Integrand size = 11, antiderivative size = 83 \[ \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x^2+x^3}}\right )-\log \left (-x+\sqrt [3]{x^2+x^3}\right )+\frac {1}{2} \log \left (x^2+x \sqrt [3]{x^2+x^3}+\left (x^2+x^3\right )^{2/3}\right ) \]
3^(1/2)*arctan(3^(1/2)*x/(x+2*(x^3+x^2)^(1/3)))-ln(-x+(x^3+x^2)^(1/3))+1/2 *ln(x^2+x*(x^3+x^2)^(1/3)+(x^3+x^2)^(2/3))
Time = 0.16 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx=\frac {x^{2/3} \sqrt [3]{1+x} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2 \sqrt [3]{1+x}}\right )-2 \log \left (-\sqrt [3]{x}+\sqrt [3]{1+x}\right )+\log \left (x^{2/3}+\sqrt [3]{x} \sqrt [3]{1+x}+(1+x)^{2/3}\right )\right )}{2 \sqrt [3]{x^2 (1+x)}} \]
(x^(2/3)*(1 + x)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) + 2*(1 + x)^(1/3))] - 2*Log[-x^(1/3) + (1 + x)^(1/3)] + Log[x^(2/3) + x^(1/3)*(1 + x)^(1/3) + (1 + x)^(2/3)]))/(2*(x^2*(1 + x))^(1/3))
Time = 0.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1917, 71}
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.
\(\displaystyle \int \frac {1}{\sqrt [3]{x^3+x^2}} \, dx\) |
\(\Big \downarrow \) 1917 |
\(\displaystyle \frac {x^{2/3} \sqrt [3]{x+1} \int \frac {1}{x^{2/3} \sqrt [3]{x+1}}dx}{\sqrt [3]{x^3+x^2}}\) |
\(\Big \downarrow \) 71 |
\(\displaystyle \frac {x^{2/3} \sqrt [3]{x+1} \left (-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x+1}}{\sqrt {3} \sqrt [3]{x}}+\frac {1}{\sqrt {3}}\right )-\frac {\log (x)}{2}-\frac {3}{2} \log \left (\frac {\sqrt [3]{x+1}}{\sqrt [3]{x}}-1\right )\right )}{\sqrt [3]{x^3+x^2}}\) |
(x^(2/3)*(1 + x)^(1/3)*(-(Sqrt[3]*ArcTan[1/Sqrt[3] + (2*(1 + x)^(1/3))/(Sq rt[3]*x^(1/3))]) - Log[x]/2 - (3*Log[-1 + (1 + x)^(1/3)/x^(1/3)])/2))/(x^2 + x^3)^(1/3)
3.12.9.3.1 Defintions of rubi rules used
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, Simp[(-Sqrt[3])*(q/d)*ArcTan[2*q*((a + b*x)^(1/3)/( Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[3]], x] + (-Simp[3*(q/(2*d))*Log[q*((a + b*x)^(1/3)/(c + d*x)^(1/3)) - 1], x] - Simp[(q/(2*d))*Log[c + d*x], x])] / ; FreeQ[{a, b, c, d}, x] && PosQ[d/b]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]) Int[ x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !Integ erQ[p] && NeQ[n, j] && PosQ[n - j]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 3.43 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.18
method | result | size |
meijerg | \(3 x^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x \right )\) | \(15\) |
pseudoelliptic | \(-\ln \left (\frac {\left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}-x}{x}\right )+\frac {\ln \left (\frac {\left (x^{2} \left (1+x \right )\right )^{\frac {2}{3}}+\left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{2} \left (1+x \right )\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )\) | \(80\) |
trager | \(\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -15 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-15 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x -19 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-9 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -9 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-9 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-5 x^{2}-2 x}{x}\right )-\ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +15 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x +11 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+17 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-24 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-20 x^{2}-15 x}{x}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+\ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +15 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {2}{3}}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}\right )^{\frac {1}{3}} x +11 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+17 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -24 \left (x^{3}+x^{2}\right )^{\frac {2}{3}}-24 x \left (x^{3}+x^{2}\right )^{\frac {1}{3}}-20 x^{2}-15 x}{x}\right )\) | \(428\) |
Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \log \left (-\frac {x - {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} + x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} + x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]
-sqrt(3)*arctan(1/3*(sqrt(3)*x + 2*sqrt(3)*(x^3 + x^2)^(1/3))/x) - log(-(x - (x^3 + x^2)^(1/3))/x) + 1/2*log((x^2 + (x^3 + x^2)^(1/3)*x + (x^3 + x^2 )^(2/3))/x^2)
\[ \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx=\int \frac {1}{\sqrt [3]{x^{3} + x^{2}}}\, dx \]
\[ \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} + x^{2}\right )}^{\frac {1}{3}}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx=-\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{2} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \log \left ({\left | {\left (\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]
-sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x + 1)^(1/3) + 1)) + 1/2*log((1/x + 1)^( 2/3) + (1/x + 1)^(1/3) + 1) - log(abs((1/x + 1)^(1/3) - 1))
Time = 5.74 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\sqrt [3]{x^2+x^3}} \, dx=\frac {3\,x\,{\left (x+1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ -x\right )}{{\left (x^3+x^2\right )}^{1/3}} \]