Integrand size = 15, antiderivative size = 94 \[ \int \frac {-1+x^3}{\sqrt [3]{2+x^3}} \, dx=\frac {1}{3} x \left (2+x^3\right )^{2/3}-\frac {5 \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{2+x^3}}\right )}{3 \sqrt {3}}+\frac {5}{9} \log \left (-x+\sqrt [3]{2+x^3}\right )-\frac {5}{18} \log \left (x^2+x \sqrt [3]{2+x^3}+\left (2+x^3\right )^{2/3}\right ) \]
1/3*x*(x^3+2)^(2/3)-5/9*arctan(3^(1/2)*x/(x+2*(x^3+2)^(1/3)))*3^(1/2)+5/9* ln(-x+(x^3+2)^(1/3))-5/18*ln(x^2+x*(x^3+2)^(1/3)+(x^3+2)^(2/3))
Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96 \[ \int \frac {-1+x^3}{\sqrt [3]{2+x^3}} \, dx=\frac {1}{18} \left (6 x \left (2+x^3\right )^{2/3}-10 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{2+x^3}}\right )+10 \log \left (-x+\sqrt [3]{2+x^3}\right )-5 \log \left (x^2+x \sqrt [3]{2+x^3}+\left (2+x^3\right )^{2/3}\right )\right ) \]
(6*x*(2 + x^3)^(2/3) - 10*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2*(2 + x^3)^(1/3 ))] + 10*Log[-x + (2 + x^3)^(1/3)] - 5*Log[x^2 + x*(2 + x^3)^(1/3) + (2 + x^3)^(2/3)])/18
Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.69, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {913, 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.
\(\displaystyle \int \frac {x^3-1}{\sqrt [3]{x^3+2}} \, dx\) |
\(\Big \downarrow \) 913 |
\(\displaystyle \frac {1}{3} x \left (x^3+2\right )^{2/3}-\frac {5}{3} \int \frac {1}{\sqrt [3]{x^3+2}}dx\) |
\(\Big \downarrow \) 769 |
\(\displaystyle \frac {1}{3} x \left (x^3+2\right )^{2/3}-\frac {5}{3} \left (\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3+2}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (\sqrt [3]{x^3+2}-x\right )\right )\) |
(x*(2 + x^3)^(2/3))/3 - (5*(ArcTan[(1 + (2*x)/(2 + x^3)^(1/3))/Sqrt[3]]/Sq rt[3] - Log[-x + (2 + x^3)^(1/3)]/2))/3
3.13.96.3.1 Defintions of rubi rules used
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]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( p + 1) + 1))/(b*(n*(p + 1) + 1)) Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b , c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 4.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\frac {x \left (x^{3}+2\right )^{\frac {2}{3}}}{3}-\frac {5 \,2^{\frac {2}{3}} x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -\frac {x^{3}}{2}\right )}{6}\) | \(29\) |
meijerg | \(-\frac {2^{\frac {2}{3}} x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -\frac {x^{3}}{2}\right )}{2}+\frac {2^{\frac {2}{3}} x^{4} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], -\frac {x^{3}}{2}\right )}{8}\) | \(38\) |
pseudoelliptic | \(\frac {6 x \left (x^{3}+2\right )^{\frac {2}{3}}+10 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x +2 \left (x^{3}+2\right )^{\frac {1}{3}}\right )}{3 x}\right )+10 \ln \left (\frac {-x +\left (x^{3}+2\right )^{\frac {1}{3}}}{x}\right )-5 \ln \left (\frac {x^{2}+x \left (x^{3}+2\right )^{\frac {1}{3}}+\left (x^{3}+2\right )^{\frac {2}{3}}}{x^{2}}\right )}{9 \left (-x +\left (x^{3}+2\right )^{\frac {1}{3}}\right ) \left (x^{2}+x \left (x^{3}+2\right )^{\frac {1}{3}}+\left (x^{3}+2\right )^{\frac {2}{3}}\right )}\) | \(119\) |
trager | \(\frac {x \left (x^{3}+2\right )^{\frac {2}{3}}}{3}+\frac {5 \ln \left (-4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{3}+6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+2\right )^{\frac {1}{3}} x^{2}-8 x^{3} \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )+3 x \left (x^{3}+2\right )^{\frac {2}{3}}-4 x^{3}-4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )-4\right )}{9}+\frac {10 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \ln \left (4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{3}-6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+2\right )^{\frac {2}{3}} x +6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+2\right )^{\frac {1}{3}} x^{2}+2 x^{3} \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )+3 \left (x^{3}+2\right )^{\frac {1}{3}} x^{2}-2 x^{3}-4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )-4\right )}{9}\) | \(226\) |
Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.91 \[ \int \frac {-1+x^3}{\sqrt [3]{2+x^3}} \, dx=\frac {1}{3} \, {\left (x^{3} + 2\right )}^{\frac {2}{3}} x + \frac {5}{9} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + 2\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \frac {5}{9} \, \log \left (-\frac {x - {\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x}\right ) - \frac {5}{18} \, \log \left (\frac {x^{2} + {\left (x^{3} + 2\right )}^{\frac {1}{3}} x + {\left (x^{3} + 2\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]
1/3*(x^3 + 2)^(2/3)*x + 5/9*sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x^3 + 2)^(1/3))/x) + 5/9*log(-(x - (x^3 + 2)^(1/3))/x) - 5/18*log((x^2 + (x^3 + 2)^(1/3)*x + (x^3 + 2)^(2/3))/x^2)
Result contains complex when optimal does not.
Time = 1.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76 \[ \int \frac {-1+x^3}{\sqrt [3]{2+x^3}} \, dx=\frac {2^{\frac {2}{3}} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {x^{3} e^{i \pi }}{2}} \right )}}{6 \Gamma \left (\frac {7}{3}\right )} - \frac {2^{\frac {2}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {x^{3} e^{i \pi }}{2}} \right )}}{6 \Gamma \left (\frac {4}{3}\right )} \]
2**(2/3)*x**4*gamma(4/3)*hyper((1/3, 4/3), (7/3,), x**3*exp_polar(I*pi)/2) /(6*gamma(7/3)) - 2**(2/3)*x*gamma(1/3)*hyper((1/3, 1/3), (4/3,), x**3*exp _polar(I*pi)/2)/(6*gamma(4/3))
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^3}{\sqrt [3]{2+x^3}} \, dx=\frac {5}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) + \frac {2 \, {\left (x^{3} + 2\right )}^{\frac {2}{3}}}{3 \, x^{2} {\left (\frac {x^{3} + 2}{x^{3}} - 1\right )}} - \frac {5}{18} \, \log \left (\frac {{\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + 2\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) + \frac {5}{9} \, \log \left (\frac {{\left (x^{3} + 2\right )}^{\frac {1}{3}}}{x} - 1\right ) \]
5/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 2)^(1/3)/x + 1)) + 2/3*(x^3 + 2)^ (2/3)/(x^2*((x^3 + 2)/x^3 - 1)) - 5/18*log((x^3 + 2)^(1/3)/x + (x^3 + 2)^( 2/3)/x^2 + 1) + 5/9*log((x^3 + 2)^(1/3)/x - 1)
\[ \int \frac {-1+x^3}{\sqrt [3]{2+x^3}} \, dx=\int { \frac {x^{3} - 1}{{\left (x^{3} + 2\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {-1+x^3}{\sqrt [3]{2+x^3}} \, dx=\int \frac {x^3-1}{{\left (x^3+2\right )}^{1/3}} \,d x \]