\(\int \frac {(1+x)^2}{\sqrt [3]{1-x^3} (1+x^3)} \, dx\) [101]

Optimal result

Integrand size = 24, antiderivative size = 135 \[ \int \frac {(1+x)^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2}}+\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{\sqrt [3]{2}} \] 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))*2^(2/3)*3^(1/2)
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07 \[ \int \frac {(1+x)^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1-x^3}}{-2 \sqrt [3]{2}+2 \sqrt [3]{2} x+\sqrt [3]{1-x^3}}\right )-2 \log \left (-\sqrt [3]{2}+\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )+\log \left (2^{2/3}-2\ 2^{2/3} x+2^{2/3} x^2+(-1+x) \sqrt [3]{2-2 x^3}+\left (1-x^3\right )^{2/3}\right )}{2 \sqrt [3]{2}} \] Input:

Integrate[(1 + x)^2/((1 - x^3)^(1/3)*(1 + x^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 + (-1 + x)*(2 - 2*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. \(383\) vs. \(2(135)=270\).

Time = 0.60 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.84, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2019, 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.

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

\(\Big \downarrow \) 2019

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

\(\Big \downarrow \) 2583

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(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^(1/3)*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)*(1 + x)^2]/(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 2019

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px 
, Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && 
 EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
 

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 = 6.32 (sec) , antiderivative size = 720, normalized size of antiderivative = 5.33

method	result	size
trager	\(\text {Expression too large to display}\)	\(720\)

Input:

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

Output:

-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(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf 
(_Z^3+4)+4*_Z^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)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf( 
_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)+x^2*RootOf(_Z^3+4)-3*x*RootOf(_Z^3+4)-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)*Roo 
tOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^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*Ro 
otOf(_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)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z 
^3+4)+x^2*RootOf(_Z^3+4)-3*x*RootOf(_Z^3+4)-2*(-x^3+1)^(2/3)+RootOf(_Z^3+4 
))/(x^2-x+1))*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)+RootOf(R 
ootOf(_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*(-x^3+1)^(1/3)*Root 
Of(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+2*(-x^3+1)^(1/3)*RootOf( 
RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)-x^2+x-1)/(x^2-x+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (101) = 202\).

Time = 5.61 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.13 \[ \int \frac {(1+x)^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=-\frac {1}{3} \cdot 2^{\frac {1}{6}} \sqrt {\frac {3}{2}} \arctan \left (\frac {2^{\frac {1}{6}} \sqrt {\frac {3}{2}} {\left (2 \cdot 2^{\frac {2}{3}} {\left (x^{4} - 4 \, x^{3} + 5 \, x^{2} - 4 \, x + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (x^{6} - 7 \, x^{5} + 10 \, x^{4} - 7 \, x^{3} + 10 \, x^{2} - 7 \, x + 1\right )} + 4 \, {\left (x^{5} - x^{4} - 3 \, x^{3} + 3 \, x^{2} + x - 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (3 \, x^{6} - 9 \, x^{5} + 6 \, x^{4} - x^{3} + 6 \, x^{2} - 9 \, x + 3\right )}}\right ) + \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (\frac {2^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} {\left (x^{2} - 3 \, x + 1\right )} - 2^{\frac {1}{3}} {\left (x^{4} - 3 \, x^{2} + 1\right )} - 4 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - x\right )}}{x^{4} - 2 \, x^{3} + 3 \, x^{2} - 2 \, x + 1}\right ) - \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (\frac {2^{\frac {2}{3}} {\left (x^{2} - x + 1\right )} - 2 \cdot 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2} - x + 1}\right ) \] Input:

integrate((1+x)^2/(-x^3+1)^(1/3)/(x^3+1),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)*lo 
g((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)*log((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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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