Integrand size = 13, antiderivative size = 84 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x} \, dx=\frac {1}{2} \left (1+x^3\right )^{2/3}+\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^3}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (-1+\sqrt [3]{1+x^3}\right )-\frac {1}{6} \log \left (1+\sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \]
1/2*(x^3+1)^(2/3)+1/3*arctan(1/3*3^(1/2)+2/3*(x^3+1)^(1/3)*3^(1/2))*3^(1/2 )+1/3*ln(-1+(x^3+1)^(1/3))-1/6*ln(1+(x^3+1)^(1/3)+(x^3+1)^(2/3))
Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.95 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x} \, dx=\frac {1}{6} \left (3 \left (1+x^3\right )^{2/3}+2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+x^3}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{1+x^3}\right )-\log \left (1+\sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right )\right ) \]
(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 = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, 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.
\(\displaystyle \int \frac {\left (x^3+1\right )^{2/3}}{x} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {\left (x^3+1\right )^{2/3}}{x^3}dx^3\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (\int \frac {1}{x^3 \sqrt [3]{x^3+1}}dx^3+\frac {3}{2} \left (x^3+1\right )^{2/3}\right )\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {1}{3} \left (-\frac {3}{2} \int \frac {1}{1-\sqrt [3]{x^3+1}}d\sqrt [3]{x^3+1}+\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{x^3+1}+1}d\sqrt [3]{x^3+1}+\frac {3}{2} \left (x^3+1\right )^{2/3}-\frac {1}{2} \log \left (x^3\right )\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{x^3+1}+1}d\sqrt [3]{x^3+1}+\frac {3}{2} \left (x^3+1\right )^{2/3}-\frac {1}{2} \log \left (x^3\right )+\frac {3}{2} \log \left (1-\sqrt [3]{x^3+1}\right )\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{3} \left (-3 \int \frac {1}{-x^6-3}d\left (2 \sqrt [3]{x^3+1}+1\right )+\frac {3}{2} \left (x^3+1\right )^{2/3}-\frac {1}{2} \log \left (x^3\right )+\frac {3}{2} \log \left (1-\sqrt [3]{x^3+1}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x^3+1}+1}{\sqrt {3}}\right )+\frac {3}{2} \left (x^3+1\right )^{2/3}-\frac {\log \left (x^3\right )}{2}+\frac {3}{2} \log \left (1-\sqrt [3]{x^3+1}\right )\right )\) |
((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
3.12.26.3.1 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 = 2.94 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.76
method | result | size |
meijerg | \(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{3} \operatorname {hypergeom}\left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], -x^{3}\right )}{3 \Gamma \left (\frac {2}{3}\right )}-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+3 \ln \left (x \right )\right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}\right )}{9 \pi }\) | \(64\) |
pseudoelliptic | \(\frac {\left (x^{3}+1\right )^{\frac {2}{3}}}{2}+\frac {\ln \left (-1+\left (x^{3}+1\right )^{\frac {1}{3}}\right )}{3}-\frac {\ln \left (1+\left (x^{3}+1\right )^{\frac {1}{3}}+\left (x^{3}+1\right )^{\frac {2}{3}}\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{3}+1\right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{3}\) | \(64\) |
trager | \(\frac {\left (x^{3}+1\right )^{\frac {2}{3}}}{2}+\frac {\ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \left (x^{3}+1\right )^{\frac {2}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+10 x^{3}-9 \left (x^{3}+1\right )^{\frac {2}{3}}-24 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-15 \left (x^{3}+1\right )^{\frac {1}{3}}+10 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+25}{x^{3}}\right )}{3}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-16 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \left (x^{3}+1\right )^{\frac {2}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-3 x^{3}+24 \left (x^{3}+1\right )^{\frac {2}{3}}+9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-15 \left (x^{3}+1\right )^{\frac {1}{3}}-19 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-4}{x^{3}}\right )}{3}\) | \(237\) |
-1/9/Pi*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)-(3/2-1/6*Pi*3^(1/2)-3/2*ln(3)+3*ln(x))*Pi*3^(1/2)/GAMMA(2 /3))
Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + \frac {1}{2} \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} - \frac {1}{6} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
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 = 0.54 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.43 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x} \, dx=- \frac {x^{2} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 \Gamma \left (\frac {1}{3}\right )} \]
Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{2} \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} - \frac {1}{6} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
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 = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.76 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{2} \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} - \frac {1}{6} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]
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 = 5.49 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x} \, dx=\frac {\ln \left ({\left (x^3+1\right )}^{1/3}-1\right )}{3}+\frac {{\left (x^3+1\right )}^{2/3}}{2}+\ln \left ({\left (x^3+1\right )}^{1/3}-9\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left ({\left (x^3+1\right )}^{1/3}-9\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]