Integrand size = 13, antiderivative size = 71 \[ \int \frac {1}{x \sqrt [3]{1+x^3}} \, dx=\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^3}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (-1+\sqrt [3]{1+x^3}\right )-\frac {1}{6} \log \left (1+\sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \]
1/3*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 = 71, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \sqrt [3]{1+x^3}} \, dx=\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^3}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (-1+\sqrt [3]{1+x^3}\right )-\frac {1}{6} \log \left (1+\sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \]
ArcTan[1/Sqrt[3] + (2*(1 + x^3)^(1/3))/Sqrt[3]]/Sqrt[3] + Log[-1 + (1 + x^ 3)^(1/3)]/3 - Log[1 + (1 + x^3)^(1/3) + (1 + x^3)^(2/3)]/6
Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.80, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {798, 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.
\(\displaystyle \int \frac {1}{x \sqrt [3]{x^3+1}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^3 \sqrt [3]{x^3+1}}dx^3\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {1}{3} \left (-\frac {3}{2} \int \frac {1}{1-\sqrt [3]{x^3+1}}d\sqrt [3]{x^3+1}+\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{x^3+1}+1}d\sqrt [3]{x^3+1}-\frac {1}{2} \log \left (x^3\right )\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{x^3+1}+1}d\sqrt [3]{x^3+1}-\frac {1}{2} \log \left (x^3\right )+\frac {3}{2} \log \left (1-\sqrt [3]{x^3+1}\right )\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{3} \left (-3 \int \frac {1}{-x^6-3}d\left (2 \sqrt [3]{x^3+1}+1\right )-\frac {1}{2} \log \left (x^3\right )+\frac {3}{2} \log \left (1-\sqrt [3]{x^3+1}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x^3+1}+1}{\sqrt {3}}\right )-\frac {\log \left (x^3\right )}{2}+\frac {3}{2} \log \left (1-\sqrt [3]{x^3+1}\right )\right )\) |
(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
3.10.33.3.1 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[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]
Time = 3.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77
method | result | size |
pseudoelliptic | \(\frac {\ln \left (-1+\left (x^{3}+1\right )^{\frac {1}{3}}\right )}{3}-\frac {\ln \left (1+\left (x^{3}+1\right )^{\frac {1}{3}}+\left (x^{3}+1\right )^{\frac {2}{3}}\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{3}+1\right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{3}\) | \(55\) |
meijerg | \(\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], -x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+3 \ln \left (x \right )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi }\) | \(63\) |
trager | \(\frac {\ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}-15 \left (x^{3}+1\right )^{\frac {2}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-10 x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+24 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+9 \left (x^{3}+1\right )^{\frac {2}{3}}-10 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+15 \left (x^{3}+1\right )^{\frac {1}{3}}-25}{x^{3}}\right )}{3}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+16 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}-15 \left (x^{3}+1\right )^{\frac {2}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+3 x^{3}-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}-24 \left (x^{3}+1\right )^{\frac {2}{3}}+19 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+15 \left (x^{3}+1\right )^{\frac {1}{3}}+4}{x^{3}}\right )}{3}\) | \(229\) |
1/3*ln(-1+(x^3+1)^(1/3))-1/6*ln(1+(x^3+1)^(1/3)+(x^3+1)^(2/3))+1/3*3^(1/2) *arctan(1/3*(2*(x^3+1)^(1/3)+1)*3^(1/2))
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x \sqrt [3]{1+x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{6} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
1/3*sqrt(3)*arctan(2/3*sqrt(3)*(x^3 + 1)^(1/3) + 1/3*sqrt(3)) - 1/6*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 = 0.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x \sqrt [3]{1+x^3}} \, dx=- \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x \Gamma \left (\frac {4}{3}\right )} \]
Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x \sqrt [3]{1+x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3) + 1)) - 1/6*log((x^3 + 1 )^(2/3) + (x^3 + 1)^(1/3) + 1) + 1/3*log((x^3 + 1)^(1/3) - 1)
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x \sqrt [3]{1+x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]
1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3) + 1)) - 1/6*log((x^3 + 1 )^(2/3) + (x^3 + 1)^(1/3) + 1) + 1/3*log(abs((x^3 + 1)^(1/3) - 1))
Time = 5.45 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x \sqrt [3]{1+x^3}} \, dx=\frac {\ln \left ({\left (x^3+1\right )}^{1/3}-1\right )}{3}+\ln \left ({\left (x^3+1\right )}^{1/3}-9\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left ({\left (x^3+1\right )}^{1/3}-9\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]