\(\int \frac {x^4}{(1-x^3)^{4/3} (1+x^3)} \, dx\) [868]

Optimal result

Integrand size = 22, antiderivative size = 274 \[ \int \frac {x^4}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {x^2}{2 \sqrt [3]{1-x^3}}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\arctan \left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{4 \sqrt [3]{2} \sqrt {3}}-\frac {1}{4} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )-\frac {\log \left ((1-x) (1+x)^2\right )}{24 \sqrt [3]{2}}-\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{12 \sqrt [3]{2}}+\frac {\log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6 \sqrt [3]{2}}+\frac {\log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{8 \sqrt [3]{2}} \] Output:

1/2*x^2/(-x^3+1)^(1/3)-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*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)-1/48*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 = 10.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.24 \[ \int \frac {x^4}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {1}{10} x^2 \left (\frac {5}{\sqrt [3]{1-x^3}}-5 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^3,-x^3\right )-x^3 \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{3},1,\frac {8}{3},x^3,-x^3\right )\right ) \] Input:

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

Output:

(x^2*(5/(1 - x^3)^(1/3) - 5*AppellF1[2/3, 1/3, 1, 5/3, x^3, -x^3] - x^3*Ap 
pellF1[5/3, 1/3, 1, 8/3, x^3, -x^3]))/10
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {971, 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.

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

\(\Big \downarrow \) 971

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

\(\Big \downarrow \) 1054

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

x^2/(2*(1 - x^3)^(1/3)) + (-(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])/2 - 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 971

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((-x^3 + 1)^(2/3)*x^4/(x^9 - x^6 - x^3 + 1), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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