Transform certain integrands so that they involve a single square root rather than two.

Question

I need--in pursuit of my problem https://mathoverflow.net/questions/342446/perform-certain-constrained-integrations-over-an-ordered-subsection-of-a-3-simpl --to integrate \begin{equation} \sqrt{(x+1) \left(2 \sqrt{-2 x^2-x+1}-x+2\right)} \end{equation} and \begin{equation} \sqrt{(1-2 x) \left(2 \sqrt{-2 x^2-x+1}-x+2\right)} \end{equation} over $x \in [\frac{1}{6},\frac{1}{4}]$, as well as \begin{equation} \sqrt{-x \left(x+2 \sqrt{-x (2 x-1)}-1\right)} \end{equation} and \begin{equation} \sqrt{(2 x-1) \left(x+2 \sqrt{-x (2 x-1)}-1\right)} \end{equation} over $x \in [\frac{1}{8} \left(2-\sqrt{2}\right),\frac{1}{6}]$.

I am rather confident--along the order of the answer of Vasily Mitch to Can a certain series of integrals over $[0,\frac{1}{16}]$ be solved using integration-by-parts? -- that these four integrands can each be simplified so that they involve only one square root, rather than their present two, thus (and products of $x^n$ with them) becoming integrable, but would like a "second opinion".

Answer 1

Yes, I believe you're right. \begin{eqnarray} 2 \sqrt{-2 x^2-x+1}-x+2&=& 2\sqrt{(1-2x)(1+x)} -x+2\\ &=&2\sqrt{(1-2x)(1+x)} +1+x+1-2x\\ &=& \left(\sqrt{1-2x}+\sqrt{1+x}\right)^2\ . \end{eqnarray} So \begin{eqnarray} \sqrt{(x+1)\left(2 \sqrt{-2 x^2-x+1}-x+2\right)}\\ =\sqrt{(1-2x)(1+x)}& +(1+x) \end{eqnarray} and \begin{eqnarray} \sqrt{(1-2x) \left(2 \sqrt{-2 x^2-x+1}-x+2\right)}\\ =\sqrt{(1-2x)(1+x)}& +(1-2x) \end{eqnarray} Also, \begin{eqnarray} 1+2 \sqrt{x(1-2x)}-x&=& 2 \sqrt{x(1-2x)} + x + 1-2x\\ &=& \left(\sqrt{x}+\sqrt{1-2x}\right)^2\ . \end{eqnarray} So \begin{eqnarray} \sqrt{x \left(1+2 \sqrt{x(1-2x)}-x\right)}&=& x+\sqrt{x(1-2x)}\ , \end{eqnarray} and \begin{eqnarray} \sqrt{(1-2x) \left(1+2 \sqrt{x(1-2x)}-x\right)}\\ =(1-2x)&+\sqrt{x(1-2x)}\ . \end{eqnarray}

Answer 2

Actually, I would up employing for my indicated original problem https://mathoverflow.net/questions/342446/perform-certain-constrained-integrations-over-an-ordered-subsection-of-a-3-simpl the five transformations, \begin{equation} \left\{\sqrt{(x+1) \left(2 \sqrt{-2 x^2-x+1}-x+2\right)}\to \sqrt{-2 x^2-x+1}+x+1,\sqrt{(1-2 x) \left(2 \sqrt{-2 x^2-x+1}-x+2\right)}\to \sqrt{-2 x^2-x+1}-2 x+1,\sqrt{(2 x-1) \left(x+2 \sqrt{-x (2 x-1)}-1\right)}\to -\sqrt{x-2 x^2}-2 x+1,\sqrt{-x (2 x-1)^2 \left(x+2 \sqrt{-x (2 x-1)}-1\right)}\to 2 x^2+\sqrt{-8 x^4+12 x^3-6 x^2+x}-x,\sqrt{(x+1) (2 x-1)^2 \left(2 \sqrt{-2 x^2-x+1}-x+2\right)}\to -2 x^2+\sqrt{-8 x^4+4 x^3+6 x^2-5 x+1}-x+1\right\}. \end{equation}

This allowed me to check the earlier calculation to high numerical accuracy (0.0036582630543034854603976004088368)--indicating the validity of the five transformations--but the last transformation still left me with an integrand $\sqrt{-8 x^4+4 x^3+6 x^2-5 x+1}$, I could not integrate--when multiplied by powers of $x$--over $x \in [\frac{1}{4},\frac{1}{3}]$. (So, I'm still frustrated in achieving my original goal.) Also, $\sqrt{-8 x^4+4 x^3+6 x^2-5 x+1}=\sqrt{(-x-1) (2 x-1)^3}$, but this doesn't seem to help.)

Also, I developed a "fun" way of deriving the rules. I computed arrays of length 50, into which I put the values of the left-hand sides (the functions to be transformed) for $x=1,\ldots,50$. I, then, got the real and imaginary parts of the resultant values, and used the Mathematica FindSequenceFunction command on the real parts and on the imaginary parts squared.