\(\int \frac {1}{\sqrt [3]{1+x} \sqrt [3]{1-x+x^2}} \, dx\) [2582]

Optimal result

Integrand size = 20, antiderivative size = 102 \[ \int \frac {1}{\sqrt [3]{1+x} \sqrt [3]{1-x+x^2}} \, dx=\frac {\sqrt [3]{1+x^3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{1+x} \sqrt [3]{1-x+x^2}}-\frac {\sqrt [3]{1+x^3} \log \left (-x+\sqrt [3]{1+x^3}\right )}{2 \sqrt [3]{1+x} \sqrt [3]{1-x+x^2}} \]

[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 = 102, 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} \[ \int \frac {1}{\sqrt [3]{1+x} \sqrt [3]{1-x+x^2}} \, dx=\frac {\sqrt [3]{x^3+1} \arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{x+1} \sqrt [3]{x^2-x+1}}-\frac {\sqrt [3]{x^3+1} \log \left (\sqrt [3]{x^3+1}-x\right )}{2 \sqrt [3]{x+1} \sqrt [3]{x^2-x+1}} \]

[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 = 20.07 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.29 \[ \int \frac {1}{\sqrt [3]{1+x} \sqrt [3]{1-x+x^2}} \, dx=\frac {3 \sqrt [3]{\frac {i+\sqrt {3}-2 i x}{3 i+\sqrt {3}}} \sqrt [3]{\frac {-i+\sqrt {3}+2 i x}{-3 i+\sqrt {3}}} (1+x)^{2/3} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},\frac {1}{3},\frac {5}{3},\frac {2 i (1+x)}{3 i+\sqrt {3}},-\frac {2 i (1+x)}{-3 i+\sqrt {3}}\right )}{2 \sqrt [3]{1-x+x^2}} \]

[In]

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

[Out]

(3*((I + Sqrt[3] - (2*I)*x)/(3*I + Sqrt[3]))^(1/3)*((-I + Sqrt[3] + (2*I)*x)/(-3*I + Sqrt[3]))^(1/3)*(1 + x)^(
2/3)*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 = 0.80 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\sqrt [3]{1+x} \sqrt [3]{1-x+x^2}} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (-\frac {4 \, \sqrt {3} {\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )}^{\frac {1}{3}} x^{2} - 2 \, \sqrt {3} {\left (x^{2} - x + 1\right )}^{\frac {2}{3}} {\left (x + 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (x^{3} + 1\right )}}{9 \, x^{3} + 1}\right ) - \frac {1}{6} \, \log \left (3 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )}^{\frac {1}{3}} x^{2} - 3 \, {\left (x^{2} - x + 1\right )}^{\frac {2}{3}} {\left (x + 1\right )}^{\frac {2}{3}} x + 1\right ) \]

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

Sympy [F]

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

[In]

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

[Out]

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

[In]

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

[Out]

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