\(\int \frac {x^4}{\sqrt [3]{1-x^3} (1+x^3)} \, dx\) [833]

Optimal result

Integrand size = 22, antiderivative size = 254 \[ \int \frac {x^4}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\arctan \left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {1}{2} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )-\frac {\log \left ((1-x) (1+x)^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6 \sqrt [3]{2}}+\frac {\log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{3 \sqrt [3]{2}}+\frac {\log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \] Output:

-1/6*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 
)+1/2*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] (verified)

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

Time = 10.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.10 \[ \int \frac {x^4}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {1}{5} x^5 \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{3},1,\frac {8}{3},x^3,-x^3\right ) \] Input:

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

Output:

(x^5*AppellF1[5/3, 1/3, 1, 8/3, x^3, -x^3])/5
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {983, 888, 991, 750, 16, 27, 1142, 25, 27, 1082, 217, 1103, 2574}

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.

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

\(\Big \downarrow \) 983

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

\(\Big \downarrow \) 888

\(\displaystyle \frac {1}{2} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )-\int \frac {x}{\sqrt [3]{1-x^3} \left (x^3+1\right )}dx\)

\(\Big \downarrow \) 991

\(\displaystyle \frac {1}{3} \int \frac {1}{(x+1) \sqrt [3]{1-x^3}}dx+\int \frac {1}{\frac {2 (1-x)^3}{1-x^3}+1}d\frac {1-x}{\sqrt [3]{1-x^3}}+\frac {1}{2} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {1}{3} \int \frac {1}{(x+1) \sqrt [3]{1-x^3}}dx+\frac {1}{3} \int \frac {\sqrt [3]{2} \left (2^{2/3}-\frac {1-x}{\sqrt [3]{1-x^3}}\right )}{\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}d\frac {1-x}{\sqrt [3]{1-x^3}}+\frac {1}{3} \int \frac {1}{\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}d\frac {1-x}{\sqrt [3]{1-x^3}}+\frac {1}{2} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )\)

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

\(\Big \downarrow \) 2574

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2)   Int[(2*Rt[a, 3] - 
 Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; 
 FreeQ[{a, b}, x]
 

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 983

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

rule 991

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> 
With[{q = Rt[b/a, 3]}, Simp[-q^2/(3*d)   Int[1/((1 - q*x)*(a + b*x^3)^(1/3) 
), x], x] + Simp[q/d   Subst[Int[1/(1 + 2*a*x^3), x], x, (1 + q*x)/(a + b*x 
^3)^(1/3)], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[b*c + 
 a*d, 0]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 2574

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[ 
Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b, 3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sq 
rt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2^(7/3 
)*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^ 
(1/3)])/(2^(7/3)*Rt[b, 3]*c), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + 
a*d^3, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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