\(\int \frac {\sqrt [3]{1-x^3}}{1+x} \, dx\) [58]

Optimal result

Integrand size = 17, antiderivative size = 482 \[ \int \frac {\sqrt [3]{1-x^3}}{1+x} \, dx=\sqrt [3]{1-x^3}+\frac {\sqrt [3]{2} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}-\frac {\arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\sqrt [3]{2} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\sqrt [3]{2} \arctan \left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \sqrt [3]{2} \log \left (1+x^3\right )+\frac {\log \left (2^{2/3}-\frac {1-x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}-\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}+\frac {1}{3} \sqrt [3]{2} \log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )-\frac {\log \left (2 \sqrt [3]{2}+\frac {(1-x)^2}{\left (1-x^3\right )^{2/3}}+\frac {2^{2/3} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6\ 2^{2/3}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2^{2/3}}-\frac {1}{2} \log \left (-x-\sqrt [3]{1-x^3}\right )+\frac {\log \left (-\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )}{2^{2/3}} \] Output:

(-x^3+1)^(1/3)-1/3*2^(1/3)*ln(x^3+1)+1/6*ln(2^(2/3)+(-1+x)/(-x^3+1)^(1/3)) 
*2^(1/3)-1/6*ln(1+2^(2/3)*(1-x)^2/(-x^3+1)^(2/3)-2^(1/3)*(1-x)/(-x^3+1)^(1 
/3))*2^(1/3)+1/3*2^(1/3)*ln(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3))-1/12*ln(2*2^(1 
/3)+(1-x)^2/(-x^3+1)^(2/3)+2^(2/3)*(1-x)/(-x^3+1)^(1/3))*2^(1/3)+1/2*ln(2^ 
(1/3)-(-x^3+1)^(1/3))*2^(1/3)-1/2*ln(-x-(-x^3+1)^(1/3))+1/2*ln(-2^(1/3)*x- 
(-x^3+1)^(1/3))*2^(1/3)+1/3*2^(1/3)*arctan(1/3*(1-2*2^(1/3)*(1-x)/(-x^3+1) 
^(1/3))*3^(1/2))*3^(1/2)+1/6*arctan(1/3*(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3))*3 
^(1/2))*2^(1/3)*3^(1/2)-1/3*arctan(1/3*(1-2*x/(-x^3+1)^(1/3))*3^(1/2))*3^( 
1/2)+1/3*2^(1/3)*arctan(1/3*(1-2*2^(1/3)*x/(-x^3+1)^(1/3))*3^(1/2))*3^(1/2 
)-1/3*2^(1/3)*arctan(1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^(1/2))*3^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 482, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2581, 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.

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

\(\Big \downarrow \) 2581

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2581

Int[(Px_.)*((c_) + (d_.)*(x_))^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] 
:> Int[ExpandIntegrand[(c^3 + d^3*x^3)^q*(a + b*x^3)^p, Px/(c^2 - c*d*x + d 
^2*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p}, x] && PolyQ[Px, x] && ILtQ[q, 0 
] && RationalQ[p] && EqQ[Denominator[p], 3]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {\sqrt [3]{1-x^3}}{1+x} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (residue poly has multiple non-linear fac 
tors)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt [3]{1-x^3}}{1+x} \, dx=\left (-x^{3}+1\right )^{\frac {1}{3}}-\left (\int \frac {\left (-x^{3}+1\right )^{\frac {1}{3}}}{x^{4}+x^{3}-x -1}d x \right )-\left (\int \frac {\left (-x^{3}+1\right )^{\frac {1}{3}} x^{2}}{x^{4}+x^{3}-x -1}d x \right ) \] Input:

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

Output:

( - x**3 + 1)**(1/3) - int(( - x**3 + 1)**(1/3)/(x**4 + x**3 - x - 1),x) - 
 int((( - x**3 + 1)**(1/3)*x**2)/(x**4 + x**3 - x - 1),x)