamma\left(\frac78\right)}\right]^{2}$$
which contained the double factorials . Are there any further identities of this type ? or probably its generalization .

A possible class of such squared-double factorial sums just comes directly from the $_2F_1$ hypergeometric expansion of $K = K(k)$, i.e., 
$$\frac{2 K}{\pi} = {}_2F_1\left(\frac12,\frac12;1;k^2\right) = \sum_{n \geq 0} \left [ \frac{(2n-1)!!}{(2n)!!} \right ]^2 k^{2n}\tag{1}$$
Applying respectively Kummer's transform and then Clausen transform gives a cubed-double factorial class
\begin{align*}\frac{4K^2}{\pi^2} & = {}_3F_2 \left( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;(2kk')^2\right) \\ & = \sum_{n \geq 0} \left [ \frac{(2n-1)!!}{(2n)!!} \right ]^3 (2kk')^{2n} \tag{2} \end{align*}
This is the key to Ramanujan's identity, and presumably this answers your question about generalizations. Another cubic class is
$$\sum_{n\geq0} \left [ \frac{(2n-1)!!}{(2n)!!} \right ]^3\!\!\!(6n+1)(2kk')^{2n}=\frac{4K(k)^2}{\pi^2}+\frac{24K(k)}{\pi^2}\left[\frac{E(k)-k' K(k)}{(1-2k^2) } \right] \tag{3}$$
An interesting consequence is the identity 
$$\frac{4}{\pi}=\sum_{n\geq 0} \frac{6n+1}{4^n} \left [ \frac{(2n-1)!!}{(2n)!!} \right ]^3$$

Proof of $(2)$ $:$ Kummer's transformation gives us, in general
$$_{2}F_{1}(2a, 2b; a+b+1/2; x) = {}_{2}F_{1}(a, b; a+b+1/2; 4x(1-x))$$
So inserting $(a, b, x) = (1/4, 1/4, k^2)$ clearly gives
$$_2F_1(1/2, 1/2; 1; k^2) = {}_2F_1(1/4, 1/4; 1; 4k^2k'^2) \tag{+}$$
The trick now is to square the right hand side of $(+)$ and use Clausen transform to note that
$$\left[_2F_1(1/4, 1/4;1;4k^2k'^2)\right]^2 = {}_3F_2(1/2,1/2,1/2;1,1;4k^2k'^2)$$
Giving the equality at last, and thus proving $(2) \;\;\; \blacksquare$
Proof of $(3)$ $:$ Differentiating $(2)$ with respect to $k$ gives 
$$\frac{4(1-2k^2)}{k'}\sum_{n \geq 0} (2kk')^{2n-1} \cdot n \cdot \left [ \frac{(2n-1)!!}{(2n)!!} \right ]^3 = \frac{8KK'}{\pi^2}$$
Where $K' = \frac{d}{dk}K(k)$. The above is clearly equivalent to 
$$\sum_{n \geq 0} (2kk')^{2n} \cdot n \cdot \left [ \frac{(2n-1)!!}{(2n)!!} \right ]^3 = \frac{4KK'}{\pi^2}\frac{kk'^2}{1-2k^2}$$
Multiplying the above with $6$ and adding to $(2)$ alongside with the $K$-differentiation formula
$$K'(k) = \frac{E(k) - k'^2K(k)}{kk'^2}$$ 
Gives the desired equality of $(3)\;\;\;\blacksquare$
