can $\int \sqrt{1 + \frac{1}{x^4}}dx$ be expressed in terms of elementary functions?

Question

I'm trying to evaluate this integral:

$$\int \sqrt{1 + \frac{1}{x^4}}dx$$

After a little substitution, the integral can be shown to be equivalent to

$$\int \sqrt{\cosh(2\alpha)}\,d\alpha,$$

which can be expressed as an elliptic integral, but I was wondering whether it can be expressed using only elementary functions?

Answer 1

I'll address the more general question of which indefinite integrals of the form $$\int \sqrt[n]{1 + x^m} \,dx$$ admit an elementary closed form, where $m$ is an integer and $n$ is a nonzero integer. This question turns out to be a special case of the work of Chebyshev (see p. 51 of Hardy's Integration of Functions of a Single Variable, which Dave L. Renfro mentioned in a comment).

First, provided $m \neq 0$, substituting $t = x^m$ transforms the integral to $$\frac1m \int t^{\frac{1-m}m} (1 + t)^{\frac1n} \,dt .$$ Now, Chebyshev's condition says that the integrand has a closed-form (in Hardy's language, "finite form") antiderivative if and only if at least one of $\frac{1 - m}m, \frac1n, \frac{1 - m}m + \frac1n$ are integers. (For the integrand in the question statement, $\sqrt{1 + \frac1{x^4}}$, these numbers are $-\frac54, \frac12, -\frac34$, so it admits no elementary antiderivative). The first is an integer if and only if $m = \pm 1$, the second is an integer if and only if $n = \pm 1$, and the third is an integer if and only if $m = n = \pm 1$, $m = n = \pm2$, or $m = -n$. So, the only elementary cases are:

• $m = -1$, when the substitution $1 + t = u^n$ rationalizes the integral, giving $$\int \sqrt[n]{1 + \frac1x} \,dx = -n \int \frac{u^n \,du}{(u^n - 1)^2},$$
• $m = 0$, when the integrand is a constant
• $m = 1$, when the integrand is a power of a linear function,
• $n = \pm 1$, when the integrand is a rational function,
• $m = n = -2$, when the integral is $$\int \frac{dx}{\sqrt{1 + \frac1{x^2}}} = \sqrt{1 + x^2} + C,$$
• $m = n = 2$, when the integral is $$\int \sqrt{1 + x^2} \,dx = \frac12 \operatorname{arsinh} x + \frac12 x \sqrt{1 + x^2} + C,$$ and
• $m = -n$, when the substitution $1 + t = t u^n$ rationalizes the integral, giving $$\int \sqrt[n]{1 + \frac{1}{x^n}} \,dx = \int \frac{u^n \,du}{u^n - 1}.$$

Remark It follows from the definition that the functions considered in this answer have a closed-form antiderivative in terms of the Gauss Hypergeometric function, ${}_2 F_1$: When $m \neq 0, n \neq 0$, $$\int \sqrt[n]{1 + x^m} \,dx = x {}_2 F_1 \left(\frac1m, -\frac1n; 1 + \frac1m; -x^m\right) + C .$$

Answer 2

Another way to arrive to @Travis Willse's more general result. $$I=\int \sqrt[n]{1 + x^m} \,dx$$ $$x^m=t \quad \implies \quad I=\frac 1m \int t^{\frac{1}{m}-1} (1+t)^{\frac{1}{n}}\,dt$$ $$(1+t)^{\frac{1}{n}}=\sum_{k=0}^\infty \binom{\frac{1}{n}}{k}\,t^k=\sum_{k=0}^\infty \frac{\Gamma \left(1+\frac{1}{n}\right)} {\Gamma (k+1)\, \Gamma\left(-k+\frac{1}{n}+1\right)}\,t^k$$ $$I=\frac{\Gamma \left(1+\frac{1}{n}\right)}{m}\int\sum_{k=0}^\infty \frac{t^{k+\frac{1}{m}-1}}{\Gamma (k+1)\,\Gamma \left(-k+\frac{1}{n}+1\right)}\,dt$$

$$I=\Gamma \left(1+\frac{1}{n}\right)\, t^{\frac 1 m}\sum_{k=0}^\infty\frac{t^{k}}{(k m+1) \,\Gamma (k+1)\, \Gamma \left(-k+\frac{1}{n}+1\right)}$$ which is the definition of a Gaussian hypergeometric function $$I=t^{\frac{1}{m}} \,_2F_1\left(\frac{1}{m},-\frac{1}{n};\frac{m+1}{m};-t\right)$$ Go back to $x$ to obtain @Travis Willse's last formula.