\(\int \frac {1}{x^9 \sqrt [3]{1-x^3} (1+x^3)} \, dx\) [831]

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 = 141 \[ \int \frac {1}{x^9 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=-\frac {\left (1-x^3\right )^{2/3}}{8 x^8}+\frac {\left (1-x^3\right )^{2/3}}{20 x^5}-\frac {17 \left (1-x^3\right )^{2/3}}{40 x^2}+\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\log \left (1+x^3\right )}{6 \sqrt [3]{2}}-\frac {\log \left (-\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}} \] Output:

-1/8*(-x^3+1)^(2/3)/x^8+1/20*(-x^3+1)^(2/3)/x^5-17/40*(-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 = 0.33 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^9 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {1}{120} \left (-\frac {3 \left (1-x^3\right )^{2/3} \left (5-2 x^3+17 x^6\right )}{x^8}+20\ 2^{2/3} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x-2^{2/3} \sqrt [3]{1-x^3}}\right )-20\ 2^{2/3} \log \left (2 x+2^{2/3} \sqrt [3]{1-x^3}\right )+10\ 2^{2/3} \log \left (-2 x^2+2^{2/3} x \sqrt [3]{1-x^3}-\sqrt [3]{2} \left (1-x^3\right )^{2/3}\right )\right ) \] Input:

Integrate[1/(x^9*(1 - x^3)^(1/3)*(1 + x^3)),x]
 

Output:

((-3*(1 - x^3)^(2/3)*(5 - 2*x^3 + 17*x^6))/x^8 + 20*2^(2/3)*Sqrt[3]*ArcTan 
[(Sqrt[3]*x)/(x - 2^(2/3)*(1 - x^3)^(1/3))] - 20*2^(2/3)*Log[2*x + 2^(2/3) 
*(1 - x^3)^(1/3)] + 10*2^(2/3)*Log[-2*x^2 + 2^(2/3)*x*(1 - x^3)^(1/3) - 2^ 
(1/3)*(1 - x^3)^(2/3)])/120
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {980, 27, 1053, 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.

\(\displaystyle \int \frac {1}{x^9 \sqrt [3]{1-x^3} \left (x^3+1\right )} \, dx\)

\(\Big \downarrow \) 980

\(\displaystyle \frac {1}{8} \int -\frac {2 \left (1-3 x^3\right )}{x^6 \sqrt [3]{1-x^3} \left (x^3+1\right )}dx-\frac {\left (1-x^3\right )^{2/3}}{8 x^8}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{4} \int \frac {1-3 x^3}{x^6 \sqrt [3]{1-x^3} \left (x^3+1\right )}dx-\frac {\left (1-x^3\right )^{2/3}}{8 x^8}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {1}{4} \left (\frac {1}{5} \int \frac {17-3 x^3}{x^3 \sqrt [3]{1-x^3} \left (x^3+1\right )}dx+\frac {\left (1-x^3\right )^{2/3}}{5 x^5}\right )-\frac {\left (1-x^3\right )^{2/3}}{8 x^8}\)

\(\Big \downarrow \) 1053

\(\displaystyle \frac {1}{4} \left (\frac {1}{5} \left (-\frac {1}{2} \int \frac {40}{\sqrt [3]{1-x^3} \left (x^3+1\right )}dx-\frac {17 \left (1-x^3\right )^{2/3}}{2 x^2}\right )+\frac {\left (1-x^3\right )^{2/3}}{5 x^5}\right )-\frac {\left (1-x^3\right )^{2/3}}{8 x^8}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {1}{5} \left (-20 \int \frac {1}{\sqrt [3]{1-x^3} \left (x^3+1\right )}dx-\frac {17 \left (1-x^3\right )^{2/3}}{2 x^2}\right )+\frac {\left (1-x^3\right )^{2/3}}{5 x^5}\right )-\frac {\left (1-x^3\right )^{2/3}}{8 x^8}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {1}{4} \left (\frac {1}{5} \left (-20 \left (-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}\right )-\frac {17 \left (1-x^3\right )^{2/3}}{2 x^2}\right )+\frac {\left (1-x^3\right )^{2/3}}{5 x^5}\right )-\frac {\left (1-x^3\right )^{2/3}}{8 x^8}\)

Input:

Int[1/(x^9*(1 - x^3)^(1/3)*(1 + x^3)),x]
 

Output:

-1/8*(1 - x^3)^(2/3)/x^8 + ((1 - x^3)^(2/3)/(5*x^5) + ((-17*(1 - x^3)^(2/3 
))/(2*x^2) - 20*(-(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)/4
 

Defintions of rubi rules used

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 = 26.48 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96

method result size
pseudoelliptic \(\frac {-20 \,2^{\frac {2}{3}} \ln \left (\frac {2^{\frac {1}{3}} x +\left (-x^{3}+1\right )^{\frac {1}{3}}}{x}\right ) x^{8}+\left (-51 x^{6}+6 x^{3}-15\right ) \left (-x^{3}+1\right )^{\frac {2}{3}}-10 \,2^{\frac {2}{3}} x^{8} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2^{\frac {2}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}}+x \right )}{3 x}\right )-\ln \left (\frac {2^{\frac {2}{3}} x^{2}-2^{\frac {1}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}} x +\left (-x^{3}+1\right )^{\frac {2}{3}}}{x^{2}}\right )\right )}{120 x^{8}}\) \(135\)
risch \(\text {Expression too large to display}\) \(941\)
trager \(\text {Expression too large to display}\) \(1168\)

Input:

int(1/x^9/(-x^3+1)^(1/3)/(x^3+1),x,method=_RETURNVERBOSE)
 

Output:

1/120*(-20*2^(2/3)*ln((2^(1/3)*x+(-x^3+1)^(1/3))/x)*x^8+(-51*x^6+6*x^3-15) 
*(-x^3+1)^(2/3)-10*2^(2/3)*x^8*(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^8
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (109) = 218\).

Time = 1.75 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.01 \[ \int \frac {1}{x^9 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {120 \cdot 2^{\frac {1}{6}} \sqrt {\frac {1}{6}} x^{8} \arctan \left (\frac {2^{\frac {1}{6}} \sqrt {\frac {1}{6}} {\left (6 \cdot 2^{\frac {2}{3}} {\left (5 \, x^{7} + 4 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (71 \, x^{9} - 111 \, x^{6} + 33 \, x^{3} - 1\right )} + 12 \, {\left (19 \, x^{8} - 16 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{109 \, x^{9} - 105 \, x^{6} + 3 \, x^{3} + 1}\right ) - 20 \cdot 2^{\frac {2}{3}} x^{8} \log \left (\frac {6 \cdot 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 2^{\frac {2}{3}} {\left (x^{3} + 1\right )} + 6 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x}{x^{3} + 1}\right ) + 10 \cdot 2^{\frac {2}{3}} x^{8} \log \left (\frac {3 \cdot 2^{\frac {2}{3}} {\left (5 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (19 \, x^{6} - 16 \, x^{3} + 1\right )} - 12 \, {\left (2 \, x^{5} - x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{6} + 2 \, x^{3} + 1}\right ) - 9 \, {\left (17 \, x^{6} - 2 \, x^{3} + 5\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{360 \, x^{8}} \] Input:

integrate(1/x^9/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="fricas")
 

Output:

1/360*(120*2^(1/6)*sqrt(1/6)*x^8*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)) - 20*2^(2/3)*x^8*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)) + 10*2^(2/3)*x^8*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)) - 9*(17*x^6 - 2*x^3 + 5)*(-x^3 + 
1)^(2/3))/x^8
 

Sympy [F]

\[ \int \frac {1}{x^9 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int \frac {1}{x^{9} \sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \] Input:

integrate(1/x**9/(-x**3+1)**(1/3)/(x**3+1),x)
 

Output:

Integral(1/(x**9*(-(x - 1)*(x**2 + x + 1))**(1/3)*(x + 1)*(x**2 - x + 1)), 
 x)
 

Maxima [F]

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

integrate(1/x^9/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="maxima")
 

Output:

integrate(1/((x^3 + 1)*(-x^3 + 1)^(1/3)*x^9), x)
 

Giac [F]

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

integrate(1/x^9/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="giac")
 

Output:

integrate(1/((x^3 + 1)*(-x^3 + 1)^(1/3)*x^9), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^9 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int \frac {1}{\left (-x^{3}+1\right )^{\frac {1}{3}} x^{12}+\left (-x^{3}+1\right )^{\frac {1}{3}} x^{9}}d x \] Input:

int(1/x^9/(-x^3+1)^(1/3)/(x^3+1),x)
 

Output:

int(1/(( - x**3 + 1)**(1/3)*x**12 + ( - x**3 + 1)**(1/3)*x**9),x)