Integrand size = 11, antiderivative size = 96 \[ \int x \sqrt [3]{1+x^3} \, dx=\frac {1}{3} x^2 \sqrt [3]{1+x^3}-\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{1+x^3}}\right )}{3 \sqrt {3}}-\frac {1}{9} \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {1}{18} \log \left (x^2+x \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \] Output:
1/3*x^2*(x^3+1)^(1/3)-1/9*arctan(3^(1/2)*x/(x+2*(x^3+1)^(1/3)))*3^(1/2)-1/ 9*ln(-x+(x^3+1)^(1/3))+1/18*ln(x^2+x*(x^3+1)^(1/3)+(x^3+1)^(2/3))
Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.94 \[ \int x \sqrt [3]{1+x^3} \, dx=\frac {1}{18} \left (6 x^2 \sqrt [3]{1+x^3}-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{1+x^3}}\right )-2 \log \left (-x+\sqrt [3]{1+x^3}\right )+\log \left (x^2+x \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right )\right ) \] Input:
Integrate[x*(1 + x^3)^(1/3),x]
Output:
(6*x^2*(1 + x^3)^(1/3) - 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)])/18
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.71, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {811, 853}
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.
\(\displaystyle \int x \sqrt [3]{x^3+1} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{3} \int \frac {x}{\left (x^3+1\right )^{2/3}}dx+\frac {1}{3} \sqrt [3]{x^3+1} x^2\) |
\(\Big \downarrow \) 853 |
\(\displaystyle \frac {1}{3} \left (-\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x-\sqrt [3]{x^3+1}\right )\right )+\frac {1}{3} \sqrt [3]{x^3+1} x^2\) |
Input:
Int[x*(1 + x^3)^(1/3),x]
Output:
(x^2*(1 + x^3)^(1/3))/3 + (-(ArcTan[(1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]]/S qrt[3]) - Log[x - (1 + x^3)^(1/3)]/2)/3
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]
Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim p[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp [Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 1.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.18
method | result | size |
meijerg | \(\frac {x^{2} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{3}\right )}{2}\) | \(17\) |
risch | \(\frac {x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}}{3}+\frac {x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{3}\right )}{6}\) | \(30\) |
pseudoelliptic | \(\frac {6 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}+2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x +2 \left (x^{3}+1\right )^{\frac {1}{3}}\right )}{3 x}\right )+\ln \left (\frac {x^{2}+x \left (x^{3}+1\right )^{\frac {1}{3}}+\left (x^{3}+1\right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-x +\left (x^{3}+1\right )^{\frac {1}{3}}}{x}\right )}{18 \left (x^{2}+x \left (x^{3}+1\right )^{\frac {1}{3}}+\left (x^{3}+1\right )^{\frac {2}{3}}\right ) \left (-x +\left (x^{3}+1\right )^{\frac {1}{3}}\right )}\) | \(119\) |
trager | \(\frac {x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}}{3}-\frac {\ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}-3 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}+x^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+1\right )}{9}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +3 \left (x^{3}+1\right )^{\frac {1}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{3}+1\right )^{\frac {2}{3}}-2 x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-1\right )}{9}\) | \(203\) |
Input:
int(x*(x^3+1)^(1/3),x,method=_RETURNVERBOSE)
Output:
1/2*x^2*hypergeom([-1/3,2/3],[5/3],-x^3)
Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.92 \[ \int x \sqrt [3]{1+x^3} \, dx=\frac {1}{3} \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \frac {1}{9} \, \log \left (-\frac {x - {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{18} \, \log \left (\frac {x^{2} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \] Input:
integrate(x*(x^3+1)^(1/3),x, algorithm="fricas")
Output:
1/3*(x^3 + 1)^(1/3)*x^2 + 1/9*sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x ^3 + 1)^(1/3))/x) - 1/9*log(-(x - (x^3 + 1)^(1/3))/x) + 1/18*log((x^2 + (x ^3 + 1)^(1/3)*x + (x^3 + 1)^(2/3))/x^2)
Result contains complex when optimal does not.
Time = 0.63 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.32 \[ \int x \sqrt [3]{1+x^3} \, dx=\frac {x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} \] Input:
integrate(x*(x**3+1)**(1/3),x)
Output:
x**2*gamma(2/3)*hyper((-1/3, 2/3), (5/3,), x**3*exp_polar(I*pi))/(3*gamma( 5/3))
Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98 \[ \int x \sqrt [3]{1+x^3} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) + \frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x {\left (\frac {x^{3} + 1}{x^{3}} - 1\right )}} + \frac {1}{18} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {1}{9} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \] Input:
integrate(x*(x^3+1)^(1/3),x, algorithm="maxima")
Output:
1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3)/x + 1)) + 1/3*(x^3 + 1)^ (1/3)/(x*((x^3 + 1)/x^3 - 1)) + 1/18*log((x^3 + 1)^(1/3)/x + (x^3 + 1)^(2/ 3)/x^2 + 1) - 1/9*log((x^3 + 1)^(1/3)/x - 1)
\[ \int x \sqrt [3]{1+x^3} \, dx=\int { {\left (x^{3} + 1\right )}^{\frac {1}{3}} x \,d x } \] Input:
integrate(x*(x^3+1)^(1/3),x, algorithm="giac")
Output:
integrate((x^3 + 1)^(1/3)*x, x)
Timed out. \[ \int x \sqrt [3]{1+x^3} \, dx=\int x\,{\left (x^3+1\right )}^{1/3} \,d x \] Input:
int(x*(x^3 + 1)^(1/3),x)
Output:
int(x*(x^3 + 1)^(1/3), x)
\[ \int x \sqrt [3]{1+x^3} \, dx=\frac {\left (x^{3}+1\right )^{\frac {1}{3}} x^{2}}{3}+\frac {\left (\int \frac {x}{\left (x^{3}+1\right )^{\frac {2}{3}}}d x \right )}{3} \] Input:
int(x*(x^3+1)^(1/3),x)
Output:
((x**3 + 1)**(1/3)*x**2 + int(((x**3 + 1)**(1/3)*x)/(x**3 + 1),x))/3