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

Optimal. Leaf size=165 \[ \frac {\log \left (3 \sqrt [3]{x^2+x+1}+\sqrt [3]{3} x-\sqrt [3]{3}\right )}{3 \sqrt [3]{3}}-\frac {\log \left (3^{2/3} x^2+9 \left (x^2+x+1\right )^{2/3}+\left (3 \sqrt [3]{3}-3 \sqrt [3]{3} x\right ) \sqrt [3]{x^2+x+1}-2\ 3^{2/3} x+3^{2/3}\right )}{6 \sqrt [3]{3}}-\frac {\tan ^{-1}\left (\frac {\frac {\sqrt [3]{x^2+x+1}}{\sqrt {3}}-\frac {2 x}{3 \sqrt [6]{3}}+\frac {2}{3 \sqrt [6]{3}}}{\sqrt [3]{x^2+x+1}}\right )}{3^{5/6}} \]

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Rubi [A]  time = 0.01, antiderivative size = 84, normalized size of antiderivative = 0.51, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {750} \begin {gather*} \frac {\log \left (-3^{2/3} \sqrt [3]{x^2+x+1}-x+1\right )}{2 \sqrt [3]{3}}-\frac {\tan ^{-1}\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 (x+2)}{2 \sqrt [3]{3}} \end {gather*}

Antiderivative was successfully verified.

[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 750

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*} \int \frac {1}{(2+x) \sqrt [3]{1+x+x^2}} \, dx &=-\frac {\tan ^{-1}\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*}

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Mathematica [C]  time = 0.06, size = 118, normalized size = 0.72 \begin {gather*} -\frac {3 \sqrt [3]{\frac {2 x-i \sqrt {3}+1}{x+2}} \sqrt [3]{\frac {2 x+i \sqrt {3}+1}{x+2}} F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};\frac {3-i \sqrt {3}}{2 x+4},\frac {3+i \sqrt {3}}{2 x+4}\right )}{2\ 2^{2/3} \sqrt [3]{x^2+x+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*((1 - I*Sqrt[3] + 2*x)/(2 + x))^(1/3)*((1 + I*Sqrt[3] + 2*x)/(2 + x))^(1/3)*AppellF1[2/3, 1/3, 1/3, 5/3, (
3 - I*Sqrt[3])/(4 + 2*x), (3 + I*Sqrt[3])/(4 + 2*x)])/(2*2^(2/3)*(1 + x + x^2)^(1/3))

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IntegrateAlgebraic [A]  time = 0.20, size = 165, normalized size = 1.00 \begin {gather*} \frac {\log \left (3 \sqrt [3]{x^2+x+1}+\sqrt [3]{3} x-\sqrt [3]{3}\right )}{3 \sqrt [3]{3}}-\frac {\log \left (3^{2/3} x^2+9 \left (x^2+x+1\right )^{2/3}+\left (3 \sqrt [3]{3}-3 \sqrt [3]{3} x\right ) \sqrt [3]{x^2+x+1}-2\ 3^{2/3} x+3^{2/3}\right )}{6 \sqrt [3]{3}}-\frac {\tan ^{-1}\left (\frac {\frac {\sqrt [3]{x^2+x+1}}{\sqrt {3}}-\frac {2 x}{3 \sqrt [6]{3}}+\frac {2}{3 \sqrt [6]{3}}}{\sqrt [3]{x^2+x+1}}\right )}{3^{5/6}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.21, size = 165, normalized size = 1.00 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x + 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 14.23, size = 1377, normalized size = 8.35 \begin {gather*} \text {Expression too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*RootOf(_Z^3-9)*ln(-(13110*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^3+43587*R
ootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^3-26220*RootOf(RootOf(_Z^3-9)^2+3*_Z*R
ootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^2-87174*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*x+52
440*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x+174348*RootOf(RootOf(_Z^3-9)^2+3*_Z
*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x+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)^2-43947*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2*x^2+132858*(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+87894*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2*x-265716*(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+43700*RootOf(_Z^3-9)*x^3+145290*R
ootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^3-43947*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2+132858*(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)-485070*RootOf(_Z^3-9)*x^2-1612719*Ro
otOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^2-794097*(x^2+x+1)^(2/3)*x-222870*RootOf(_Z^3-9)*x-740979*
RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x+794097*(x^2+x+1)^(2/3)-397670*RootOf(_Z^3-9)-1322139*Roo
tOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2))/(2+x)^3)+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+39330*RootOf(RootOf(_Z
^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^3-29058*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9
*_Z^2)*RootOf(_Z^3-9)^3*x^2-78660*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^2-1
31841*(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+58116*RootOf(Root
Of(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x+157320*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)
+9*_Z^2)^2*RootOf(_Z^3-9)^2*x+131841*(x^2+x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootO
f(_Z^3-9)^2-43947*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2*x^2-264699*(x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*Roo
tOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x^2+87894*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2*x+529398*(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-4843*RootOf(_Z^3-9)*x^3-13110*RootOf(RootOf(_Z^3
-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^3-43947*(x^2+x+1)^(1/3)*RootOf(_Z^3-9)^2-264699*(x^2+x+1)^(1/3)*RootOf(Roo
tOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)+450399*RootOf(_Z^3-9)*x^2+1219230*RootOf(RootOf(_Z^3-
9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^2+398574*(x^2+x+1)^(2/3)*x+421341*RootOf(_Z^3-9)*x+1140570*RootOf(RootOf(_Z
^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x-398574*(x^2+x+1)^(2/3)+440713*RootOf(_Z^3-9)+1193010*RootOf(RootOf(_Z^3-
9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2))/(2+x)^3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x + 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (x+2\right )\,{\left (x^2+x+1\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (x + 2\right ) \sqrt [3]{x^{2} + x + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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