\(\int \frac {1}{x^6 \sqrt [3]{1-x^3} (1+x^3)} \, dx\) [830]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 22, antiderivative size = 124 \[ \int \frac {1}{x^6 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=-\frac {\left (1-x^3\right )^{2/3}}{5 x^5}+\frac {\left (1-x^3\right )^{2/3}}{5 x^2}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log \left (1+x^3\right )}{6 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}} \] Output:

-1/5*(-x^3+1)^(2/3)/x^5+1/5*(-x^3+1)^(2/3)/x^2-1/6*arctan(1/3*(1-2*2^(1/3) 
*x/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)-1/12*ln(x^3+1)*2^(2/3)+1/4*ln( 
-2^(1/3)*x-(-x^3+1)^(1/3))*2^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^6 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {1}{60} \left (-\frac {12 \left (1-x^3\right )^{5/3}}{x^5}-10\ 2^{2/3} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x-2^{2/3} \sqrt [3]{1-x^3}}\right )+10\ 2^{2/3} \log \left (2 x+2^{2/3} \sqrt [3]{1-x^3}\right )-5\ 2^{2/3} \log \left (-2 x^2+2^{2/3} x \sqrt [3]{1-x^3}-\sqrt [3]{2} \left (1-x^3\right )^{2/3}\right )\right ) \] Input:

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

Output:

((-12*(1 - x^3)^(5/3))/x^5 - 10*2^(2/3)*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x - 2^ 
(2/3)*(1 - x^3)^(1/3))] + 10*2^(2/3)*Log[2*x + 2^(2/3)*(1 - x^3)^(1/3)] - 
5*2^(2/3)*Log[-2*x^2 + 2^(2/3)*x*(1 - x^3)^(1/3) - 2^(1/3)*(1 - x^3)^(2/3) 
])/60
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {980, 25, 1053, 27, 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.

\(\displaystyle \int \frac {1}{x^6 \sqrt [3]{1-x^3} \left (x^3+1\right )} \, dx\)

\(\Big \downarrow \) 980

\(\displaystyle \frac {1}{5} \int -\frac {2-3 x^3}{x^3 \sqrt [3]{1-x^3} \left (x^3+1\right )}dx-\frac {\left (1-x^3\right )^{2/3}}{5 x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{5} \int \frac {2-3 x^3}{x^3 \sqrt [3]{1-x^3} \left (x^3+1\right )}dx-\frac {\left (1-x^3\right )^{2/3}}{5 x^5}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {1}{5} \left (\frac {1}{2} \int \frac {10}{\sqrt [3]{1-x^3} \left (x^3+1\right )}dx+\frac {\left (1-x^3\right )^{2/3}}{x^2}\right )-\frac {\left (1-x^3\right )^{2/3}}{5 x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (5 \int \frac {1}{\sqrt [3]{1-x^3} \left (x^3+1\right )}dx+\frac {\left (1-x^3\right )^{2/3}}{x^2}\right )-\frac {\left (1-x^3\right )^{2/3}}{5 x^5}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {1}{5} \left (5 \left (-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}\right )+\frac {\left (1-x^3\right )^{2/3}}{x^2}\right )-\frac {\left (1-x^3\right )^{2/3}}{5 x^5}\)

Input:

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

Output:

-1/5*(1 - x^3)^(2/3)/x^5 + ((1 - x^3)^(2/3)/x^2 + 5*(-(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))))/5
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[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 980
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) 
)^(q_), x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q 
 + 1)/(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^n*(m + 1))   Int[(e*x)^(m + n)*( 
a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) 
 + b*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, 
q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, 
b, c, d, e, m, n, p, q, x]
 

rule 1053
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_ 
))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b 
*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^n*( 
m + 1))   Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) 
- e*(b*c + a*d)*(m + n + 1) - e*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2 
) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 
0] && LtQ[m, -1]
 
Maple [A] (verified)

Time = 26.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02

method result size
pseudoelliptic \(\frac {10 \,2^{\frac {2}{3}} \ln \left (\frac {2^{\frac {1}{3}} x +\left (-x^{3}+1\right )^{\frac {1}{3}}}{x}\right ) x^{5}+12 \left (x^{3}-1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}}-5 \,2^{\frac {2}{3}} x^{5} \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2^{\frac {2}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}}+x \right )}{3 x}\right )+\ln \left (\frac {2^{\frac {2}{3}} x^{2}-2^{\frac {1}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}} x +\left (-x^{3}+1\right )^{\frac {2}{3}}}{x^{2}}\right )\right )}{60 x^{5}}\) \(127\)
risch \(\text {Expression too large to display}\) \(640\)
trager \(\text {Expression too large to display}\) \(1163\)

Input:

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

Output:

1/60*(10*2^(2/3)*ln((2^(1/3)*x+(-x^3+1)^(1/3))/x)*x^5+12*(x^3-1)*(-x^3+1)^ 
(2/3)-5*2^(2/3)*x^5*(-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)))/x^ 
5
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (95) = 190\).

Time = 1.80 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.23 \[ \int \frac {1}{x^6 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=-\frac {60 \cdot 2^{\frac {1}{6}} \sqrt {\frac {1}{6}} x^{5} \arctan \left (\frac {2^{\frac {1}{6}} \sqrt {\frac {1}{6}} {\left (6 \cdot 2^{\frac {2}{3}} {\left (5 \, x^{7} + 4 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (71 \, x^{9} - 111 \, x^{6} + 33 \, x^{3} - 1\right )} + 12 \, {\left (19 \, x^{8} - 16 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{109 \, x^{9} - 105 \, x^{6} + 3 \, x^{3} + 1}\right ) - 10 \cdot 2^{\frac {2}{3}} x^{5} \log \left (\frac {6 \cdot 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 2^{\frac {2}{3}} {\left (x^{3} + 1\right )} + 6 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x}{x^{3} + 1}\right ) + 5 \cdot 2^{\frac {2}{3}} x^{5} \log \left (\frac {3 \cdot 2^{\frac {2}{3}} {\left (5 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (19 \, x^{6} - 16 \, x^{3} + 1\right )} - 12 \, {\left (2 \, x^{5} - x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{6} + 2 \, x^{3} + 1}\right ) - 36 \, {\left (x^{3} - 1\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{180 \, x^{5}} \] Input:

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

Output:

-1/180*(60*2^(1/6)*sqrt(1/6)*x^5*arctan(2^(1/6)*sqrt(1/6)*(6*2^(2/3)*(5*x^ 
7 + 4*x^4 - x)*(-x^3 + 1)^(2/3) - 2^(1/3)*(71*x^9 - 111*x^6 + 33*x^3 - 1) 
+ 12*(19*x^8 - 16*x^5 + x^2)*(-x^3 + 1)^(1/3))/(109*x^9 - 105*x^6 + 3*x^3 
+ 1)) - 10*2^(2/3)*x^5*log((6*2^(1/3)*(-x^3 + 1)^(1/3)*x^2 + 2^(2/3)*(x^3 
+ 1) + 6*(-x^3 + 1)^(2/3)*x)/(x^3 + 1)) + 5*2^(2/3)*x^5*log((3*2^(2/3)*(5* 
x^4 - x)*(-x^3 + 1)^(2/3) + 2^(1/3)*(19*x^6 - 16*x^3 + 1) - 12*(2*x^5 - x^ 
2)*(-x^3 + 1)^(1/3))/(x^6 + 2*x^3 + 1)) - 36*(x^3 - 1)*(-x^3 + 1)^(2/3))/x 
^5
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^6 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int \frac {1}{\left (-x^{3}+1\right )^{\frac {1}{3}} x^{9}+\left (-x^{3}+1\right )^{\frac {1}{3}} x^{6}}d x \] Input:

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

Output:

int(1/(( - x**3 + 1)**(1/3)*x**9 + ( - x**3 + 1)**(1/3)*x**6),x)