Integrand size = 16, antiderivative size = 140 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-1+x^3}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \text {arctanh}\left (\frac {1+x}{-1+x-2 \sqrt [3]{-1+x^3}}\right )-\frac {1}{6} \log \left (1-\sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )+\frac {1}{6} \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right ) \]
1/3*arctan(3^(1/2)*x/(x+2*(x^3-1)^(1/3)))*3^(1/2)-1/3*arctan(-1/3*3^(1/2)+ 2/3*(x^3-1)^(1/3)*3^(1/2))*3^(1/2)-2/3*arctanh((1+x)/(-1+x-2*(x^3-1)^(1/3) ))-1/6*ln(1-(x^3-1)^(1/3)+(x^3-1)^(2/3))+1/6*ln(x^2+x*(x^3-1)^(1/3)+(x^3-1 )^(2/3))
Time = 4.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.98 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {1}{6} \left (2 \sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{-1+x^3}}{\sqrt {3}}\right )+2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )+4 \text {arctanh}\left (\frac {1+x}{1-x+2 \sqrt [3]{-1+x^3}}\right )-\log \left (1-\sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )+\log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )\right ) \]
(2*Sqrt[3]*ArcTan[(1 - 2*(-1 + x^3)^(1/3))/Sqrt[3]] + 2*Sqrt[3]*ArcTan[(Sq rt[3]*x)/(x + 2*(-1 + x^3)^(1/3))] + 4*ArcTanh[(1 + x)/(1 - x + 2*(-1 + x^ 3)^(1/3))] - Log[1 - (-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)] + Log[x^2 + x*(- 1 + x^3)^(1/3) + (-1 + x^3)^(2/3)])/6
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.67, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2383, 2009}
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.
\(\displaystyle \int \frac {x-1}{x \sqrt [3]{x^3-1}} \, dx\) |
\(\Big \downarrow \) 2383 |
\(\displaystyle \int \left (\frac {1}{\sqrt [3]{x^3-1}}-\frac {1}{x \sqrt [3]{x^3-1}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {1-2 \sqrt [3]{x^3-1}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (\sqrt [3]{x^3-1}+1\right )-\frac {1}{2} \log \left (\sqrt [3]{x^3-1}-x\right )-\frac {\log (x)}{2}\) |
ArcTan[(1 + (2*x)/(-1 + x^3)^(1/3))/Sqrt[3]]/Sqrt[3] + ArcTan[(1 - 2*(-1 + x^3)^(1/3))/Sqrt[3]]/Sqrt[3] - Log[x]/2 + Log[1 + (-1 + x^3)^(1/3)]/2 - L og[-x + (-1 + x^3)^(1/3)]/2
3.20.72.3.1 Defintions of rubi rules used
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> I nt[ExpandIntegrand[(c*x)^m*Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n , p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) && !IGtQ[m, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.84 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.81
method | result | size |
meijerg | \(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} \left (\frac {2 \pi \sqrt {3}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+3 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}+\frac {{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{\operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}}}\) | \(113\) |
trager | \(\text {Expression too large to display}\) | \(2489\) |
-1/6/Pi*3^(1/2)*GAMMA(2/3)/signum(x^3-1)^(1/3)*(-signum(x^3-1))^(1/3)*(2/9 *Pi*3^(1/2)/GAMMA(2/3)*x^3*hypergeom([1,1,4/3],[2,2],x^3)+2/3*(-1/6*Pi*3^( 1/2)-3/2*ln(3)+3*ln(x)+I*Pi)*Pi*3^(1/2)/GAMMA(2/3))+1/signum(x^3-1)^(1/3)* (-signum(x^3-1))^(1/3)*x*hypergeom([1/3,1/3],[4/3],x^3)
Time = 0.52 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.40 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2 \, \sqrt {3} {\left (x^{4} + 2 \, x^{3} + x^{2} - x - 1\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}} - 2 \, \sqrt {3} {\left (x^{5} + x^{4} - x^{3} - 2 \, x^{2} - x\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}} + \sqrt {3} {\left (x^{5} + 2 \, x^{4} + 2 \, x^{3} - x^{2} - 2 \, x - 1\right )}}{3 \, {\left (2 \, x^{6} + 3 \, x^{5} - 4 \, x^{3} - 3 \, x^{2} + 1\right )}}\right ) + \frac {1}{3} \, \log \left (-x^{3} - x^{2} - {\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x + 1\right )} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x^{2} + x\right )} + 1\right ) - \frac {1}{6} \, \log \left (-x^{3} + {\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x + 1\right )} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + x + 1\right ) \]
1/3*sqrt(3)*arctan(1/3*(2*sqrt(3)*(x^4 + 2*x^3 + x^2 - x - 1)*(x^3 - 1)^(2 /3) - 2*sqrt(3)*(x^5 + x^4 - x^3 - 2*x^2 - x)*(x^3 - 1)^(1/3) + sqrt(3)*(x ^5 + 2*x^4 + 2*x^3 - x^2 - 2*x - 1))/(2*x^6 + 3*x^5 - 4*x^3 - 3*x^2 + 1)) + 1/3*log(-x^3 - x^2 - (x^3 - 1)^(2/3)*(x + 1) - (x^3 - 1)^(1/3)*(x^2 + x) + 1) - 1/6*log(-x^3 + (x^3 - 1)^(2/3)*(x + 1) - (x^3 - 1)^(1/3)*(x + 1) + x + 1)
Result contains complex when optimal does not.
Time = 1.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.43 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {x e^{- \frac {i \pi }{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{3}}} \right )}}{3 x \Gamma \left (\frac {4}{3}\right )} \]
x*exp(-I*pi/3)*gamma(1/3)*hyper((1/3, 1/3), (4/3,), x**3)/(3*gamma(4/3)) + gamma(1/3)*hyper((1/3, 1/3), (4/3,), exp_polar(2*I*pi)/x**3)/(3*x*gamma(4 /3))
Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.89 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) - \frac {1}{6} \, \log \left ({\left (x^{3} - 1\right )}^{\frac {2}{3}} - {\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) - \frac {1}{3} \, \log \left (\frac {{\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \]
-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 - 1)^(1/3) - 1)) - 1/3*sqrt(3)*arc tan(1/3*sqrt(3)*(2*(x^3 - 1)^(1/3)/x + 1)) - 1/6*log((x^3 - 1)^(2/3) - (x^ 3 - 1)^(1/3) + 1) + 1/3*log((x^3 - 1)^(1/3) + 1) + 1/6*log((x^3 - 1)^(1/3) /x + (x^3 - 1)^(2/3)/x^2 + 1) - 1/3*log((x^3 - 1)^(1/3)/x - 1)
\[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\int { \frac {x - 1}{{\left (x^{3} - 1\right )}^{\frac {1}{3}} x} \,d x } \]
Time = 5.81 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.71 \[ \int \frac {-1+x}{x \sqrt [3]{-1+x^3}} \, dx=\frac {\ln \left ({\left (x^3-1\right )}^{1/3}+1\right )}{3}+\ln \left (9\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+{\left (x^3-1\right )}^{1/3}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (9\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+{\left (x^3-1\right )}^{1/3}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {x\,{\left (1-x^3\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ x^3\right )}{{\left (x^3-1\right )}^{1/3}} \]