How to integrate $$\int \frac{1}{\sqrt{x+1}-\sqrt{x+3}+\sqrt{x+5}}dx$$.
In my school life , I learnt how to integrate $$\int \frac{1}{\sqrt{x+a}-\sqrt{x+b}}dx$$ or $$\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}}dx$$.
To integrate these types of expressions , we generally used to rationalize these expressions and then we used to integrate this expressions. For e.g.- to integrate $$\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}}dx$$, we used to multiply the numerator and the denominator by $(\sqrt{x+a}-\sqrt{x+b})$. Therefore finally the above integral will become $$\int \frac{\sqrt{x-a}-\sqrt{x-b}}{a-b}dx$$. Now after this step , it becomes very easy to integrate the above expression.
Similarly to integrate $$\int \frac{1}{\sqrt{x+a}-\sqrt{x+b}}dx$$, we have to multiply the numerator and denominator by $(\sqrt{x+a}+\sqrt{x+b})$ and the integral will become easy.
Suddenly yesterday , it came in my mind that how to integrate $$\int \frac{1}{\sqrt{x+a}-\sqrt{x+b}+\sqrt{x+c}}dx$$, where $a,b,c$ are in A.P.(Arithmetic Progression)?
I don't think that the method of rationalization will work here. Also I noticed one important thing here that Wolfram Alpha is not showing any result for $$\int \frac{1}{\sqrt{x+1}-\sqrt{x+3}+\sqrt{x+5}}dx$$ but Wolfram Alpha is giving result for $$\int_{0}^{1}\frac{1}{\sqrt{x+1}-\sqrt{x+3}+\sqrt{x+5}}dx$$ and the value is around $0.59$. Please help me out with this integral.
It is a very tough integral for me.
SOLUTION OUTLINE :
Let middle term be $\sqrt{x+b}=y$
We get $x=y^2-b$
Substitute that in the other terms to get $\sqrt{y^2+a-b}$ & $\sqrt{y^2+c-b}$
We can Partially rationalize Denominator : multiply $y+[\sqrt{\square}-\sqrt{\blacksquare}]$ by $y-[\sqrt{\square}-\sqrt{\blacksquare}]$
Denominator will be $y^2-[\square-2\sqrt{{\square}{\blacksquare}}+\blacksquare]$
We can then rationalize the Denominator which includes Single $\sqrt{\cdot}$ term.
We will have Numerator in terms of $y^2$ & $\sqrt{\cdot}$ while Denominator will be in terms of $y^2$ & higher Powers.
Substitute $z=y^2$ & Integrate each "numerator term / Denominator terms" to get the whole Integral.
Over-all , $x=z-b$ , which will give Original Integral.
With wolfram , thanks to comment by user "pie" , we can see that it will be a Monster. It will be very large & cumbersome to write out.
Numerical Integration is not cumbersome , though we will get approximate values.