\(\int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx\) [718]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 20, antiderivative size = 109 \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=-\frac {\sqrt [3]{1-x^3} \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{1-x} \sqrt [3]{1+x+x^2}}+\frac {\sqrt [3]{1-x^3} \log \left (x+\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \] Output:

-1/3*(-x^3+1)^(1/3)*arctan(1/3*(1-2*x/(-x^3+1)^(1/3))*3^(1/2))*3^(1/2)/(1- 
x)^(1/3)/(x^2+x+1)^(1/3)+1/2*(-x^3+1)^(1/3)*ln(x+(-x^3+1)^(1/3))/(1-x)^(1/ 
3)/(x^2+x+1)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 10.06 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=-\frac {3 (1-x)^{2/3} \sqrt [3]{\frac {-i+\sqrt {3}-2 i x}{-3 i+\sqrt {3}}} \sqrt [3]{\frac {i+\sqrt {3}+2 i x}{3 i+\sqrt {3}}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},\frac {1}{3},\frac {5}{3},-\frac {2 i (-1+x)}{3 i+\sqrt {3}},\frac {2 i (-1+x)}{-3 i+\sqrt {3}}\right )}{2 \sqrt [3]{1+x+x^2}} \] Input:

Integrate[1/((1 - x)^(1/3)*(1 + x + x^2)^(1/3)),x]
 

Output:

(-3*(1 - x)^(2/3)*((-I + Sqrt[3] - (2*I)*x)/(-3*I + Sqrt[3]))^(1/3)*((I + 
Sqrt[3] + (2*I)*x)/(3*I + Sqrt[3]))^(1/3)*AppellF1[2/3, 1/3, 1/3, 5/3, ((- 
2*I)*(-1 + x))/(3*I + Sqrt[3]), ((2*I)*(-1 + x))/(-3*I + Sqrt[3])])/(2*(1 
+ x + x^2)^(1/3))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1151, 769}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{x^2+x+1}} \, dx\)

\(\Big \downarrow \) 1151

\(\displaystyle \frac {\sqrt [3]{1-x^3} \int \frac {1}{\sqrt [3]{1-x^3}}dx}{\sqrt [3]{1-x} \sqrt [3]{x^2+x+1}}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {\sqrt [3]{1-x^3} \left (\frac {1}{2} \log \left (\sqrt [3]{1-x^3}+x\right )-\frac {\arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}\right )}{\sqrt [3]{1-x} \sqrt [3]{x^2+x+1}}\)

Input:

Int[1/((1 - x)^(1/3)*(1 + x + x^2)^(1/3)),x]
 

Output:

((1 - x^3)^(1/3)*(-(ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3]]/Sqrt[3]) + 
 Log[x + (1 - x^3)^(1/3)]/2))/((1 - x)^(1/3)*(1 + x + x^2)^(1/3))
 

Defintions of rubi rules used

rule 769
Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
 

rule 1151
Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[(d + e*x)^FracPart[p]*((a + b*x + c*x^2)^FracPart[p]/(a*d + c 
*e*x^3)^FracPart[p])   Int[(d + e*x)^(m - p)*(a*d + c*e*x^3)^p, x], x] /; F 
reeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && 
 IGtQ[m - p + 1, 0] &&  !IntegerQ[p]
 
Maple [F]

\[\int \frac {1}{\left (1-x \right )^{\frac {1}{3}} \left (x^{2}+x +1\right )^{\frac {1}{3}}}d x\]

Input:

int(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x)
 

Output:

int(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} {\left (x^{2} + x + 1\right )}^{\frac {1}{3}} x^{2} {\left (-x + 1\right )}^{\frac {1}{3}} + 2 \, \sqrt {3} {\left (x^{2} + x + 1\right )}^{\frac {2}{3}} x {\left (-x + 1\right )}^{\frac {2}{3}} - \sqrt {3} {\left (x^{3} - 1\right )}}{9 \, x^{3} - 1}\right ) + \frac {1}{6} \, \log \left (3 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{3}} x^{2} {\left (-x + 1\right )}^{\frac {1}{3}} + 3 \, {\left (x^{2} + x + 1\right )}^{\frac {2}{3}} x {\left (-x + 1\right )}^{\frac {2}{3}} + 1\right ) \] Input:

integrate(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x, algorithm="fricas")
 

Output:

-1/3*sqrt(3)*arctan((4*sqrt(3)*(x^2 + x + 1)^(1/3)*x^2*(-x + 1)^(1/3) + 2* 
sqrt(3)*(x^2 + x + 1)^(2/3)*x*(-x + 1)^(2/3) - sqrt(3)*(x^3 - 1))/(9*x^3 - 
 1)) + 1/6*log(3*(x^2 + x + 1)^(1/3)*x^2*(-x + 1)^(1/3) + 3*(x^2 + x + 1)^ 
(2/3)*x*(-x + 1)^(2/3) + 1)
 

Sympy [F]

\[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\int \frac {1}{\sqrt [3]{1 - x} \sqrt [3]{x^{2} + x + 1}}\, dx \] Input:

integrate(1/(1-x)**(1/3)/(x**2+x+1)**(1/3),x)
 

Output:

Integral(1/((1 - x)**(1/3)*(x**2 + x + 1)**(1/3)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x, algorithm="maxima")
 

Output:

integrate(1/((x^2 + x + 1)^(1/3)*(-x + 1)^(1/3)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x, algorithm="giac")
 

Output:

integrate(1/((x^2 + x + 1)^(1/3)*(-x + 1)^(1/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\int \frac {1}{{\left (1-x\right )}^{1/3}\,{\left (x^2+x+1\right )}^{1/3}} \,d x \] Input:

int(1/((1 - x)^(1/3)*(x + x^2 + 1)^(1/3)),x)
 

Output:

int(1/((1 - x)^(1/3)*(x + x^2 + 1)^(1/3)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\int \frac {1}{\left (1-x \right )^{\frac {1}{3}} \left (x^{2}+x +1\right )^{\frac {1}{3}}}d x \] Input:

int(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x)
 

Output:

int(1/(( - x + 1)**(1/3)*(x**2 + x + 1)**(1/3)),x)