I found this integral from a friend of mine
$$I = \int_0^{\frac{\sqrt3}2} {\frac{1}{{{x^{3/4}}{{\left( {1 + x} \right)}^{3/4}}}}dx} $$
Which its closed-form is :$$\frac{{2\sqrt 2 {\pi ^{3/2}}}}{{3{\Gamma ^2}( {\frac{3}{4}} )}}$$
Look at this closed form I have a feeling that the given integral can be involved with the complete elliptic integral of the first kind $K( {\frac{1}{2}} )$. But I don't know how to convert this integral to get that result.
I tried to use sub:$$\eqalign{
& x = \frac{{1 – t}}{{1 + t}} \Rightarrow t = \frac{{1 – x}}{{1 + x}} \Rightarrow dx = – \frac{2}{{{{\left( {1 + t} \right)}^2}}}dt \cr
& \Rightarrow I = \frac{2}{{{2^{3/4}}}}\int\limits_{7 – 4\sqrt 3 }^1 {\frac{1}{{\sqrt {1 + t} {{\left( {1 – t} \right)}^{3/4}}}}} dt,{\text{ since }}7 – 4\sqrt 3 = {\left( {2 – \sqrt 3 } \right)^2}{\text{ then let}}:t = {u^2} \cr
& \Rightarrow I = \frac{4}{{{2^{3/4}}}}\int\limits_{2 – \sqrt 3 }^1 {\frac{u}{{\sqrt {1 + {u^2}} {{\left( {1 – {u^2}} \right)}^{3/4}}}}} du \cr} $$
So I get stuck here. May I ask for help? Or give me a hint about substitution. Thank you very much.
Edit #1: After using generalized binomial theorem and changing order of summation and integration, with some manipulation with the last sum, I arrived at:$$I=2\ 2^{3/4} \sqrt[8]{3} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};-\frac{\sqrt{3}}{2}\right)$$
May be, there is a transformation with $_2F_1$ can link this result with the elliptic integral. I am still trying to figure out.
Best Answer
This is to answer the OP how to convert it to the elliptic integral. Let $u=2x+1$
$$I=\int_1^{1+\sqrt3} \frac1{(u^2-1)^{3/4}}du$$
Let $u=\csc \theta$
$$I=\sqrt2\int_{\theta_0}^{\pi/2} \frac1{\sqrt{\sin\theta\cos\theta}}d\theta$$
where $\theta_0=\arcsin\frac{\sqrt3-1}2$, and note that
$$2\sin\theta\cos\theta=(\sin\theta+\cos\theta)^2-1=2\cos^2(\theta-\frac\pi4)-1=1-2\sin^2(\theta-\frac\pi4)$$
hence $$I=2\int_{\theta_0}^{\pi/2} \frac1{\sqrt{1-2\sin^2(\theta-\frac\pi4)}}d\theta$$
Let $\phi=\theta-\frac\pi4$
$$I=2\int_{\theta_0-\frac\pi4}^{\frac\pi4} \frac1{\sqrt{1-2\sin^2 \phi}}d\phi$$
Use the defintion of elliptic integral of the first kind
$$F(a,k)=\int_0^a \frac1{\sqrt{1-k^2\sin^2\phi}}d\phi$$
We get
$$\int_0^{\frac{\sqrt3}2} {\frac{1}{{{x^{3/4}}{{\left( {1 + x} \right)}^{3/4}}}}dx}=2F\left(\frac\pi4,\sqrt2\right)-2F\left(\arcsin\left(\frac{\sqrt3-1}2\right)-\frac\pi4,\sqrt2\right)\tag{1}$$
Using the Reciprocal-Modulus Transformation formula as suggested by @Tyma Gaidash, the first half of eq.(1) is converted to
$$2F\left(\frac\pi4,\sqrt2\right)=2\cdot \frac1{\sqrt2} F\left(\frac\pi2, \frac1{\sqrt2}\right)=\sqrt2 K\left(\frac1{\sqrt2}\right)$$
where $K(k)=\int_0^\frac\pi2 \frac1{\sqrt{1-k^2\sin^2\phi}}d\phi$ is complete elliptic integral of the first kind, hence
$$\boxed{\int_0^{\frac{\sqrt3}2} {\frac{1}{{{x^{3/4}}{{\left( {1 + x} \right)}^{3/4}}}}dx}=\sqrt2 K\left(\frac1{\sqrt2}\right)-2F\left(\arcsin\left(\frac{\sqrt3-1}2\right)-\frac\pi4,\sqrt2\right)}$$
Remarks:
For the closed form, the integral limit $\sqrt3+1=\frac2{\sqrt3-1}$ reminds me maybe it is related to the elliptic curve, where I ever asked a question here. But I have very limited knowledge on this topic.