How to evaluate $\int_0^1 \frac{dx}{\sqrt{x + \sqrt{x^2 + \sqrt{x^3}}}}?$

Question

$$\int_0^1 \frac{dx}{\sqrt{x + \sqrt{x^2 + \sqrt{x^3}}}}$$

How to evaluate the above integral? I am trying this question by substituting $x = \tan^{2 / 3} y$. Then the denominator will become $$\sqrt{\tan^{2/3} y+\sqrt{\tan^{4/3} y+ \tan y}} .$$ But how to proceed further?

Answer 1

Using the rewrite from @user170231, consider the substitution $\frac{1}{\sqrt{x}} = \sinh^2 t$

$$I = \int_{\sinh^{-1}(1)}^\infty \frac{4\cosh t\:dt}{\sinh^3 t\sqrt{1+\cosh t}}$$

$$ = \frac{1}{\sqrt{2}}\int_{\sinh^{-1}(1)}^\infty\frac{\cosh\left(\frac{t}{2}\right)}{\sinh^3\left(\frac{t}{2}\right)\cosh^3\left(\frac{t}{2}\right)}+\frac{\sinh\left(\frac{t}{2}\right)}{\sinh^2\left(\frac{t}{2}\right)\cosh^4\left(\frac{t}{2}\right)}\:\frac{dt}{2}$$

Notice that the integrand is almost a product rule. Using the interchange $t\leftrightarrow \frac{t}{2}$ we can split the integral up as

$$I = \color{green}{\frac{1}{2\sqrt{2}}\left[\frac{-1}{\sinh^2t \cosh^3t}\right]_{\cosh^{-1}\sqrt{\frac{1+\sqrt{2}}{2}}}^\infty}$$

$$ - \frac{1}{2\sqrt{2}}\int_{\cosh^{-1}\sqrt{\frac{1+\sqrt{2}}{2}}}^\infty \frac{(1-\tanh^2 t)\operatorname{sech}^2t}{\sinh t}dt$$

$$= \color{green}{2\sqrt{\sqrt{2}-1}}+\color{red}{\frac{\operatorname{sech} t}{6\sqrt{2}}\left(2+\operatorname{sech}^2 t\right)\Biggr|_{t=\operatorname{sech}^{-1}\sqrt{2\sqrt{2}-2}}}$$

$$-\color{blue}{\frac{1}{6\sqrt{2}}\int_{\cosh^{-1}\sqrt{\frac{1+\sqrt{2}}{2}}}^\infty \frac{3\sinh t}{\cosh^2 t - 1}-\tanh t \operatorname{sech}t\:dt}$$

$$= \color{green}{2\sqrt{\sqrt{2}-1}} + \color{red}{\frac{\sqrt{2\sqrt{2}-2}}{3}} + \color{blue}{\frac{1}{6\sqrt{2}}\Biggr[3\coth^{-1}(\cosh t)-\operatorname{sech}t\Biggr]_{\operatorname{sech}^{-1}\sqrt{2\sqrt{2}-2}}^\infty}$$

Crunching the numbers, we get a final analytic answer of

$$I = \boxed{\color{green}{2\sqrt{\sqrt{2}-1}} + \color{purple}{\frac{\sqrt{5\sqrt{2}-1}}{6}} - \color{blue}{\frac{1}{2\sqrt{2}}\tanh^{-1}\sqrt{2\sqrt{2}-2}}\approx 1.157416}$$

Answer 2

A few substitutions pave the way for a fairly simple partial fraction expansion:

$$\begin{align*} & \int_0^1 \frac{dx}{\sqrt{x + \sqrt{x^2 + \sqrt{x^3}}}} \\ &= \int_0^1 \frac{dx}{\sqrt x \sqrt{1 + \sqrt{1 + \frac1{\sqrt x}}}} \\ &= 2 \int_0^1 \frac{dy}{\sqrt{1+\sqrt{1+\frac1y}}} \tag1 \\ &= 4 \int_{\sqrt2}^\infty \frac{z}{\left(z^2-1\right)^2 \sqrt{z+1}} \, dz \tag2 \\ &= 8 \int_{\sqrt{\sqrt2+1}}^\infty \frac{w^2-1}{w^4 \left(w^2-2\right)^2} \, dw \tag3 \end{align*}$$

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• $(1)$ $y = \sqrt x$
• $(2)$ $z=\sqrt{\dfrac{y+1}y}$
• $(3)$ $w=\sqrt{z+1}$

Answer 3

Here is yet an other solution, obtained by rewriting the integral as an integral on the appropriate curve $C$. The reason why we "can integrate" becomes transparent, this curve $C$ turns out to have "by chance" genus zero.

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Let us denote by $y$ the radical in the denominator, then exhibit an algebraic relation between $x$ and $y$. We have after successive transformations: $$ \begin{aligned} y &= \sqrt{x + \sqrt{x^2 + \sqrt{x^3}}}\ ,\\ y^2 - x &= \sqrt{x^2 + \sqrt{x^3}}\ ,\\ (y^2 - x)^2 &= x^2 + \sqrt{x^3}\ ,\\ y^4 - 2y^2x &= \sqrt{x^3}\ ,\\ y^4(y^2 - 2x)^2 - x^3 &= 0\ . \end{aligned} $$ So we have to integrate $dx/y$ on the curve $C$ with algebraic equation given by the last line above from the point $(x,y)=(0,0)$ to the point $(x,y)=(1,\sqrt{1+\sqrt 2})$. Since $\sqrt2$ and its cousin will appear relatively often, let us introduce the notations: $$ \bbox[lightyellow]{\qquad \begin{aligned} a &=\sqrt 2\ ,\\ b &=\sqrt{a-1}=\sqrt{\sqrt2-1}\ . \end{aligned} \qquad } $$ It turns out that the curve $C$ is rational, (it has genus zero,) and we can ask the computer for a rational parametrization, code is postponed. A version of the offered parametrization is $$ \bbox[lightblue]{ \qquad x = \frac{u^8}{(1-2u^2)^2}\ ,\qquad y = \frac {u^3}{1-2u^2}\ ,\qquad u\in[0, b]\ , \qquad } $$ and it is easy to check that $y = \sqrt{x + \sqrt{x^2 + \sqrt{x^3}}}$, the one needed intermediate step showing this, starting from the formula for $x$, being $\sqrt{x^2 + \sqrt{x^3}}= \frac{u^6(1-u^2)}{(1-2u^2)^2}$. Performing the above $x$-substitution we have for the given integral $J$: $$ \begin{aligned} J &:=\int_0^1\frac{dx}{\sqrt{x + \sqrt{x^2 + \sqrt{x^3}}}} =\int_ {\substack{ C\\ \text{from }(x,y)=(0,\ 0)\\ \text{to }(x,y)=(1, \ 1/b)}} \ \frac{dx}y \\ &= \int_0^b\frac{8u^4(1-u^2)}{(1-2u^2)^2}\; du = \int_0^b\left[\frac1{(1-2u^2)^2} - \frac1{1-2u^2}-2u^2\right]\; du \\ &= \bbox[lightyellow]{\ \frac 16(13+2a)b +\frac a4\log(a+1) +\frac a4\log(1-ab) \ } \ . \end{aligned} $$ $\square$

(The above handy formula will be checked numerically in the second note.)

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Note: Computer aid:

sage: x, y = QQ['x,y'].gens()
....: C = Curve(y^4*(y^2 - 2*x)^2 - x^3)
sage: 
sage: C.genus()
0
sage: C.rational_parameterization()
Scheme morphism:
  From: Affine Space of dimension 1 over Rational Field
  To:   Affine Plane Curve over Rational Field defined by y^8 - 4*x*y^6 + 4*x^2*y^4 - x^3
  Defn: Defined on coordinates by sending (t) to
        (1/(t^8 - 4*t^6 + 4*t^4), (-1)/(t^3 - 2*t))

And we took $u=1/t$ instead.

Note: Computer check, pari/gp:

? intnum(x=0, 1, 1/sqrt(x + sqrt(x^2 + sqrt(x^3))))
%1 = 1.1574160236587956204374053042986128629
? a = sqrt(2);
? b = sqrt(a - 1);
? (13 + 2*a)*b/6 + a/4*log(a + 1) + a/4*log(1 - a*b)
%6 = 1.1574160236587956204374053042986128635