Integrand size = 15, antiderivative size = 122 \[ \int \frac {1}{\sqrt [3]{-1-x+x^2+x^3}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1-x+x^2+x^3}}{2+2 x+\sqrt [3]{-1-x+x^2+x^3}}\right )-\log \left (-1-x+\sqrt [3]{-1-x+x^2+x^3}\right )+\frac {1}{2} \log \left (1+2 x+x^2+(1+x) \sqrt [3]{-1-x+x^2+x^3}+\left (-1-x+x^2+x^3\right )^{2/3}\right ) \]
-3^(1/2)*arctan(3^(1/2)*(x^3+x^2-x-1)^(1/3)/(2+2*x+(x^3+x^2-x-1)^(1/3)))-l n(-1-x+(x^3+x^2-x-1)^(1/3))+1/2*ln(1+2*x+x^2+(1+x)*(x^3+x^2-x-1)^(1/3)+(x^ 3+x^2-x-1)^(2/3))
Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\sqrt [3]{-1-x+x^2+x^3}} \, dx=\frac {\sqrt [3]{-1+x} (1+x)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1+x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{1+x}}\right )-2 \log \left (\sqrt [3]{-1+x}-\sqrt [3]{1+x}\right )+\log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{1+x}+(1+x)^{2/3}\right )\right )}{2 \sqrt [3]{(-1+x) (1+x)^2}} \]
((-1 + x)^(1/3)*(1 + x)^(2/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*(1 + x)^(1/3))/(2 *(-1 + x)^(1/3) + (1 + x)^(1/3))] - 2*Log[(-1 + x)^(1/3) - (1 + x)^(1/3)] + Log[(-1 + x)^(2/3) + (-1 + x)^(1/3)*(1 + x)^(1/3) + (1 + x)^(2/3)]))/(2* ((-1 + x)*(1 + x)^2)^(1/3))
Time = 0.22 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2477, 473, 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.
\(\displaystyle \int \frac {1}{\sqrt [3]{x^3+x^2-x-1}} \, dx\) |
\(\Big \downarrow \) 2477 |
\(\displaystyle \frac {\sqrt [3]{x+1} \sqrt [3]{x^2-1} \int \frac {1}{\sqrt [3]{x+1} \sqrt [3]{x^2-1}}dx}{\sqrt [3]{x^3+x^2-x-1}}\) |
\(\Big \downarrow \) 473 |
\(\displaystyle \frac {\left (x^2-1\right ) \int \frac {1}{\sqrt [3]{x-1} (x+1)^{2/3}}dx}{(x-1)^{2/3} \sqrt [3]{x+1} \sqrt [3]{x^3+x^2-x-1}}\) |
\(\Big \downarrow \) 71 |
\(\displaystyle \frac {\left (x^2-1\right ) \left (-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x-1}}{\sqrt {3} \sqrt [3]{x+1}}+\frac {1}{\sqrt {3}}\right )-\frac {1}{2} \log (x+1)-\frac {3}{2} \log \left (\frac {\sqrt [3]{x-1}}{\sqrt [3]{x+1}}-1\right )\right )}{(x-1)^{2/3} \sqrt [3]{x+1} \sqrt [3]{x^3+x^2-x-1}}\) |
((-1 + x^2)*(-(Sqrt[3]*ArcTan[1/Sqrt[3] + (2*(-1 + x)^(1/3))/(Sqrt[3]*(1 + x)^(1/3))]) - Log[1 + x]/2 - (3*Log[-1 + (-1 + x)^(1/3)/(1 + x)^(1/3)])/2 ))/((-1 + x)^(2/3)*(1 + x)^(1/3)*(-1 - x + x^2 + x^3)^(1/3))
3.18.99.3.1 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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 1))) Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) && !Gt Q[a, 0] && !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
Int[(Px_)^(p_), x_Symbol] :> With[{a = Coeff[Px, x, 0], b = Coeff[Px, x, 1] , c = Coeff[Px, x, 2], d = Coeff[Px, x, 3]}, Simp[Px^p/((c + d*x)^p*(b + d* x^2)^p) Int[(c + d*x)^p*(b + d*x^2)^p, x], x] /; EqQ[b*c - a*d, 0]] /; Fr eeQ[p, x] && PolyQ[Px, x, 3] && !IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.29 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.89
method | result | size |
trager | \(-\ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}} x +4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}+3 x \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +x^{2}+3 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-1}{1+x}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-3 \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}+3 x \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}-6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +2 x^{2}+3 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+4 x +2}{1+x}\right )\) | \(352\) |
-ln(-(4*RootOf(_Z^2-_Z+1)^2*x^2+3*RootOf(_Z^2-_Z+1)*(x^3+x^2-x-1)^(2/3)-3* RootOf(_Z^2-_Z+1)*(x^3+x^2-x-1)^(1/3)*x+4*RootOf(_Z^2-_Z+1)^2*x-4*RootOf(_ Z^2-_Z+1)*x^2-3*RootOf(_Z^2-_Z+1)*(x^3+x^2-x-1)^(1/3)+3*x*(x^3+x^2-x-1)^(1 /3)-2*RootOf(_Z^2-_Z+1)*x+x^2+3*(x^3+x^2-x-1)^(1/3)+2*RootOf(_Z^2-_Z+1)-1) /(1+x))+RootOf(_Z^2-_Z+1)*ln((2*RootOf(_Z^2-_Z+1)^2*x^2+3*RootOf(_Z^2-_Z+1 )*(x^3+x^2-x-1)^(2/3)+2*RootOf(_Z^2-_Z+1)^2*x-5*RootOf(_Z^2-_Z+1)*x^2-3*(x ^3+x^2-x-1)^(2/3)+3*x*(x^3+x^2-x-1)^(1/3)-6*RootOf(_Z^2-_Z+1)*x+2*x^2+3*(x ^3+x^2-x-1)^(1/3)-RootOf(_Z^2-_Z+1)+4*x+2)/(1+x))
Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\sqrt [3]{-1-x+x^2+x^3}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x + 1\right )} + 2 \, \sqrt {3} {\left (x^{3} + x^{2} - x - 1\right )}^{\frac {1}{3}}}{3 \, {\left (x + 1\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} + x^{2} - x - 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + 2 \, x + {\left (x^{3} + x^{2} - x - 1\right )}^{\frac {2}{3}} + 1}{x^{2} + 2 \, x + 1}\right ) - \log \left (-\frac {x - {\left (x^{3} + x^{2} - x - 1\right )}^{\frac {1}{3}} + 1}{x + 1}\right ) \]
-sqrt(3)*arctan(1/3*(sqrt(3)*(x + 1) + 2*sqrt(3)*(x^3 + x^2 - x - 1)^(1/3) )/(x + 1)) + 1/2*log((x^2 + (x^3 + x^2 - x - 1)^(1/3)*(x + 1) + 2*x + (x^3 + x^2 - x - 1)^(2/3) + 1)/(x^2 + 2*x + 1)) - log(-(x - (x^3 + x^2 - x - 1 )^(1/3) + 1)/(x + 1))
\[ \int \frac {1}{\sqrt [3]{-1-x+x^2+x^3}} \, dx=\int \frac {1}{\sqrt [3]{x^{3} + x^{2} - x - 1}}\, dx \]
\[ \int \frac {1}{\sqrt [3]{-1-x+x^2+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} + x^{2} - x - 1\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {1}{\sqrt [3]{-1-x+x^2+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} + x^{2} - x - 1\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt [3]{-1-x+x^2+x^3}} \, dx=\int \frac {1}{{\left (x^3+x^2-x-1\right )}^{1/3}} \,d x \]