\(\int \frac {1}{x^5 \sqrt [3]{1-x^3} (1+x^3)} \, dx\) [836]

Optimal result

Integrand size = 22, antiderivative size = 289 \[ \int \frac {1}{x^5 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=-\frac {\left (1-x^3\right )^{2/3}}{4 x^4}+\frac {\left (1-x^3\right )^{2/3}}{2 x}+\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\arctan \left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {1}{4} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )+\frac {\log \left ((1-x) (1+x)^2\right )}{12 \sqrt [3]{2}}+\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6 \sqrt [3]{2}}-\frac {\log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{3 \sqrt [3]{2}}-\frac {\log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \] Output:

-1/4*(-x^3+1)^(2/3)/x^4+1/2*(-x^3+1)^(2/3)/x+1/6*arctan(1/3*(1-2*2^(1/3)*( 
1-x)/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)+1/12*arctan(1/3*(1+2^(1/3)*( 
1-x)/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)+1/4*x^2*hypergeom([1/3, 2/3] 
,[5/3],x^3)+1/24*ln((1-x)*(1+x)^2)*2^(2/3)+1/12*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)-1/6*ln(1+2^(1/3)*(1-x)/(-x 
^3+1)^(1/3))*2^(2/3)-1/8*ln(-1+x+2^(2/3)*(-x^3+1)^(1/3))*2^(2/3)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 11.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^5 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {5 \left (1-x^3\right )^{2/3} \left (-1+2 x^3\right )+15 x^6 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^3,-x^3\right )+2 x^9 \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{3},1,\frac {8}{3},x^3,-x^3\right )}{20 x^4} \] Input:

Integrate[1/(x^5*(1 - x^3)^(1/3)*(1 + x^3)),x]
 

Output:

(5*(1 - x^3)^(2/3)*(-1 + 2*x^3) + 15*x^6*AppellF1[2/3, 1/3, 1, 5/3, x^3, - 
x^3] + 2*x^9*AppellF1[5/3, 1/3, 1, 8/3, x^3, -x^3])/(20*x^4)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {980, 27, 975, 25, 1054, 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.

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

\(\Big \downarrow \) 980

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 975

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1054

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {2^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {1}{2} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )+\frac {\left (1-x^3\right )^{2/3}}{x}+\frac {\log \left (\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{3 \sqrt [3]{2}}-\frac {1}{3} 2^{2/3} \log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )-\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}+x-1\right )}{2 \sqrt [3]{2}}+\frac {\log \left ((1-x) (x+1)^2\right )}{6 \sqrt [3]{2}}\right )-\frac {\left (1-x^3\right )^{2/3}}{4 x^4}\)

Input:

Int[1/(x^5*(1 - x^3)^(1/3)*(1 + x^3)),x]
 

Output:

-1/4*(1 - x^3)^(2/3)/x^4 + ((1 - x^3)^(2/3)/x + (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]) + (x^2*Hypergeometric2 
F1[1/3, 2/3, 5/3, x^3])/2 + Log[(1 - x)*(1 + x)^2]/(6*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[- 
1 + x + 2^(2/3)*(1 - x^3)^(1/3)]/(2*2^(1/3)))/2
 

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 975

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/ 
(a*e*(m + 1))), x] - Simp[1/(a*e^n*(m + 1))   Int[(e*x)^(m + n)*(a + b*x^n) 
^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m 
 + 1) + b*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && 
 NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] && IntBinomi 
alQ[a, b, c, d, e, m, n, p, q, x]
 

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 1054

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, p}, x] && IGtQ[n, 0]
 

rule 2009

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

Maple [F]

\[\int \frac {1}{x^{5} \left (-x^{3}+1\right )^{\frac {1}{3}} \left (x^{3}+1\right )}d x\]

Input:

int(1/x^5/(-x^3+1)^(1/3)/(x^3+1),x)
 

Output:

int(1/x^5/(-x^3+1)^(1/3)/(x^3+1),x)
 

Fricas [F]

\[ \int \frac {1}{x^5 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int { \frac {1}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{5}} \,d x } \] Input:

integrate(1/x^5/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="fricas")
 

Output:

integral(-(-x^3 + 1)^(2/3)/(x^11 - x^5), x)
 

Sympy [F]

\[ \int \frac {1}{x^5 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int \frac {1}{x^{5} \sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \] Input:

integrate(1/x**5/(-x**3+1)**(1/3)/(x**3+1),x)
 

Output:

Integral(1/(x**5*(-(x - 1)*(x**2 + x + 1))**(1/3)*(x + 1)*(x**2 - x + 1)), 
 x)
 

Maxima [F]

\[ \int \frac {1}{x^5 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int { \frac {1}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{5}} \,d x } \] Input:

integrate(1/x^5/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="maxima")
 

Output:

integrate(1/((x^3 + 1)*(-x^3 + 1)^(1/3)*x^5), x)
 

Giac [F]

\[ \int \frac {1}{x^5 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int { \frac {1}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{5}} \,d x } \] Input:

integrate(1/x^5/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="giac")
 

Output:

integrate(1/((x^3 + 1)*(-x^3 + 1)^(1/3)*x^5), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int \frac {1}{x^5\,{\left (1-x^3\right )}^{1/3}\,\left (x^3+1\right )} \,d x \] Input:

int(1/(x^5*(1 - x^3)^(1/3)*(x^3 + 1)),x)
 

Output:

int(1/(x^5*(1 - x^3)^(1/3)*(x^3 + 1)), x)
 

Reduce [F]

\[ \int \frac {1}{x^5 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int \frac {1}{\left (-x^{3}+1\right )^{\frac {1}{3}} x^{8}+\left (-x^{3}+1\right )^{\frac {1}{3}} x^{5}}d x \] Input:

int(1/x^5/(-x^3+1)^(1/3)/(x^3+1),x)
                                                                                    
                                                                                    
 

Output:

int(1/(( - x**3 + 1)**(1/3)*x**8 + ( - x**3 + 1)**(1/3)*x**5),x)