Integrand size = 31, antiderivative size = 225 \[ \int \frac {(-2+x) \sqrt [3]{x-x^2+x^3}}{(-1+x) \left (-1+x+x^2\right )} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x-x^2+x^3}}\right )+\sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{x-x^2+x^3}}\right )-\log \left (-x+\sqrt [3]{x-x^2+x^3}\right )+\sqrt [3]{2} \log \left (-2 x+2^{2/3} \sqrt [3]{x-x^2+x^3}\right )+\frac {1}{2} \log \left (x^2+x \sqrt [3]{x-x^2+x^3}+\left (x-x^2+x^3\right )^{2/3}\right )-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{x-x^2+x^3}+\sqrt [3]{2} \left (x-x^2+x^3\right )^{2/3}\right )}{2^{2/3}} \]
-3^(1/2)*arctan(3^(1/2)*x/(x+2*(x^3-x^2+x)^(1/3)))+2^(1/3)*3^(1/2)*arctan( 3^(1/2)*x/(x+2^(2/3)*(x^3-x^2+x)^(1/3)))-ln(-x+(x^3-x^2+x)^(1/3))+2^(1/3)* ln(-2*x+2^(2/3)*(x^3-x^2+x)^(1/3))+1/2*ln(x^2+x*(x^3-x^2+x)^(1/3)+(x^3-x^2 +x)^(2/3))-1/2*ln(2*x^2+2^(2/3)*x*(x^3-x^2+x)^(1/3)+2^(1/3)*(x^3-x^2+x)^(2 /3))*2^(1/3)
Time = 1.78 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.24 \[ \int \frac {(-2+x) \sqrt [3]{x-x^2+x^3}}{(-1+x) \left (-1+x+x^2\right )} \, dx=\frac {x^{2/3} \left (1-x+x^2\right )^{2/3} \left (-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1-x+x^2}}\right )+2 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2^{2/3} \sqrt [3]{1-x+x^2}}\right )-2 \log \left (-x^{2/3}+\sqrt [3]{1-x+x^2}\right )+2 \sqrt [3]{2} \log \left (-2 x^{2/3}+2^{2/3} \sqrt [3]{1-x+x^2}\right )+\log \left (x^{4/3}+x^{2/3} \sqrt [3]{1-x+x^2}+\left (1-x+x^2\right )^{2/3}\right )-\sqrt [3]{2} \log \left (2 x^{4/3}+2^{2/3} x^{2/3} \sqrt [3]{1-x+x^2}+\sqrt [3]{2} \left (1-x+x^2\right )^{2/3}\right )\right )}{2 \left (x \left (1-x+x^2\right )\right )^{2/3}} \]
(x^(2/3)*(1 - x + x^2)^(2/3)*(-2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(2/3))/(x^(2/3) + 2*(1 - x + x^2)^(1/3))] + 2*2^(1/3)*Sqrt[3]*ArcTan[(Sqrt[3]*x^(2/3))/(x ^(2/3) + 2^(2/3)*(1 - x + x^2)^(1/3))] - 2*Log[-x^(2/3) + (1 - x + x^2)^(1 /3)] + 2*2^(1/3)*Log[-2*x^(2/3) + 2^(2/3)*(1 - x + x^2)^(1/3)] + Log[x^(4/ 3) + x^(2/3)*(1 - x + x^2)^(1/3) + (1 - x + x^2)^(2/3)] - 2^(1/3)*Log[2*x^ (4/3) + 2^(2/3)*x^(2/3)*(1 - x + x^2)^(1/3) + 2^(1/3)*(1 - x + x^2)^(2/3)] ))/(2*(x*(1 - x + x^2))^(2/3))
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.
| \(\displaystyle \int \frac {(x-2) \sqrt [3]{x^3-x^2+x}}{(x-1) \left (x^2+x-1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x^3-x^2+x} \int -\frac {(2-x) \sqrt [3]{x} \sqrt [3]{x^2-x+1}}{(1-x) \left (-x^2-x+1\right )}dx}{\sqrt [3]{x} \sqrt [3]{x^2-x+1}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x^3-x^2+x} \int \frac {(2-x) \sqrt [3]{x} \sqrt [3]{x^2-x+1}}{(1-x) \left (-x^2-x+1\right )}dx}{\sqrt [3]{x} \sqrt [3]{x^2-x+1}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 \sqrt [3]{x^3-x^2+x} \int \frac {(2-x) x \sqrt [3]{x^2-x+1}}{(1-x) \left (-x^2-x+1\right )}d\sqrt [3]{x}}{\sqrt [3]{x} \sqrt [3]{x^2-x+1}}\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle -\frac {3 \sqrt [3]{x^3-x^2+x} \int \frac {(2-x) x \sqrt [3]{x^2-x+1}}{x^3-2 x+1}d\sqrt [3]{x}}{\sqrt [3]{x} \sqrt [3]{x^2-x+1}}\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle -\frac {3 \sqrt [3]{x^3-x^2+x} \int \left (\frac {(2-x) \sqrt [3]{x^2-x+1} x}{3 \left (\sqrt [3]{x}-1\right )}+\frac {\left (-\sqrt [3]{x}-2\right ) (2-x) \sqrt [3]{x^2-x+1} x}{3 \left (x^{2/3}+\sqrt [3]{x}+1\right )}+\frac {(-x-2) (2-x) \sqrt [3]{x^2-x+1} x}{x^2+x-1}\right )d\sqrt [3]{x}}{\sqrt [3]{x} \sqrt [3]{x^2-x+1}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt [3]{x^3-x^2+x} \left (\frac {1}{3} \int \frac {\sqrt [3]{x^2-x+1}}{\sqrt [3]{x}-1}d\sqrt [3]{x}-\frac {1}{3} \left (1-i \sqrt {3}\right ) \int \frac {\sqrt [3]{x^2-x+1}}{2 \sqrt [3]{x}-i \sqrt {3}+1}d\sqrt [3]{x}-\frac {1}{3} \left (1+i \sqrt {3}\right ) \int \frac {\sqrt [3]{x^2-x+1}}{2 \sqrt [3]{x}+i \sqrt {3}+1}d\sqrt [3]{x}-2 \int \frac {\sqrt [3]{x^2-x+1}}{2 x-\sqrt {5}+1}d\sqrt [3]{x}-2 \int \frac {\sqrt [3]{x^2-x+1}}{2 x+\sqrt {5}+1}d\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{x^2-x+1}}\) |
3.26.93.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 6.45 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.89
| method | result | size |
| pseudoelliptic | \(\frac {\ln \left (\frac {{\left (x \left (x^{2}-x +1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}-x +1\right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\left (2 {\left (x \left (x^{2}-x +1\right )\right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )+2^{\frac {1}{3}} \ln \left (\frac {-2^{\frac {1}{3}} x +{\left (x \left (x^{2}-x +1\right )\right )}^{\frac {1}{3}}}{x}\right )-\frac {2^{\frac {1}{3}} \ln \left (\frac {2^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} {\left (x \left (x^{2}-x +1\right )\right )}^{\frac {1}{3}} x +{\left (x \left (x^{2}-x +1\right )\right )}^{\frac {2}{3}}}{x^{2}}\right )}{2}-2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} {\left (x \left (x^{2}-x +1\right )\right )}^{\frac {1}{3}}+x \right )}{3 x}\right )-\ln \left (\frac {{\left (x \left (x^{2}-x +1\right )\right )}^{\frac {1}{3}}-x}{x}\right )\) | \(201\) |
| trager | \(\text {Expression too large to display}\) | \(2000\) |
1/2*ln(((x*(x^2-x+1))^(2/3)+(x*(x^2-x+1))^(1/3)*x+x^2)/x^2)+3^(1/2)*arctan (1/3*(2*(x*(x^2-x+1))^(1/3)+x)*3^(1/2)/x)+2^(1/3)*ln((-2^(1/3)*x+(x*(x^2-x +1))^(1/3))/x)-1/2*2^(1/3)*ln((2^(2/3)*x^2+2^(1/3)*(x*(x^2-x+1))^(1/3)*x+( x*(x^2-x+1))^(2/3))/x^2)-2^(1/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(x*(x ^2-x+1))^(1/3)+x)/x)-ln(((x*(x^2-x+1))^(1/3)-x)/x)
Exception generated. \[ \int \frac {(-2+x) \sqrt [3]{x-x^2+x^3}}{(-1+x) \left (-1+x+x^2\right )} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (residue poly has multiple non-linear fac tors)
\[ \int \frac {(-2+x) \sqrt [3]{x-x^2+x^3}}{(-1+x) \left (-1+x+x^2\right )} \, dx=\int \frac {\sqrt [3]{x \left (x^{2} - x + 1\right )} \left (x - 2\right )}{\left (x - 1\right ) \left (x^{2} + x - 1\right )}\, dx \]
\[ \int \frac {(-2+x) \sqrt [3]{x-x^2+x^3}}{(-1+x) \left (-1+x+x^2\right )} \, dx=\int { \frac {{\left (x^{3} - x^{2} + x\right )}^{\frac {1}{3}} {\left (x - 2\right )}}{{\left (x^{2} + x - 1\right )} {\left (x - 1\right )}} \,d x } \]
\[ \int \frac {(-2+x) \sqrt [3]{x-x^2+x^3}}{(-1+x) \left (-1+x+x^2\right )} \, dx=\int { \frac {{\left (x^{3} - x^{2} + x\right )}^{\frac {1}{3}} {\left (x - 2\right )}}{{\left (x^{2} + x - 1\right )} {\left (x - 1\right )}} \,d x } \]
Timed out. \[ \int \frac {(-2+x) \sqrt [3]{x-x^2+x^3}}{(-1+x) \left (-1+x+x^2\right )} \, dx=\int \frac {\left (x-2\right )\,{\left (x^3-x^2+x\right )}^{1/3}}{\left (x-1\right )\,\left (x^2+x-1\right )} \,d x \]