\(\int \frac {1}{x^5 (1-x^3)^{4/3} (1+x^3)} \, dx\) [871]

Optimal result

Integrand size = 22, antiderivative size = 308 \[ \int \frac {1}{x^5 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {1}{2 x^4 \sqrt [3]{1-x^3}}-\frac {3 \left (1-x^3\right )^{2/3}}{4 x^4}-\frac {\left (1-x^3\right )^{2/3}}{x}+\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {\arctan \left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{4 \sqrt [3]{2} \sqrt {3}}-\frac {1}{2} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )+\frac {\log \left ((1-x) (1+x)^2\right )}{24 \sqrt [3]{2}}+\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6 \sqrt [3]{2}}-\frac {\log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{8 \sqrt [3]{2}} \] Output:

1/2/x^4/(-x^3+1)^(1/3)-3/4*(-x^3+1)^(2/3)/x^4-(-x^3+1)^(2/3)/x+1/12*arctan 
(1/3*(1-2*2^(1/3)*(1-x)/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)+1/24*arct 
an(1/3*(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)-1/2*x^2*h 
ypergeom([1/3, 2/3],[5/3],x^3)+1/48*ln((1-x)*(1+x)^2)*2^(2/3)+1/24*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)-1/12*ln 
(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3))*2^(2/3)-1/16*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 = 11.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^5 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=-\frac {\frac {5 \left (1+x^3-4 x^6\right )}{\sqrt [3]{1-x^3}}+5 x^6 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^3,-x^3\right )+4 x^9 \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{3},1,\frac {8}{3},x^3,-x^3\right )}{20 x^4} \] Input:

Integrate[1/(x^5*(1 - x^3)^(4/3)*(1 + x^3)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 308, 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 = {972, 1053, 27, 1053, 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.

\(\displaystyle \int \frac {1}{x^5 \left (1-x^3\right )^{4/3} \left (x^3+1\right )} \, dx\)

\(\Big \downarrow \) 972

\(\displaystyle \frac {1}{2} \int \frac {5 x^3+6}{x^5 \sqrt [3]{1-x^3} \left (x^3+1\right )}dx+\frac {1}{2 x^4 \sqrt [3]{1-x^3}}\)

\(\Big \downarrow \) 1053

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1053

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

\(\Big \downarrow \) 1054

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\arctan \left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+x^2 \left (-\operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )\right )-\frac {2 \left (1-x^3\right )^{2/3}}{x}+\frac {\log \left (\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{6 \sqrt [3]{2}}-\frac {\log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{3 \sqrt [3]{2}}-\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}+x-1\right )}{4 \sqrt [3]{2}}-\frac {3 \left (1-x^3\right )^{2/3}}{2 x^4}+\frac {\log \left ((1-x) (x+1)^2\right )}{12 \sqrt [3]{2}}\right )+\frac {1}{2 x^4 \sqrt [3]{1-x^3}}\)

Input:

Int[1/(x^5*(1 - x^3)^(4/3)*(1 + x^3)),x]
 

Output:

1/(2*x^4*(1 - x^3)^(1/3)) + ((-3*(1 - x^3)^(2/3))/(2*x^4) - (2*(1 - x^3)^( 
2/3))/x + ArcTan[(1 - (2*2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]]/(2^(1/ 
3)*Sqrt[3]) + ArcTan[(1 + (2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]]/(2*2 
^(1/3)*Sqrt[3]) - x^2*Hypergeometric2F1[1/3, 2/3, 5/3, x^3] + Log[(1 - x)* 
(1 + x)^2]/(12*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)]/(6*2^(1/3)) - Log[1 + (2^(1/3)*(1 - x))/( 
1 - x^3)^(1/3)]/(3*2^(1/3)) - Log[-1 + x + 2^(2/3)*(1 - x^3)^(1/3)]/(4*2^( 
1/3)))/2
 

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 972

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

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]

\[\int \frac {1}{x^{5} \left (-x^{3}+1\right )^{\frac {4}{3}} \left (x^{3}+1\right )}d x\]

Input:

int(1/x^5/(-x^3+1)^(4/3)/(x^3+1),x)
 

Output:

int(1/x^5/(-x^3+1)^(4/3)/(x^3+1),x)
 

Fricas [F]

\[ \int \frac {1}{x^5 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int { \frac {1}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {4}{3}} x^{5}} \,d x } \] Input:

integrate(1/x^5/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="fricas")
 

Output:

integral((-x^3 + 1)^(2/3)/(x^14 - x^11 - x^8 + x^5), x)
 

Sympy [F]

\[ \int \frac {1}{x^5 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int \frac {1}{x^{5} \left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {4}{3}} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \] Input:

integrate(1/x**5/(-x**3+1)**(4/3)/(x**3+1),x)
 

Output:

Integral(1/(x**5*(-(x - 1)*(x**2 + x + 1))**(4/3)*(x + 1)*(x**2 - x + 1)), 
 x)
 

Maxima [F]

\[ \int \frac {1}{x^5 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int { \frac {1}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {4}{3}} x^{5}} \,d x } \] Input:

integrate(1/x^5/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="maxima")
 

Output:

integrate(1/((x^3 + 1)*(-x^3 + 1)^(4/3)*x^5), x)
 

Giac [F]

\[ \int \frac {1}{x^5 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int { \frac {1}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {4}{3}} x^{5}} \,d x } \] Input:

integrate(1/x^5/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="giac")
 

Output:

integrate(1/((x^3 + 1)*(-x^3 + 1)^(4/3)*x^5), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int \frac {1}{x^5\,{\left (1-x^3\right )}^{4/3}\,\left (x^3+1\right )} \,d x \] Input:

int(1/(x^5*(1 - x^3)^(4/3)*(x^3 + 1)),x)
 

Output:

int(1/(x^5*(1 - x^3)^(4/3)*(x^3 + 1)), x)
 

Reduce [F]

\[ \int \frac {1}{x^5 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=-\left (\int \frac {1}{\left (-x^{3}+1\right )^{\frac {1}{3}} x^{11}-\left (-x^{3}+1\right )^{\frac {1}{3}} x^{5}}d x \right ) \] Input:

int(1/x^5/(-x^3+1)^(4/3)/(x^3+1),x)
 

Output:

 - int(1/(( - x**3 + 1)**(1/3)*x**11 - ( - x**3 + 1)**(1/3)*x**5),x)