Integrand size = 15, antiderivative size = 70 \[ \int \frac {\left (1-x^3\right )^{2/3}}{x} \, dx=\frac {1}{2} \left (1-x^3\right )^{2/3}+\frac {\arctan \left (\frac {1+2 \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\log (x)}{2}+\frac {1}{2} \log \left (1-\sqrt [3]{1-x^3}\right ) \]
1/2*(-x^3+1)^(2/3)-1/2*ln(x)+1/2*ln(1-(-x^3+1)^(1/3))+1/3*arctan(1/3*(1+2* (-x^3+1)^(1/3))*3^(1/2))*3^(1/2)
Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.29 \[ \int \frac {\left (1-x^3\right )^{2/3}}{x} \, dx=\frac {1}{6} \left (3 \left (1-x^3\right )^{2/3}+2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1-x^3}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{1-x^3}\right )-\log \left (1+\sqrt [3]{1-x^3}+\left (1-x^3\right )^{2/3}\right )\right ) \]
(3*(1 - x^3)^(2/3) + 2*Sqrt[3]*ArcTan[(1 + 2*(1 - x^3)^(1/3))/Sqrt[3]] + 2 *Log[-1 + (1 - x^3)^(1/3)] - Log[1 + (1 - x^3)^(1/3) + (1 - x^3)^(2/3)])/6
Time = 0.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {798, 60, 67, 16, 1083, 217}
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.
| \(\displaystyle \int \frac {\left (1-x^3\right )^{2/3}}{x} \, dx\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (1-x^3\right )^{2/3}}{x^3}dx^3\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {1}{x^3 \sqrt [3]{1-x^3}}dx^3+\frac {3}{2} \left (1-x^3\right )^{2/3}\right )\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {1}{3} \left (-\frac {3}{2} \int \frac {1}{1-\sqrt [3]{1-x^3}}d\sqrt [3]{1-x^3}+\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{1-x^3}+1}d\sqrt [3]{1-x^3}+\frac {3}{2} \left (1-x^3\right )^{2/3}-\frac {1}{2} \log \left (x^3\right )\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{1-x^3}+1}d\sqrt [3]{1-x^3}+\frac {3}{2} \left (1-x^3\right )^{2/3}-\frac {1}{2} \log \left (x^3\right )+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^3}\right )\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{3} \left (-3 \int \frac {1}{-x^6-3}d\left (2 \sqrt [3]{1-x^3}+1\right )+\frac {3}{2} \left (1-x^3\right )^{2/3}-\frac {1}{2} \log \left (x^3\right )+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^3}\right )\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )+\frac {3}{2} \left (1-x^3\right )^{2/3}-\frac {\log \left (x^3\right )}{2}+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^3}\right )\right )\) |
((3*(1 - x^3)^(2/3))/2 + Sqrt[3]*ArcTan[(1 + 2*(1 - x^3)^(1/3))/Sqrt[3]] - Log[x^3]/2 + (3*Log[1 - (1 - x^3)^(1/3)])/2)/3
3.2.7.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[3/(2*b*q) Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / ; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
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])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 1.93 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94
| method | result | size |
| meijerg | \(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+3 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}+\frac {2 \pi \sqrt {3}\, x^{3} {}_{3}^{}{\moversetsp {}{\mundersetsp {}{F_{2}^{}}}}\left (\frac {1}{3},1,1;2,2;x^{3}\right )}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{9 \pi }\) | \(66\) |
| pseudoelliptic | \(\frac {\left (-x^{3}+1\right )^{\frac {2}{3}}}{2}-\frac {\ln \left (\left (-x^{3}+1\right )^{\frac {2}{3}}+\left (-x^{3}+1\right )^{\frac {1}{3}}+1\right )}{6}+\frac {\arctan \left (\frac {\left (1+2 \left (-x^{3}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {\ln \left (\left (-x^{3}+1\right )^{\frac {1}{3}}-1\right )}{3}\) | \(74\) |
| trager | \(\frac {\left (-x^{3}+1\right )^{\frac {2}{3}}}{2}+\frac {\ln \left (\frac {-211 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-3126 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+5502 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}}-11543 x^{3}-14247 \left (-x^{3}+1\right )^{\frac {2}{3}}-19749 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}+1688 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-5502 \left (-x^{3}+1\right )^{\frac {1}{3}}+15935 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+21437}{x^{3}}\right )}{3}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-\frac {-1649 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+9683 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+5502 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}}+1266 x^{3}+19749 \left (-x^{3}+1\right )^{\frac {2}{3}}+14247 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}+13192 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-5502 \left (-x^{3}+1\right )^{\frac {1}{3}}-6557 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-1055}{x^{3}}\right )}{3}\) | \(257\) |
-1/9/Pi*3^(1/2)*GAMMA(2/3)*(-(3/2-1/6*Pi*3^(1/2)-3/2*ln(3)+3*ln(x)+I*Pi)*P i*3^(1/2)/GAMMA(2/3)+2/3*Pi*3^(1/2)/GAMMA(2/3)*x^3*hypergeom([1/3,1,1],[2, 2],x^3))
Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.07 \[ \int \frac {\left (1-x^3\right )^{2/3}}{x} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + \frac {1}{2} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
1/3*sqrt(3)*arctan(2/3*sqrt(3)*(-x^3 + 1)^(1/3) + 1/3*sqrt(3)) + 1/2*(-x^3 + 1)^(2/3) - 1/6*log((-x^3 + 1)^(2/3) + (-x^3 + 1)^(1/3) + 1) + 1/3*log(( -x^3 + 1)^(1/3) - 1)
Result contains complex when optimal does not.
Time = 0.58 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.59 \[ \int \frac {\left (1-x^3\right )^{2/3}}{x} \, dx=- \frac {x^{2} e^{\frac {2 i \pi }{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {1}{x^{3}}} \right )}}{3 \Gamma \left (\frac {1}{3}\right )} \]
Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04 \[ \int \frac {\left (1-x^3\right )^{2/3}}{x} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{2} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-x^3 + 1)^(1/3) + 1)) + 1/2*(-x^3 + 1)^ (2/3) - 1/6*log((-x^3 + 1)^(2/3) + (-x^3 + 1)^(1/3) + 1) + 1/3*log((-x^3 + 1)^(1/3) - 1)
Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.06 \[ \int \frac {\left (1-x^3\right )^{2/3}}{x} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{2} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - \frac {1}{6} \, \log \left ({\left (-x^{3} + 1\right )}^{\frac {2}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]
1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-x^3 + 1)^(1/3) + 1)) + 1/2*(-x^3 + 1)^ (2/3) - 1/6*log((-x^3 + 1)^(2/3) + (-x^3 + 1)^(1/3) + 1) + 1/3*log(abs((-x ^3 + 1)^(1/3) - 1))
Time = 0.45 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.30 \[ \int \frac {\left (1-x^3\right )^{2/3}}{x} \, dx=\frac {\ln \left ({\left (1-x^3\right )}^{1/3}-1\right )}{3}+\ln \left ({\left (1-x^3\right )}^{1/3}-9\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left ({\left (1-x^3\right )}^{1/3}-9\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {{\left (1-x^3\right )}^{2/3}}{2} \]
log((1 - x^3)^(1/3) - 1)/3 + log((1 - x^3)^(1/3) - 9*((3^(1/2)*1i)/6 - 1/6 )^2)*((3^(1/2)*1i)/6 - 1/6) - log((1 - x^3)^(1/3) - 9*((3^(1/2)*1i)/6 + 1/ 6)^2)*((3^(1/2)*1i)/6 + 1/6) + (1 - x^3)^(2/3)/2
\[ \int \frac {\left (1-x^3\right )^{2/3}}{x} \, dx=\frac {\left (-x^{3}+1\right )^{\frac {2}{3}}}{2}-\left (\int \frac {\left (-x^{3}+1\right )^{\frac {2}{3}}}{x^{4}-x}d x \right ) \]