Automated Substitution Methods for Symbolic Integration

The Maxima CAS package incomplete_gamma_int attempts to find a change of variable that converts a given integrand into either the form $\int \mathrm{e}^{-t} t^a \mathrm{d}t$ or the form $\int t^a (1-t)^b \mathrm{d}t$. When successful, the package will return the antiderivative in terms of either the incomplete gamma function or the Gauss hypergeometric function.

The method is akin to an automatic derivative divide (integration by substitution) with a seed function involving a so-called special function. This project started with only the incomplete gamma function as a seed, and then I extended it to other seed functions. Likely, I should rename the project.

Example Integrals

Below are some examples:

(%i1) my_int(sqrt(x-1) * sqrt(x) * (2*x-1) * %e^(x-x^2), x);
(%o1) -gamma_incomplete(3/2, (x-1)*x)

(%i2) my_int(x^2*sqrt(1-x^2)^(1/3),x);
(%o2) (hypergeometric([-(1/6),3/2],[5/2],x^2)*x^3)/3

(%i3) my_int(((x-1)*(x+1)*(x^2+1)^(2/3))/x^(8/3),x);
(%o3) (3*(x^2+1)*(x^4+2*x^2+1)^(1/3))/(5*x^(5/3))

(%i4) my_int((x-1)^(2/3)*x^(2/3)*(2*x-1)*sqrt(-x^2+x+1),x);
(%o4) (3*hypergeometric([-(1/2),5/3],[8/3],(x-1)*x)*(x-1)^(5/3)*x^(5/3))/5

(%i5) my_int((sqrt(-x^2+x-1)*(x^2-1)*(x^2+1)^(1/3))/x^(17/6),x);
(%o5) (3*hypergeometric([-(1/2),4/3],[7/3],(x^2+1)/x)*(x^2+1)*(x^8-2*x^6+2*x^2-1)^(1/3)) 
      / (4*x^(4/3)*(x^6-3*x^4+3*x^2-1)^(1/3))

Mathematica® Integration Example

Here, we use both plain Mathematica (version 14.1) as well as the Rubi integrator (version V4.17.3.0) for Mathematica on example (%i5) from above. Neither gives a satisfactory antiderivative. For details about Rubi, see Rule-Based Integration System.

This example shows that the method used in the incomplete_gamma_int package can find some antiderivatives that other packages are not able to find. But I am not suggesting that the package incomplete_gamma_int is in any way competitive to Mathematica or Rubi.

In[14]:= Integrate[(Sqrt[-x^2 + x - 1] (x^2 - 1) (x^2 + 1)^(1/3))/x^(17/6), x]

Out[14]= ∫(Sqrt[-1 + x - x^2] (-1 + x^2) (1 + x^2)^(1/3))/x^(17/6) dx

In[15]:= Get["Rubi`"]

In[16]:= Int[(Sqrt[-x^2 + x - 1] (x^2 - 1) (x^2 + 1)^(1/3))/x^(17/6), x]

Out[16]= 6 Subst[Int[Sqrt[-1 + x^6 - x^12] (1 + x^12)^(1/3), x], x, x^(1/6)] 
           - 6 Subst[Int[(Sqrt[-1 + x^6 - x^12] (1 + x^12)^(1/3))/x^12, x], x, x^(1/6)]

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