\(\int x^3 (1+x^3)^{2/3} \, dx\) [1439]

Optimal result

Integrand size = 13, antiderivative size = 102 \[ \int x^3 \left (1+x^3\right )^{2/3} \, dx=\frac {1}{18} \left (1+x^3\right )^{2/3} \left (2 x+3 x^4\right )-\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{1+x^3}}\right )}{9 \sqrt {3}}+\frac {1}{27} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{54} \log \left (x^2+x \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \] Output:

1/18*(x^3+1)^(2/3)*(3*x^4+2*x)-1/27*arctan(3^(1/2)*x/(x+2*(x^3+1)^(1/3)))* 
3^(1/2)+1/27*ln(-x+(x^3+1)^(1/3))-1/54*ln(x^2+x*(x^3+1)^(1/3)+(x^3+1)^(2/3 
))
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.95 \[ \int x^3 \left (1+x^3\right )^{2/3} \, dx=\frac {1}{54} \left (3 x \left (1+x^3\right )^{2/3} \left (2+3 x^3\right )-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{1+x^3}}\right )+2 \log \left (-x+\sqrt [3]{1+x^3}\right )-\log \left (x^2+x \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right )\right ) \] Input:

Integrate[x^3*(1 + x^3)^(2/3),x]
 

Output:

(3*x*(1 + x^3)^(2/3)*(2 + 3*x^3) - 2*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2*(1 
+ x^3)^(1/3))] + 2*Log[-x + (1 + x^3)^(1/3)] - Log[x^2 + x*(1 + x^3)^(1/3) 
 + (1 + x^3)^(2/3)])/54
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {811, 843, 769}

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.

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

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 769

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

Input:

Int[x^3*(1 + x^3)^(2/3),x]
 

Output:

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

Defintions of rubi rules used

rule 769

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

rule 811

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

rule 843

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 1.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.17

method	result	size
meijerg	\(\frac {x^{4} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], -x^{3}\right )}{4}\)	\(17\)
risch	\(\frac {x \left (3 x^{3}+2\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{18}-\frac {x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{3}\right )}{9}\)	\(33\)
pseudoelliptic	\(\frac {-\ln \left (\frac {x^{2}+x \left (x^{3}+1\right )^{\frac {1}{3}}+\left (x^{3}+1\right )^{\frac {2}{3}}}{x^{2}}\right )+2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x +2 \left (x^{3}+1\right )^{\frac {1}{3}}\right )}{3 x}\right )+2 \ln \left (\frac {-x +\left (x^{3}+1\right )^{\frac {1}{3}}}{x}\right )+\left (9 x^{4}+6 x \right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{54 {\left (\left (x^{3}+1\right )^{\frac {2}{3}}+x \left (x +\left (x^{3}+1\right )^{\frac {1}{3}}\right )\right )}^{2} {\left (-\left (x^{3}+1\right )^{\frac {1}{3}}+x \right )}^{2}}\)	\(125\)
trager	\(\frac {x \left (3 x^{3}+2\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{18}+\frac {\ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}} x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}+1\right )^{\frac {2}{3}}+3 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}+4 x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{27}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+1\right )}{27}-\frac {\ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+1\right )}{27}\)	\(289\)

Input:

int(x^3*(x^3+1)^(2/3),x,method=_RETURNVERBOSE)
 

Output:

1/4*x^4*hypergeom([-2/3,4/3],[7/3],-x^3)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.92 \[ \int x^3 \left (1+x^3\right )^{2/3} \, dx=\frac {1}{18} \, {\left (3 \, x^{4} + 2 \, x\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}} + \frac {1}{27} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {1}{27} \, \log \left (-\frac {x - {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{54} \, \log \left (\frac {x^{2} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \] Input:

integrate(x^3*(x^3+1)^(2/3),x, algorithm="fricas")
 

Output:

1/18*(3*x^4 + 2*x)*(x^3 + 1)^(2/3) + 1/27*sqrt(3)*arctan(1/3*(sqrt(3)*x + 
2*sqrt(3)*(x^3 + 1)^(1/3))/x) + 1/27*log(-(x - (x^3 + 1)^(1/3))/x) - 1/54* 
log((x^2 + (x^3 + 1)^(1/3)*x + (x^3 + 1)^(2/3))/x^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.30 \[ \int x^3 \left (1+x^3\right )^{2/3} \, dx=\frac {x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \] Input:

integrate(x**3*(x**3+1)**(2/3),x)
 

Output:

x**4*gamma(4/3)*hyper((-2/3, 4/3), (7/3,), x**3*exp_polar(I*pi))/(3*gamma( 
7/3))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.19 \[ \int x^3 \left (1+x^3\right )^{2/3} \, dx=\frac {1}{27} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \frac {\frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}} + \frac {2 \, {\left (x^{3} + 1\right )}^{\frac {5}{3}}}{x^{5}}}{18 \, {\left (\frac {2 \, {\left (x^{3} + 1\right )}}{x^{3}} - \frac {{\left (x^{3} + 1\right )}^{2}}{x^{6}} - 1\right )}} - \frac {1}{54} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) + \frac {1}{27} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \] Input:

integrate(x^3*(x^3+1)^(2/3),x, algorithm="maxima")
 

Output:

1/27*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3)/x + 1)) - 1/18*((x^3 + 
1)^(2/3)/x^2 + 2*(x^3 + 1)^(5/3)/x^5)/(2*(x^3 + 1)/x^3 - (x^3 + 1)^2/x^6 - 
 1) - 1/54*log((x^3 + 1)^(1/3)/x + (x^3 + 1)^(2/3)/x^2 + 1) + 1/27*log((x^ 
3 + 1)^(1/3)/x - 1)
 

Giac [F]

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

integrate(x^3*(x^3+1)^(2/3),x, algorithm="giac")
 

Output:

integrate((x^3 + 1)^(2/3)*x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (1+x^3\right )^{2/3} \, dx=\int x^3\,{\left (x^3+1\right )}^{2/3} \,d x \] Input:

int(x^3*(x^3 + 1)^(2/3),x)
 

Output:

int(x^3*(x^3 + 1)^(2/3), x)
 

Reduce [F]

\[ \int x^3 \left (1+x^3\right )^{2/3} \, dx=\frac {\left (x^{3}+1\right )^{\frac {2}{3}} x^{4}}{6}+\frac {\left (x^{3}+1\right )^{\frac {2}{3}} x}{9}-\frac {\left (\int \frac {1}{\left (x^{3}+1\right )^{\frac {1}{3}}}d x \right )}{9} \] Input:

int(x^3*(x^3+1)^(2/3),x)
 

Output:

(3*(x**3 + 1)**(2/3)*x**4 + 2*(x**3 + 1)**(2/3)*x - 2*int((x**3 + 1)**(2/3 
)/(x**3 + 1),x))/18