\(\int \frac {1-x}{(1+x+x^2) \sqrt [3]{1+x^3}} \, dx\) [102]

Optimal result

Integrand size = 23, antiderivative size = 119 \[ \int \frac {1-x}{\left (1+x+x^2\right ) \sqrt [3]{1+x^3}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1+x)}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2}}-\frac {\log \left (1+\frac {2^{2/3} (1+x)^2}{\left (1+x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1+x)}{\sqrt [3]{1+x^3}}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (1+\frac {\sqrt [3]{2} (1+x)}{\sqrt [3]{1+x^3}}\right )}{\sqrt [3]{2}} \] Output:

-1/4*ln(1+2^(2/3)*(1+x)^2/(x^3+1)^(2/3)-2^(1/3)*(1+x)/(x^3+1)^(1/3))*2^(2/ 
3)+1/2*ln(1+2^(1/3)*(1+x)/(x^3+1)^(1/3))*2^(2/3)-1/2*arctan(1/3*(1-2*2^(1/ 
3)*(1+x)/(x^3+1)^(1/3))*3^(1/2))*3^(1/2)*2^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.17 \[ \int \frac {1-x}{\left (1+x+x^2\right ) \sqrt [3]{1+x^3}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1+x^3}}{-2 \sqrt [3]{2}-2 \sqrt [3]{2} x+\sqrt [3]{1+x^3}}\right )+2 \log \left (\sqrt [3]{2}+\sqrt [3]{2} x+\sqrt [3]{1+x^3}\right )-\log \left (2^{2/3}+2\ 2^{2/3} x+2^{2/3} x^2-\sqrt [3]{2} (1+x) \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right )}{2 \sqrt [3]{2}} \] Input:

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

Output:

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(357\) vs. \(2(119)=238\).

Time = 0.57 (sec) , antiderivative size = 357, normalized size of antiderivative = 3.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2583, 2009}

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.

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

\(\Big \downarrow \) 2583

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\arctan \left (\frac {\frac {2 \sqrt [3]{2} x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {2^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (x+1)}{\sqrt [3]{x^3+1}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {\frac {\sqrt [3]{2} (x+1)}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\arctan \left (\frac {2^{2/3} \sqrt [3]{x^3+1}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\log \left (1-x^3\right )}{3 \sqrt [3]{2}}-\frac {\log \left (\frac {2^{2/3} (x+1)^2}{\left (x^3+1\right )^{2/3}}-\frac {\sqrt [3]{2} (x+1)}{\sqrt [3]{x^3+1}}+1\right )}{3 \sqrt [3]{2}}+\frac {1}{3} 2^{2/3} \log \left (\frac {\sqrt [3]{2} (x+1)}{\sqrt [3]{x^3+1}}+1\right )-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{x^3+1}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (\sqrt [3]{2} x-\sqrt [3]{x^3+1}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2^{2/3} \sqrt [3]{x^3+1}+x+1\right )}{2 \sqrt [3]{2}}-\frac {\log \left ((1-x)^2 (x+1)\right )}{6 \sqrt [3]{2}}\)

Input:

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

Output:

ArcTan[(1 + (2*2^(1/3)*x)/(1 + x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]) - (2 
^(2/3)*ArcTan[(1 - (2*2^(1/3)*(1 + x))/(1 + x^3)^(1/3))/Sqrt[3]])/Sqrt[3] 
- ArcTan[(1 + (2^(1/3)*(1 + x))/(1 + x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3] 
) - ArcTan[(1 + 2^(2/3)*(1 + x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]) - Log[ 
(1 - x)^2*(1 + x)]/(6*2^(1/3)) + Log[1 - x^3]/(3*2^(1/3)) - Log[1 + (2^(2/ 
3)*(1 + x)^2)/(1 + x^3)^(2/3) - (2^(1/3)*(1 + x))/(1 + x^3)^(1/3)]/(3*2^(1 
/3)) + (2^(2/3)*Log[1 + (2^(1/3)*(1 + x))/(1 + x^3)^(1/3)])/3 - Log[2^(1/3 
) - (1 + x^3)^(1/3)]/(2*2^(1/3)) - Log[2^(1/3)*x - (1 + x^3)^(1/3)]/(2*2^( 
1/3)) + Log[1 + x - 2^(2/3)*(1 + x^3)^(1/3)]/(2*2^(1/3))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2583

Int[(Px_.)*((c_) + (d_.)*(x_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^3)^(p 
_.), x_Symbol] :> Simp[1/c^q   Int[ExpandIntegrand[(c^3 - d^3*x^3)^q*(a + b 
*x^3)^p, Px/(c - d*x)^q, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Poly 
Q[Px, x] && EqQ[d^2 - c*e, 0] && ILtQ[q, 0] && RationalQ[p] && EqQ[Denomina 
tor[p], 3]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 7.41 (sec) , antiderivative size = 687, normalized size of antiderivative = 5.77

method	result	size
trager	\(\text {Expression too large to display}\)	\(687\)

Input:

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

Output:

RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*ln(-(RootOf(RootOf(_Z^ 
3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^2*x+RootOf(RootOf(_Z^3-4 
)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)*(x^3+1)^(2/3)-2*RootOf(Root 
Of(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*(x^3+1)^(1/3)*x-2*RootOf(RootOf(_ 
Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*(x^3+1)^(1/3)-x^2-x-1)/(x^2+x+1))-1/2 
*ln((RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^3* 
x+(x^3+1)^(2/3)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf 
(_Z^3-4)^2-(x^3+1)^(1/3)*RootOf(_Z^3-4)^2*x-2*(x^3+1)^(1/3)*RootOf(RootOf( 
_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)*x-(x^3+1)^(1/3)*RootO 
f(_Z^3-4)^2-2*(x^3+1)^(1/3)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4* 
_Z^2)*RootOf(_Z^3-4)+RootOf(_Z^3-4)*x^2+3*RootOf(_Z^3-4)*x+2*(x^3+1)^(2/3) 
+RootOf(_Z^3-4))/(x^2+x+1))*RootOf(_Z^3-4)-ln((RootOf(RootOf(_Z^3-4)^2+2*_ 
Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^3*x+(x^3+1)^(2/3)*RootOf(RootOf(_Z 
^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^2-(x^3+1)^(1/3)*RootOf( 
_Z^3-4)^2*x-2*(x^3+1)^(1/3)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4* 
_Z^2)*RootOf(_Z^3-4)*x-(x^3+1)^(1/3)*RootOf(_Z^3-4)^2-2*(x^3+1)^(1/3)*Root 
Of(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)+RootOf(_Z^3 
-4)*x^2+3*RootOf(_Z^3-4)*x+2*(x^3+1)^(2/3)+RootOf(_Z^3-4))/(x^2+x+1))*Root 
Of(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (93) = 186\).

Time = 5.24 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.23 \[ \int \frac {1-x}{\left (1+x+x^2\right ) \sqrt [3]{1+x^3}} \, dx=\frac {1}{3} \cdot 2^{\frac {1}{6}} \sqrt {\frac {3}{2}} \arctan \left (\frac {2^{\frac {1}{6}} \sqrt {\frac {3}{2}} {\left (2 \cdot 2^{\frac {2}{3}} {\left (x^{4} + 4 \, x^{3} + 5 \, x^{2} + 4 \, x + 1\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (x^{6} + 7 \, x^{5} + 10 \, x^{4} + 7 \, x^{3} + 10 \, x^{2} + 7 \, x + 1\right )} - 4 \, {\left (x^{5} + x^{4} - 3 \, x^{3} - 3 \, x^{2} + x + 1\right )} {\left (x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (3 \, x^{6} + 9 \, x^{5} + 6 \, x^{4} + x^{3} + 6 \, x^{2} + 9 \, x + 3\right )}}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (\frac {2^{\frac {2}{3}} {\left (x^{3} + 1\right )}^{\frac {2}{3}} {\left (x^{2} + 3 \, x + 1\right )} - 2^{\frac {1}{3}} {\left (x^{4} - 3 \, x^{2} + 1\right )} - 4 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{2} + x\right )}}{x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (\frac {2^{\frac {2}{3}} {\left (x^{2} + x + 1\right )} + 2 \cdot 2^{\frac {1}{3}} {\left (x^{3} + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + 2 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2} + x + 1}\right ) \] Input:

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

Output:

1/3*2^(1/6)*sqrt(3/2)*arctan(1/3*2^(1/6)*sqrt(3/2)*(2*2^(2/3)*(x^4 + 4*x^3 
 + 5*x^2 + 4*x + 1)*(x^3 + 1)^(2/3) + 2^(1/3)*(x^6 + 7*x^5 + 10*x^4 + 7*x^ 
3 + 10*x^2 + 7*x + 1) - 4*(x^5 + x^4 - 3*x^3 - 3*x^2 + x + 1)*(x^3 + 1)^(1 
/3))/(3*x^6 + 9*x^5 + 6*x^4 + x^3 + 6*x^2 + 9*x + 3)) - 1/12*2^(2/3)*log(( 
2^(2/3)*(x^3 + 1)^(2/3)*(x^2 + 3*x + 1) - 2^(1/3)*(x^4 - 3*x^2 + 1) - 4*(x 
^3 + 1)^(1/3)*(x^2 + x))/(x^4 + 2*x^3 + 3*x^2 + 2*x + 1)) + 1/6*2^(2/3)*lo 
g((2^(2/3)*(x^2 + x + 1) + 2*2^(1/3)*(x^3 + 1)^(1/3)*(x + 1) + 2*(x^3 + 1) 
^(2/3))/(x^2 + x + 1))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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