\(\int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx\) [2585]

Optimal result

Integrand size = 20, antiderivative size = 109 \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=-\frac {\sqrt [3]{1-x^3} \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{1-x} \sqrt [3]{1+x+x^2}}+\frac {\sqrt [3]{1-x^3} \log \left (x+\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \]

[Out]

1/2*(-x^3+1)^(1/3)*ln(x+(-x^3+1)^(1/3))/(1-x)^(1/3)/(x^2+x+1)^(1/3)-1/3*(-x^3+1)^(1/3)*arctan(1/3*(1-2*x/(-x^3
+1)^(1/3))*3^(1/2))/(1-x)^(1/3)/(x^2+x+1)^(1/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {727, 245} \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\frac {\sqrt [3]{1-x^3} \log \left (\sqrt [3]{1-x^3}+x\right )}{2 \sqrt [3]{1-x} \sqrt [3]{x^2+x+1}}-\frac {\sqrt [3]{1-x^3} \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{1-x} \sqrt [3]{x^2+x+1}} \]

[In]

Int[1/((1 - x)^(1/3)*(1 + x + x^2)^(1/3)),x]

[Out]

-(((1 - x^3)^(1/3)*ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*(1 - x)^(1/3)*(1 + x + x^2)^(1/3))) +
((1 - x^3)^(1/3)*Log[x + (1 - x^3)^(1/3)])/(2*(1 - x)^(1/3)*(1 + x + x^2)^(1/3))

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 727

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d + e*x)^FracPart[p]
*((a + b*x + c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p]), Int[(d + e*x)^(m - p)*(a*d + c*e*x^3)^p, x], x]
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && IGtQ[m - p + 1, 0] &&  !Intege
rQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{1-x^3} \int \frac {1}{\sqrt [3]{1-x^3}} \, dx}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \\ & = -\frac {\sqrt [3]{1-x^3} \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{1-x} \sqrt [3]{1+x+x^2}}+\frac {\sqrt [3]{1-x^3} \log \left (x+\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 10.05 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=-\frac {3 (1-x)^{2/3} \sqrt [3]{\frac {-i+\sqrt {3}-2 i x}{-3 i+\sqrt {3}}} \sqrt [3]{\frac {i+\sqrt {3}+2 i x}{3 i+\sqrt {3}}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},\frac {1}{3},\frac {5}{3},-\frac {2 i (-1+x)}{3 i+\sqrt {3}},\frac {2 i (-1+x)}{-3 i+\sqrt {3}}\right )}{2 \sqrt [3]{1+x+x^2}} \]

[In]

Integrate[1/((1 - x)^(1/3)*(1 + x + x^2)^(1/3)),x]

[Out]

(-3*(1 - x)^(2/3)*((-I + Sqrt[3] - (2*I)*x)/(-3*I + Sqrt[3]))^(1/3)*((I + Sqrt[3] + (2*I)*x)/(3*I + Sqrt[3]))^
(1/3)*AppellF1[2/3, 1/3, 1/3, 5/3, ((-2*I)*(-1 + x))/(3*I + Sqrt[3]), ((2*I)*(-1 + x))/(-3*I + Sqrt[3])])/(2*(
1 + x + x^2)^(1/3))

Maple [F]

\[\int \frac {1}{\left (1-x \right )^{\frac {1}{3}} \left (x^{2}+x +1\right )^{\frac {1}{3}}}d x\]

[In]

int(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x)

[Out]

int(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x)

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} {\left (x^{2} + x + 1\right )}^{\frac {1}{3}} x^{2} {\left (-x + 1\right )}^{\frac {1}{3}} + 2 \, \sqrt {3} {\left (x^{2} + x + 1\right )}^{\frac {2}{3}} x {\left (-x + 1\right )}^{\frac {2}{3}} - \sqrt {3} {\left (x^{3} - 1\right )}}{9 \, x^{3} - 1}\right ) + \frac {1}{6} \, \log \left (3 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{3}} x^{2} {\left (-x + 1\right )}^{\frac {1}{3}} + 3 \, {\left (x^{2} + x + 1\right )}^{\frac {2}{3}} x {\left (-x + 1\right )}^{\frac {2}{3}} + 1\right ) \]

[In]

integrate(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x, algorithm="fricas")

[Out]

-1/3*sqrt(3)*arctan((4*sqrt(3)*(x^2 + x + 1)^(1/3)*x^2*(-x + 1)^(1/3) + 2*sqrt(3)*(x^2 + x + 1)^(2/3)*x*(-x +
1)^(2/3) - sqrt(3)*(x^3 - 1))/(9*x^3 - 1)) + 1/6*log(3*(x^2 + x + 1)^(1/3)*x^2*(-x + 1)^(1/3) + 3*(x^2 + x + 1
)^(2/3)*x*(-x + 1)^(2/3) + 1)

Sympy [F]

\[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\int \frac {1}{\sqrt [3]{1 - x} \sqrt [3]{x^{2} + x + 1}}\, dx \]

[In]

integrate(1/(1-x)**(1/3)/(x**2+x+1)**(1/3),x)

[Out]

Integral(1/((1 - x)**(1/3)*(x**2 + x + 1)**(1/3)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x, algorithm="maxima")

[Out]

integrate(1/((x^2 + x + 1)^(1/3)*(-x + 1)^(1/3)), x)

Giac [F]

\[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate(1/(1-x)^(1/3)/(x^2+x+1)^(1/3),x, algorithm="giac")

[Out]

integrate(1/((x^2 + x + 1)^(1/3)*(-x + 1)^(1/3)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{1+x+x^2}} \, dx=\int \frac {1}{{\left (1-x\right )}^{1/3}\,{\left (x^2+x+1\right )}^{1/3}} \,d x \]

[In]

int(1/((1 - x)^(1/3)*(x + x^2 + 1)^(1/3)),x)

[Out]

int(1/((1 - x)^(1/3)*(x + x^2 + 1)^(1/3)), x)