As has been mentioned many times on this site, anytime you have an algebraic function containing the square root of a cubic or a quartic, you are bound to bump into an elliptic integral.
Usually, such things are handled by using Jacobian elliptic functions for substitutions (in a manner similar to using substitution with trigonometric or hyperbolic functions when you have the square root of a quadratic in an integral).
I'll skip the tedious details of figuring out the proper substitution, since Byrd and Friedman give a formula for handling your integral (formula 212.00 in their handbook):
$$\int_y^\infty\frac{\mathrm dt}{\sqrt{(t^2+a^2)(t^2-b^2)}}=\frac1{\sqrt{a^2+b^2}}F\left(\arcsin\left(\sqrt{\frac{a^2+b^2}{a^2+y^2}}\right) \mid\frac{a^2}{a^2+b^2}\right)$$
where $F(\phi|m)$ is the incomplete elliptic integral of the first kind.
Coming back to your integral, we let $u=2\sqrt{3}-3$ and $v=2\sqrt{3}+3$ such that
$$\int_1^\infty\frac{\mathrm dx}{\sqrt{3x^4+6x^2-1}}=\frac1{\sqrt{uv}}\int_1^\infty \frac{\mathrm dx}{\sqrt{(x^2+1/u)(x^2-1/v)}}$$
Using the quoted formula, the integral reduces to
$$\frac1{\sqrt{u+v}}F\left(\arcsin\left(\sqrt{\frac{u+v}{v+uv}}\right)\mid\frac{v}{u+v}\right)$$
Substituting the values of $u$ and $v$ into this expression and simplifying, we have the result
$$\frac1{\sqrt[4]{48}}F\left(\arcsin\left(\sqrt{\sqrt{3}-1}\right)\mid\frac{2+\sqrt{3}}{4}\right)$$
which agrees with the numerical result in the comments.
As an aside, I consider it a capital annoyance that Mathematica often returns results with complex amplitudes even for real results...
This particular integral definitely looks simple. What changes when we try to evaluate it?
To start with, a simple substitution like $u=e^x, du=e^xdx$ goes like this:
$$\int\frac{dx}{1+2^x+3^x}=\int\frac {e^xdx}{e^x+e^{x(\ln 2+1)}+e^{x(\ln 3+1)}}=\int\frac{du}{u+u^{\ln 2+1}+u^{\ln 3+1}}$$
Already we have left the carefree world of simple functions behind, and this new function does not look easy... Perhaps a different approach would help:
$$3^x+2^x=(3^{\frac x2}+2^{\frac x2})^2-\sqrt{6^x}$$
No, that looks even worse.
This is part of the difficulty with integrals of this type; the common transformations that we are familiar with often fail to line up with certain problems that we would like to solve.
Since $1+2^x+3^x$ is strictly positive for $x\in\Bbb R$, we could try a substitution like $\cosh u=1+2^x+3^x,\sinh udu=(2^x\ln 2+3^x\ln 3)dx,$ but this fails to produce a viable substitution as well since we would be left with
$$\int\frac {\sinh u}{(2^x\ln 2+3^x\ln 3)\cosh u}du$$
as $\cosh u=1+2^x+3^x$ is not nicely soluble for $x,$ nor does any other obvious transformation present itself for placement as $f(u)=2^x\ln 2+3^x\ln 3$.
So indefinite integration by substitution is out (at least as far as the possibilities I am aware of), and integration by parts does not appear to yield any useful results. What about comparisons with definite integrals? What options are available here?
If the value of the integrals $\int\frac 1{1+3^x+3^x}dx=\int\frac 1{1+2\cdot 3^x}dx$ and $\int\frac 1{1+3^{\frac x2}+3^x}dx$ were known, this might be easier as we might be able to limit the possibilities...
In order to have a solvable integral, it is imperative that the denominator be factored fully, or at least to the point where partial fractions can take over and each part can be solved individually. The choice of $\int\frac 1{1+3^{\frac x2}+3^x}dx$ is one that is "near" to the original, and also happens to be cyclotomic, so that all the linear (complex) factors are well-known, and thus partial fractions is "easily" applicable.
In general, let $p_n(x)$ be the $n$th cyclotomic polynomial, and let $q_n(x)$ be the polynomial such that $p_n(x)\cdot q_n(x)=x^n-1.$ (For example, $p_3(2^x)=1+2^x+2^{2x}=1+2^x+4^x$, and $q_3(2^x)=2^x-1$.) Then the integral $\int\frac 1{p_n(\alpha^x)}dx$ can be expressed as
$$\int\frac{q_n(\alpha^x)}{\alpha^{nx}-1}dx=\int\frac{\sum\limits_{i=0}^{\deg(q)+1}a_i\alpha^{ix}}{\alpha^{nx}-1}dx=\sum_{i=0}^{\deg(q)+1}a_i\int\frac{\alpha^{ix}}{\alpha^{nx}-1}dx$$
With a substitution like $u=\alpha^x, du=(\ln\alpha)\alpha^xdx$ this becomes
$$\frac 1{\ln\alpha}\sum_{i=0}^{\deg(q)+1}a_i\int\frac{u^{i-1}}{u^n-1}du\tag 1$$
which is almost a direct translation to an integral with a cyclotomic polynomial; in fact, $(1)$ becomes
$$\frac 1{\ln\alpha}\int\frac{q_n(u)}{u(u^n-1)}du=\frac 1{\ln\alpha}\int\frac{q_n(u)}{u(u-1)\sum_{i=0}^{n-1}u^i}du=\frac1{\ln\alpha}\int\frac 1{up_n(u)}du\tag 2$$
After all that, we know that one of the integrals of interest above, namely, $\int\frac 1{1+3^{\frac x2}+3^x}dx$ can be written with transformation $u=(\sqrt 3)^x$ as
$$\frac2{\ln 3}\int\frac 1{u(u^2+u+1)}du\to\\x-\frac 1{\ln 3}\ln(3^x+3^{\frac x2}+1)-\frac 2{\sqrt 3\ln 3}\arctan\left(\frac 1{\sqrt 3}(2\cdot 3^{\frac x2}+1)\right)+c\tag 3$$
$$\int\frac 1{1+2\cdot 3^x}dx=x-\ln(2\cdot 3^x+1))+c\tag 4$$
Taking integration limits as $x\in[1,+\inf)$ from the question, we get the following values:
$$(3)|_1^\infty=-1+\frac{-\sqrt 3 \pi+2 \sqrt 3 \arctan(2+\frac 1{\sqrt 3})+3 \ln(4+\sqrt 3)}{\ln 27}\approx 0.2003338$$
$$(4)|_1^\infty=\frac{\ln\frac 76}{\ln 3}\approx 0.140314$$
This is only mildly useful as an upper and lower bound. While the value of the posted integral is definitely between these bounds, this is not a very accurate result and could certainly do with some improvement. Note that other integrals that fall outside these bounds (or do not supply sufficient information to be specified as bounds) include $\int\frac 1{1+2^{x+1}}dx,\int\frac 1{4^x+2^x+1}dx,\int\frac 1{4^x+1}dx,\int\frac 1{3^x+1}dx$.
Fortunately, the choice to integrate $\int\frac 1{3^x+3^{\frac x2}+1}dx$ suggests a route to solve the integral completely:
- Find a monotonic sequence of rationals $x_n=\frac ab$ whose limit is $\log_3 2$
- Find the value of $$I_n=\int\frac1{3^x+3^{x_n\cdot x}+1}dx$$ (this will be a three-term polynomial of some degree that will be factorable, possibly nicely depending on the sequence $x_n$ used)
- Find the limit $\lim_{n\to\infty}I_n$
This is not to say that these steps will be easy or computable in reasonable time frames...
Best Answer
Let $$f(x,t) = \frac{2 x-1}{x^{2/3} \sqrt{(x+1) \left((x+1)^3-t x\right)}} \qquad I(t) = \int_0^\infty f(x,t) dx$$ with $g(x,t) = \frac{3 \left(x^2+x\right)}{2 x-1}f$, one checks $$3t \frac{\partial f}{\partial t} + f+\frac{\partial g}{\partial x}= 0$$ so $$3tI'(t)+I(t) = -\int_0^\infty \frac{\partial g}{\partial x} dx = -g(\infty,t)+g(0,t) = 0$$ solving this ODE gives $I(t) = Ct^{-1/3}$. The function $t^{-1/3}$ blows up near $t=0$, but our $I(t)$ is continuous at $t=0$, this forces $C=0$. So $I(t)\equiv 0$ for $t$ near $0$. QED.