\(\int \frac {x^6}{(1-x^3)^{4/3} (1+x^3)} \, dx\) [860]

Optimal result

Integrand size = 22, antiderivative size = 153 \[ \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {x}{2 \sqrt [3]{1-x^3}}+\frac {\arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}-\frac {1}{2} \log \left (x+\sqrt [3]{1-x^3}\right ) \] Output:

1/2*x/(-x^3+1)^(1/3)+1/3*arctan(1/3*(1-2*x/(-x^3+1)^(1/3))*3^(1/2))*3^(1/2 
)-1/12*arctan(1/3*(1-2*2^(1/3)*x/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)- 
1/24*ln(x^3+1)*2^(2/3)+1/8*ln(-2^(1/3)*x-(-x^3+1)^(1/3))*2^(2/3)-1/2*ln(x+ 
(-x^3+1)^(1/3))
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.44 \[ \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {1}{24} \left (\frac {12 x}{\sqrt [3]{1-x^3}}+8 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x-2 \sqrt [3]{1-x^3}}\right )-2\ 2^{2/3} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x-2^{2/3} \sqrt [3]{1-x^3}}\right )-8 \log \left (x+\sqrt [3]{1-x^3}\right )+2\ 2^{2/3} \log \left (2 x+2^{2/3} \sqrt [3]{1-x^3}\right )+4 \log \left (x^2-x \sqrt [3]{1-x^3}+\left (1-x^3\right )^{2/3}\right )-2^{2/3} \log \left (-2 x^2+2^{2/3} x \sqrt [3]{1-x^3}-\sqrt [3]{2} \left (1-x^3\right )^{2/3}\right )\right ) \] Input:

Integrate[x^6/((1 - x^3)^(4/3)*(1 + x^3)),x]
 

Output:

((12*x)/(1 - x^3)^(1/3) + 8*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x - 2*(1 - x^3)^(1 
/3))] - 2*2^(2/3)*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x - 2^(2/3)*(1 - x^3)^(1/3)) 
] - 8*Log[x + (1 - x^3)^(1/3)] + 2*2^(2/3)*Log[2*x + 2^(2/3)*(1 - x^3)^(1/ 
3)] + 4*Log[x^2 - x*(1 - x^3)^(1/3) + (1 - x^3)^(2/3)] - 2^(2/3)*Log[-2*x^ 
2 + 2^(2/3)*x*(1 - x^3)^(1/3) - 2^(1/3)*(1 - x^3)^(2/3)])/24
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {970, 1026, 769, 901}

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.

\(\displaystyle \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (x^3+1\right )} \, dx\)

\(\Big \downarrow \) 970

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

\(\Big \downarrow \) 1026

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

\(\Big \downarrow \) 769

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

\(\Big \downarrow \) 901

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

Input:

Int[x^6/((1 - x^3)^(4/3)*(1 + x^3)),x]
 

Output:

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

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 901

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit 
h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S 
qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] 
 + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0]
 

rule 970

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[(-a)*e^(2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n) 
^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Simp[e^(2*n) 
/(b*n*(b*c - a*d)*(p + 1))   Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d 
*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^ 
n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[ 
n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, 
n, p, q, x]
 

rule 1026

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)* 
(x_)^(n_)), x_Symbol] :> Simp[f/d   Int[(a + b*x^n)^p, x], x] + Simp[(d*e - 
 c*f)/d   Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, 
 p, n}, x]
 

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.43

method	result	size
pseudoelliptic	\(-\frac {-2 \,2^{\frac {2}{3}} \ln \left (\frac {2^{\frac {1}{3}} x +\left (-x^{3}+1\right )^{\frac {1}{3}}}{x}\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}+8 \ln \left (\frac {x +\left (-x^{3}+1\right )^{\frac {1}{3}}}{x}\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}+\left (\left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2^{\frac {2}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}}+x \right )}{3 x}\right )+\ln \left (\frac {2^{\frac {2}{3}} x^{2}-2^{\frac {1}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}} x +\left (-x^{3}+1\right )^{\frac {2}{3}}}{x^{2}}\right )\right ) 2^{\frac {2}{3}}+8 \sqrt {3}\, \arctan \left (\frac {\left (-2 \left (-x^{3}+1\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )-4 \ln \left (\frac {\left (-x^{3}+1\right )^{\frac {2}{3}}-\left (-x^{3}+1\right )^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}-12 x}{24 \left (-x^{3}+1\right )^{\frac {1}{3}}}\)	\(219\)

Input:

int(x^6/(-x^3+1)^(4/3)/(x^3+1),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(-2*2^(2/3)*ln((2^(1/3)*x+(-x^3+1)^(1/3))/x)*(-x^3+1)^(1/3)+8*ln((x+ 
(-x^3+1)^(1/3))/x)*(-x^3+1)^(1/3)+((-2*3^(1/2)*arctan(1/3*3^(1/2)*(-2^(2/3 
)*(-x^3+1)^(1/3)+x)/x)+ln((2^(2/3)*x^2-2^(1/3)*(-x^3+1)^(1/3)*x+(-x^3+1)^( 
2/3))/x^2))*2^(2/3)+8*3^(1/2)*arctan(1/3*(-2*(-x^3+1)^(1/3)+x)*3^(1/2)/x)- 
4*ln(((-x^3+1)^(2/3)-(-x^3+1)^(1/3)*x+x^2)/x^2))*(-x^3+1)^(1/3)-12*x)/(-x^ 
3+1)^(1/3)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.54 \[ \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=-\frac {12 \cdot 2^{\frac {1}{6}} \sqrt {\frac {1}{6}} {\left (x^{3} - 1\right )} \arctan \left (-\frac {2^{\frac {1}{6}} \sqrt {\frac {1}{6}} {\left (2^{\frac {1}{3}} x - 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{x}\right ) - 2 \cdot 2^{\frac {2}{3}} {\left (x^{3} - 1\right )} \log \left (\frac {2^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) + 2^{\frac {2}{3}} {\left (x^{3} - 1\right )} \log \left (\frac {2^{\frac {2}{3}} x^{2} - 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 8 \, \sqrt {3} {\left (x^{3} - 1\right )} \arctan \left (-\frac {\sqrt {3} x - 2 \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) + 8 \, {\left (x^{3} - 1\right )} \log \left (\frac {x + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) - 4 \, {\left (x^{3} - 1\right )} \log \left (\frac {x^{2} - {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 12 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x}{24 \, {\left (x^{3} - 1\right )}} \] Input:

integrate(x^6/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="fricas")
 

Output:

-1/24*(12*2^(1/6)*sqrt(1/6)*(x^3 - 1)*arctan(-2^(1/6)*sqrt(1/6)*(2^(1/3)*x 
 - 2*(-x^3 + 1)^(1/3))/x) - 2*2^(2/3)*(x^3 - 1)*log((2^(1/3)*x + (-x^3 + 1 
)^(1/3))/x) + 2^(2/3)*(x^3 - 1)*log((2^(2/3)*x^2 - 2^(1/3)*(-x^3 + 1)^(1/3 
)*x + (-x^3 + 1)^(2/3))/x^2) - 8*sqrt(3)*(x^3 - 1)*arctan(-1/3*(sqrt(3)*x 
- 2*sqrt(3)*(-x^3 + 1)^(1/3))/x) + 8*(x^3 - 1)*log((x + (-x^3 + 1)^(1/3))/ 
x) - 4*(x^3 - 1)*log((x^2 - (-x^3 + 1)^(1/3)*x + (-x^3 + 1)^(2/3))/x^2) + 
12*(-x^3 + 1)^(2/3)*x)/(x^3 - 1)
 

Sympy [F]

\[ \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int \frac {x^{6}}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {4}{3}} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \] Input:

integrate(x**6/(-x**3+1)**(4/3)/(x**3+1),x)
 

Output:

Integral(x**6/((-(x - 1)*(x**2 + x + 1))**(4/3)*(x + 1)*(x**2 - x + 1)), x 
)
 

Maxima [F]

\[ \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int { \frac {x^{6}}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(x^6/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="maxima")
 

Output:

integrate(x^6/((x^3 + 1)*(-x^3 + 1)^(4/3)), x)
 

Giac [F]

\[ \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int { \frac {x^{6}}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(x^6/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="giac")
 

Output:

integrate(x^6/((x^3 + 1)*(-x^3 + 1)^(4/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int \frac {x^6}{{\left (1-x^3\right )}^{4/3}\,\left (x^3+1\right )} \,d x \] Input:

int(x^6/((1 - x^3)^(4/3)*(x^3 + 1)),x)
 

Output:

int(x^6/((1 - x^3)^(4/3)*(x^3 + 1)), x)
 

Reduce [F]

\[ \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=-\left (\int \frac {x^{6}}{\left (-x^{3}+1\right )^{\frac {1}{3}} x^{6}-\left (-x^{3}+1\right )^{\frac {1}{3}}}d x \right ) \] Input:

int(x^6/(-x^3+1)^(4/3)/(x^3+1),x)
                                                                                    
                                                                                    
 

Output:

 - int(x**6/(( - x**3 + 1)**(1/3)*x**6 - ( - x**3 + 1)**(1/3)),x)