Symbolic Integration of Polynomials

Question

Apply an indefinite integral to a given string. The only rules you will be using are defined as such:

∫cx^(n)dx = (c/(n+1))x^(n+1) + C, n ≠ -1
c, C, and n are all constants.

Specifications:

• You must be able to integrate polynomials with any of the possible features:
  • A coefficient, possibly a fraction in the format (numerator/denominator).
  • Recognition that e and π are constants, and in their use, be able to form fractions or expressions containing them (can be held in a fraction like (e/denominator) or (numerator/e), or, if in exponents, x^(e+1))
    • Aside of these two special constants, all coefficients will be rational, real numbers.
  • An exponent, possibly a fraction, in the format x^(exponent)
    • Expressions with e or π in them, aside of themselves, will not be in exponents. (you will not have to integrate stuff like x^(e+1), but you might integrate x^(e))
  • Can use non-x 1-char variables (i.e. f)
    • This is only for ASCII ranges 65-90 and 97-122.
  • You do not have to use chain rule or integrate x^(-1).
• Output must have padding (separation between terms, i.e. x^2 + x + C.
• If it is unknown how to integrate with the above features, the program should print out "Cannot integrate "+input.
• It must be a full program.

Bonuses:

• -10% if you print out the "pretty" exponents formatted for markdown (instead of x^2, x<sup>2</sup>).
• -10% if you print out the equation (i.e. ∫xdx = (1/2)x^2 + C)

Examples:

Input:

x

Output:

(1/2)x^(2) + C

---

Input:

-f^(-2)

Output:

f^(-1) + C

---

Input:

(1/7)x^(1/7) + 5

Output:

(1/56)x^(8/7) + 5x + C

---

Input:

πx^e

Output:

(π/(e+1))x^(e+1) + C

---

Input:

(f+1)^(-1)

Output:

Cannot integrate (f+1)^(-1)

Answer 1

Mathematica 478 * 0.9 = 430.2

φ=(α=ToExpression;Π=StringReplace;σ="Cannot integrate "<>#1;Λ=DeleteDuplicates@StringCases[#1,RegularExpression["[a-df-zA-Z]+"]];μ=Length@Λ;If[μ>1,σ,If[μ<1,Λ="x",Λ=Λ[[1]]];Ψ=α@Π[#1,{"e"->" E ","π"->" π "}];Φ=α@Λ;Θ=α@Π[#1,{"e"->" 2 ","π"->" 2 "}];λ=Exponent[Θ,Φ,List];Θ=Simplify[Θ*Φ^Max@@Abs@λ];Θ=PowerExpand[Θ/.Φ->Φ^LCM@@Denominator@λ];If[Coefficient[Ψ,Φ,-1]==0&&PolynomialQ[Θ,Φ],"∫("<>#1<>")d"<>Λ<>" = "<>Π[ToString[Integrate[Ψ,Φ],InputForm],{"E"->"e","Pi"->"π"}]<>" + C",σ]])&

This creates a true function φ that takes one String as Input. (Does that count as complete program for Mathematica?)

The ungolfed version would be:

φ=(
    σ="Cannot integrate "<>#1;
    Λ=DeleteDuplicates@StringCases[#1,RegularExpression["[a-df-zA-Z]+"]];
    If[Length@Λ>1,σ,
        If[Length@Λ<1,Λ="x",Λ=Λ[[1]]];
        Ψ=ToExpression@StringReplace[#1,{"e"->" E ","π"->" π "}];
        Φ=ToExpression@Λ;
        Θ=ToExpression@StringReplace[#1,{"e"->" 2 ","π"->" 2 "}];
        λ=Exponent[Θ,Φ,List];
        Θ=Simplify[Θ*Φ^Max@@Abs@λ];
        Θ=PowerExpand[Θ/.Φ->Φ^LCM@@Denominator@λ];
        If[Coefficient[Ψ,Φ,-1]==0&&PolynomialQ[Θ,Φ],
            "∫("<>#1<>")d"<>Λ<>" = "<>StringReplace[ToString[Integrate[Ψ,Φ],InputForm],{"E"->"e","Pi"->"π"}]<>" + C",
            σ
        ]
    ]
)&

Note that the Greek letters are necessary to be able to use all the other letters in the input.

Answer 2

MATLAB, 646 x 0.9 = 581.4 bytes

t=input('','s');p=char(960);s=regexprep(t,{p,'pi([a-zA-Z])','([a-zA-Z])pi','([\)e\d])([a-zA-Z])','([a-zA-Z])(([\(\d]|pi))','e^(\(.+?\))','e'},{'pi','pi*$1','$1*pi','$1*$2','$1*$2','exp($1)','exp(1)'});r=[s(regexp(s,'\<[a-zA-Z]\>')),'x'];r=r(1);e=0;try
I=int(sym(strsplit(s,' + ')),r);S=[];for i=I
S=[S char(i) ' + '];end
b=0;o=[];for i=1:nnz(S)
c=S(i);b=b+(c==40)-(c==41);if(c==42&&S(i+1)==r)||(b&&c==32)
c='';end
o=[o c];end
o=regexprep(char([8747 40 t ')d' r ' = ' o 67]),{'pi','exp\(1\)','exp','\^([^\(])',['1/' r]},{p,'e','e^','^($1)',[r '^(-1)']});catch
e=1;end
if e||~isempty(strfind(o,'log'))
disp(['Cannot integrate ' t]);else
disp(o);end

This is currently a work-in-progress using MATLABs built in symbolic integration capabilities. Currently the requirements have been updated so the format now matches the requirements. It also does qualify for the second -10% bonus.

If anyone wants to pitch in and suggest ways of correcting the output, or use this code as a basis for another answer, feel free :). If I can find the time, I'll keep playing with it and see if I can think how to reformat the output.

Update: Ok, so after a bit more work, here is how the code currently stands. It is still a work in progress, but now getting closer to matching the required output.

t=input('','s'); %Get input as a string
p=char(960); %Pi character
s=regexprep(t,{p,'pi([a-zA-Z])','([a-zA-Z])pi','([\)e\d])([a-zA-Z])','([a-zA-Z])(([\(\d]|pi))','e^(\(.+?\))','e'},{'pi','pi*$1','$1*pi','$1*$2','$1*$2','exp($1)','exp(1)'}); %Reformat input to work with built in symbolic integration
r=[s(regexp(s,'\<[a-zA-Z]\>')),'x'];r=r(1); %determine the variable we are integrating
e=0; %Assume success
try
    I=int(sym(strsplit(s,' + ')),r); %Integrate each term seperately to avoid unwanted simplificaiton
    S=[];
    for i=I
        S=[S char(i) ' + ']; %Recombine integrated terms
    end
    %Now postprocess the output to try and match the requirements
    b=0;o=[];
    for i=1:nnz(S)
        %Work through the integrated string character by character
        c=S(i);
        b=b+(c=='(')-(c==')'); %Keep track of how many layers deep of brackets we are in
        if(c=='*'&&S(i+1)==r)||(b&&c==' ') %If a '*' sign preceeds a variable. Also deblank string.
            c=''; %Delete this character
        end
        o=[o c]; %merge into new output string.
    end
    o=regexprep([char(8747) '(' t ')d' r ' = ' o 'C'],{'pi','exp\(1\)','exp','\^([^\(])',['1/' r]},{p,'e','e^','^($1)',[r '^(-1)']});
catch
    e=1; %failed to integrate
end
if e||~isempty(strfind(o,'log'))
    disp(['Cannot integrate ' t])  %bit of a hack - matlab can integrate 1/x, so if we get a log, we pretend it didn't work.
else
    disp(o)% Display it.
end

Here are some examples of what it currently produces. As you can see, it's not quite right, but getting closer.

Inputs:

x
-f^(-2)
(1/7)x^(1/7) + 5
πx^e
(f+1)^(-1)

Outputs:

∫(x)dx = x^(2)/2 + C
∫(-f^(-2))df = f^(-1) + C
∫((1/7)x^(1/7) + 5)dx = x^(8/7)/8 + 5x + C
∫(πx^(e))dx = (πx^(e+1))/(e+1) + C
Cannot integrate (f+1)^(-1)