Integral involving a fraction of a cubic root of a third degree polynomial.

Question

I can't seem to find a way to solve this integral:

$$∫\frac{73}{\sqrt[3]{x^3-73x+2}}dx$$

I tried many substitutions but couldn't find anything useful.

Answer 1

Integrals of this type can be expressed as hypergeometic series. In the present case, the answer given by wolfram alpha expresses the result as a function of Appell's hypergeometic series. This series has an integral representations:

$$ \frac{\Gamma(\alpha)\Gamma(\gamma-\alpha)}{\Gamma(\gamma)}F_1(\alpha;\beta,\beta';\gamma;x,y) = \int_0^1 u^{\alpha-1}(1-u)^{\gamma-\alpha-1}(1-ux)^{-\beta}(1-uy)^{-\beta'}$$

The present integral may be expressed in this from if the cubic expression is broken down into its roots:

$$ \int \frac{73dx}{(x-a)^{\frac{1}{3}}(x-b)^{\frac{1}{3}}(x-c)^{\frac{1}{3}}}$$

Substitute $y=x-a$

$$ \int \frac{73dy}{(y)^{\frac{1}{3}}(y+a-b)^{\frac{1}{3}}(y+a-c)^{\frac{1}{3}}}$$

Express as:

$$ 73\int y^{\frac{-1}{3}}(y+a-b)^{\frac{-1}{3}}(y+a-c)^{\frac{-1}{3}}dy$$

Bring $a-b$ and $a-c$ out:

$$ 73(a-b)^{\frac{-1}{3}}(a-c)^{\frac{-1}{3}}\int y^{\frac{-1}{3}}(1+\frac{y}{a-b})^{\frac{-1}{3}}(1+\frac{y}{a-c})^{\frac{-1}{3}}dy$$

Now if the limits of integration are simply 0 to 1 then the parameters of the series may be found by inspection. If a more general integral is needed then one must consider the integral:

$$ 73(a-b)^{\frac{-1}{3}}(a-c)^{\frac{-1}{3}}\int_0^t y^{\frac{-1}{3}}(1+\frac{y}{a-b})^{\frac{-1}{3}}(1+\frac{y}{a-c})^{\frac{-1}{3}}dy$$

Substituting $y=tz$ will then change the bounds of integration back to 0 and 1 and the parameters of the series can then again be found by inspection.