\(\int \frac {1}{(2+x) \sqrt [3]{1+x+x^2}} \, dx\) [2215]

Optimal result

Integrand size = 16, antiderivative size = 165 \[ \int \frac {1}{(2+x) \sqrt [3]{1+x+x^2}} \, dx=-\frac {\arctan \left (\frac {\frac {2}{3 \sqrt [6]{3}}-\frac {2 x}{3 \sqrt [6]{3}}+\frac {\sqrt [3]{1+x+x^2}}{\sqrt {3}}}{\sqrt [3]{1+x+x^2}}\right )}{3^{5/6}}+\frac {\log \left (-\sqrt [3]{3}+\sqrt [3]{3} x+3 \sqrt [3]{1+x+x^2}\right )}{3 \sqrt [3]{3}}-\frac {\log \left (3^{2/3}-2\ 3^{2/3} x+3^{2/3} x^2+\left (3 \sqrt [3]{3}-3 \sqrt [3]{3} x\right ) \sqrt [3]{1+x+x^2}+9 \left (1+x+x^2\right )^{2/3}\right )}{6 \sqrt [3]{3}} \]

[Out]

-1/3*arctan((2/9*3^(5/6)-2/9*x*3^(5/6)+1/3*(x^2+x+1)^(1/3)*3^(1/2))/(x^2+x+1)^(1/3))*3^(1/6)+1/9*ln(-3^(1/3)+3
^(1/3)*x+3*(x^2+x+1)^(1/3))*3^(2/3)-1/18*ln(3^(2/3)-2*3^(2/3)*x+3^(2/3)*x^2+(3*3^(1/3)-3*3^(1/3)*x)*(x^2+x+1)^
(1/3)+9*(x^2+x+1)^(2/3))*3^(2/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.51, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {764} \[ \int \frac {1}{(2+x) \sqrt [3]{1+x+x^2}} \, dx=-\frac {\arctan \left (\frac {2 (1-x)}{3 \sqrt [6]{3} \sqrt [3]{x^2+x+1}}+\frac {1}{\sqrt {3}}\right )}{3^{5/6}}+\frac {\log \left (-3^{2/3} \sqrt [3]{x^2+x+1}-x+1\right )}{2 \sqrt [3]{3}}-\frac {\log (x+2)}{2 \sqrt [3]{3}} \]

[In]

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

[Out]

-(ArcTan[1/Sqrt[3] + (2*(1 - x))/(3*3^(1/6)*(1 + x + x^2)^(1/3))]/3^(5/6)) - Log[2 + x]/(2*3^(1/3)) + Log[1 -
x - 3^(2/3)*(1 + x + x^2)^(1/3)]/(2*3^(1/3))

Rule 764

Int[1/(((d_.) + (e_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(1/3)), x_Symbol] :> With[{q = Rt[3*c*e^2*(2*c*
d - b*e), 3]}, Simp[(-Sqrt[3])*c*e*(ArcTan[1/Sqrt[3] + 2*((c*d - b*e - c*e*x)/(Sqrt[3]*q*(a + b*x + c*x^2)^(1/
3)))]/q^2), x] + (-Simp[3*c*e*(Log[d + e*x]/(2*q^2)), x] + Simp[3*c*e*(Log[c*d - b*e - c*e*x - q*(a + b*x + c*
x^2)^(1/3)]/(2*q^2)), x])] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && EqQ[c^2*d^2 - b*c*d*e + b^2*
e^2 - 3*a*c*e^2, 0] && PosQ[c*e^2*(2*c*d - b*e)]

Rubi steps \begin{align*} \text {integral}& = -\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 (1-x)}{3 \sqrt [6]{3} \sqrt [3]{1+x+x^2}}\right )}{3^{5/6}}-\frac {\log (2+x)}{2 \sqrt [3]{3}}+\frac {\log \left (1-x-3^{2/3} \sqrt [3]{1+x+x^2}\right )}{2 \sqrt [3]{3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(2+x) \sqrt [3]{1+x+x^2}} \, dx=\frac {-6 \arctan \left (\frac {2\ 3^{5/6}-2\ 3^{5/6} x+3 \sqrt {3} \sqrt [3]{1+x+x^2}}{9 \sqrt [3]{1+x+x^2}}\right )+\sqrt {3} \left (2 \log \left (-\sqrt [3]{3}+\sqrt [3]{3} x+3 \sqrt [3]{1+x+x^2}\right )-\log \left (3^{2/3}-2\ 3^{2/3} x+3^{2/3} x^2-3 \sqrt [3]{3} (-1+x) \sqrt [3]{1+x+x^2}+9 \left (1+x+x^2\right )^{2/3}\right )\right )}{6\ 3^{5/6}} \]

[In]

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

[Out]

(-6*ArcTan[(2*3^(5/6) - 2*3^(5/6)*x + 3*Sqrt[3]*(1 + x + x^2)^(1/3))/(9*(1 + x + x^2)^(1/3))] + Sqrt[3]*(2*Log
[-3^(1/3) + 3^(1/3)*x + 3*(1 + x + x^2)^(1/3)] - Log[3^(2/3) - 2*3^(2/3)*x + 3^(2/3)*x^2 - 3*3^(1/3)*(-1 + x)*
(1 + x + x^2)^(1/3) + 9*(1 + x + x^2)^(2/3)]))/(6*3^(5/6))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 10.66 (sec) , antiderivative size = 2074, normalized size of antiderivative = 12.57

method	result	size
trager	\(\text {Expression too large to display}\)	\(2074\)

[In]

int(1/(2+x)/(x^2+x+1)^(1/3),x,method=_RETURNVERBOSE)

[Out]

-1/9*ln(-(13110*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^3-4257*RootOf(RootOf(_Z
^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^3+131841*(x^2+x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z
*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^2*x-26220*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z
^3-9)^3*x^2+8514*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^2-131841*(x^2+x+1)^(
2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^2+44286*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)
^2*x^2-131841*(x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x^2+52440*Roo
tOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x-17028*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(
_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x-88572*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2*x+263682*(x^2+x+1)^(1/3)*RootOf(Ro
otOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x-4370*RootOf(_Z^3-9)*x^3+1419*RootOf(RootOf(_Z^3-9)
^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^3-398574*x*(x^2+x+1)^(2/3)+44286*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2-131841*(x^2
+x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)+406410*RootOf(_Z^3-9)*x^2-13196
7*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^2+398574*(x^2+x+1)^(2/3)+380190*RootOf(_Z^3-9)*x-12345
3*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x+397670*RootOf(_Z^3-9)-129129*RootOf(RootOf(_Z^3-9)^2+3
*_Z*RootOf(_Z^3-9)+9*_Z^2))/(2+x)^3)*RootOf(_Z^3-9)-1/3*ln(-(13110*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)
+9*_Z^2)*RootOf(_Z^3-9)^3*x^3-4257*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^3+
131841*(x^2+x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^2*x-26220*RootOf(Roo
tOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^2+8514*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9
)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^2-131841*(x^2+x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*Ro
otOf(_Z^3-9)^2+44286*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2*x^2-131841*(x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*
RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x^2+52440*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^
3-9)^3*x-17028*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x-88572*(x^2+x+1)^(1/3)*
RootOf(_Z^3-9)^2*x+263682*(x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x
-4370*RootOf(_Z^3-9)*x^3+1419*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^3-398574*x*(x^2+x+1)^(2/3)
+44286*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2-131841*(x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_
Z^2)*RootOf(_Z^3-9)+406410*RootOf(_Z^3-9)*x^2-131967*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^2+3
98574*(x^2+x+1)^(2/3)+380190*RootOf(_Z^3-9)*x-123453*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x+397
670*RootOf(_Z^3-9)-129129*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2))/(2+x)^3)*RootOf(RootOf(_Z^3-9)^
2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)+1/3*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*ln((14529*RootOf(RootOf(
_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^3+4257*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*
_Z^2)^2*RootOf(_Z^3-9)^2*x^3+131841*(x^2+x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf
(_Z^3-9)^2*x-29058*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^2-8514*RootOf(RootOf
(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^2-131841*(x^2+x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3
*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^2-88233*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2*x^2-131841*(x^2+x+1)^(1/3)*
RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x^2+58116*RootOf(RootOf(_Z^3-9)^2+3*_Z*Root
Of(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x+17028*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-
9)^2*x+176466*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2*x+263682*(x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z
^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x+48430*RootOf(_Z^3-9)*x^3+14190*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z
^2)*x^3+794097*x*(x^2+x+1)^(2/3)-88233*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2-131841*(x^2+x+1)^(1/3)*RootOf(RootOf(_
Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)-537573*RootOf(_Z^3-9)*x^2-157509*RootOf(RootOf(_Z^3-9)^2+3
*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^2-794097*(x^2+x+1)^(2/3)-246993*RootOf(_Z^3-9)*x-72369*RootOf(RootOf(_Z^3-9)^2+3*
_Z*RootOf(_Z^3-9)+9*_Z^2)*x-440713*RootOf(_Z^3-9)-129129*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2))/
(2+x)^3)

Fricas [A] (verification not implemented)

none

Time = 1.41 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(2+x) \sqrt [3]{1+x+x^2}} \, dx=-\frac {1}{18} \cdot 3^{\frac {2}{3}} \log \left (\frac {3 \cdot 3^{\frac {2}{3}} {\left (x^{2} + x + 1\right )}^{\frac {2}{3}} + 3^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} - 3 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )}}{x^{2} + 4 \, x + 4}\right ) + \frac {1}{9} \cdot 3^{\frac {2}{3}} \log \left (\frac {3^{\frac {1}{3}} {\left (x - 1\right )} + 3 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{3}}}{x + 2}\right ) - \frac {1}{3} \cdot 3^{\frac {1}{6}} \arctan \left (\frac {3^{\frac {1}{6}} {\left (6 \cdot 3^{\frac {2}{3}} {\left (x^{2} + x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )} + 3^{\frac {1}{3}} {\left (x^{3} + 6 \, x^{2} + 12 \, x + 8\right )} + 6 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )}\right )}}{3 \, {\left (x^{3} - 12 \, x^{2} - 6 \, x - 10\right )}}\right ) \]

[In]

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

[Out]

-1/18*3^(2/3)*log((3*3^(2/3)*(x^2 + x + 1)^(2/3) + 3^(1/3)*(x^2 - 2*x + 1) - 3*(x^2 + x + 1)^(1/3)*(x - 1))/(x
^2 + 4*x + 4)) + 1/9*3^(2/3)*log((3^(1/3)*(x - 1) + 3*(x^2 + x + 1)^(1/3))/(x + 2)) - 1/3*3^(1/6)*arctan(1/3*3
^(1/6)*(6*3^(2/3)*(x^2 + x + 1)^(2/3)*(x - 1) + 3^(1/3)*(x^3 + 6*x^2 + 12*x + 8) + 6*(x^2 + x + 1)^(1/3)*(x^2
- 2*x + 1))/(x^3 - 12*x^2 - 6*x - 10))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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