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

Optimal. Leaf size=177 \[ \frac {\log \left (3 \sqrt [3]{x^2-x+1}+\sqrt [3]{3} x-2 \sqrt [3]{3}\right )}{3 \sqrt [3]{3}}-\frac {\log \left (3^{2/3} x^2+9 \left (x^2-x+1\right )^{2/3}+\left (6 \sqrt [3]{3}-3 \sqrt [3]{3} x\right ) \sqrt [3]{x^2-x+1}-4\ 3^{2/3} x+4\ 3^{2/3}\right )}{6 \sqrt [3]{3}}-\frac {\tan ^{-1}\left (\frac {\frac {\sqrt [3]{x^2-x+1}}{\sqrt {3}}-\frac {2 x}{3 \sqrt [6]{3}}+\frac {4}{3 \sqrt [6]{3}}}{\sqrt [3]{x^2-x+1}}\right )}{3^{5/6}} \]

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Rubi [A]  time = 0.02, antiderivative size = 88, normalized size of antiderivative = 0.50, number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {750} \begin {gather*} \frac {\log \left (-3^{2/3} \sqrt [3]{x^2-x+1}-x+2\right )}{2 \sqrt [3]{3}}-\frac {\tan ^{-1}\left (\frac {2 (2-x)}{3 \sqrt [6]{3} \sqrt [3]{x^2-x+1}}+\frac {1}{\sqrt {3}}\right )}{3^{5/6}}-\frac {\log (x+1)}{2 \sqrt [3]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[1/Sqrt[3] + (2*(2 - x))/(3*3^(1/6)*(1 - x + x^2)^(1/3))]/3^(5/6)) - Log[1 + x]/(2*3^(1/3)) + Log[2 -
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}{(1+x) \sqrt [3]{1-x+x^2}} \, dx &=-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 (2-x)}{3 \sqrt [6]{3} \sqrt [3]{1-x+x^2}}\right )}{3^{5/6}}-\frac {\log (1+x)}{2 \sqrt [3]{3}}+\frac {\log \left (2-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 = 120, normalized size = 0.68 \begin {gather*} -\frac {3 \sqrt [3]{\frac {2 x-i \sqrt {3}-1}{x+1}} \sqrt [3]{\frac {2 x+i \sqrt {3}-1}{x+1}} F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};\frac {3-i \sqrt {3}}{2 x+2},\frac {3+i \sqrt {3}}{2 x+2}\right )}{2\ 2^{2/3} \sqrt [3]{x^2-x+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.21, size = 177, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\frac {4}{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 (-2 \sqrt [3]{3}+\sqrt [3]{3} x+3 \sqrt [3]{1-x+x^2}\right )}{3 \sqrt [3]{3}}-\frac {\log \left (4\ 3^{2/3}-4\ 3^{2/3} x+3^{2/3} x^2+\left (6 \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(4/(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
[-2*3^(1/3) + 3^(1/3)*x + 3*(1 - x + x^2)^(1/3)]/(3*3^(1/3)) - Log[4*3^(2/3) - 4*3^(2/3)*x + 3^(2/3)*x^2 + (6*
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 = 2.31, size = 175, normalized size = 0.99 \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} - 4 \, x + 4\right )} - 3 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x - 2\right )}}{x^{2} + 2 \, x + 1}\right ) + \frac {1}{9} \cdot 3^{\frac {2}{3}} \log \left (\frac {3^{\frac {1}{3}} {\left (x - 2\right )} + 3 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{3}}}{x + 1}\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 - 2\right )} + 3^{\frac {1}{3}} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} + 6 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 4 \, x + 4\right )}\right )}}{3 \, {\left (x^{3} - 15 \, x^{2} + 21 \, x - 17\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+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 - 4*x + 4) - 3*(x^2 - x + 1)^(1/3)*(x - 2))/(x
^2 + 2*x + 1)) + 1/9*3^(2/3)*log((3^(1/3)*(x - 2) + 3*(x^2 - x + 1)^(1/3))/(x + 1)) - 1/3*3^(1/6)*arctan(1/3*3
^(1/6)*(6*3^(2/3)*(x^2 - x + 1)^(2/3)*(x - 2) + 3^(1/3)*(x^3 + 3*x^2 + 3*x + 1) + 6*(x^2 - x + 1)^(1/3)*(x^2 -
4*x + 4))/(x^3 - 15*x^2 + 21*x - 17))

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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 + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 14.08, size = 1416, normalized size = 8.00

method	result	size
trager	\(\text {Expression too large to display}\)	\(1416\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*RootOf(_Z^3-9)*ln((61590*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^3+65997*Ro
otOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^3+123180*RootOf(RootOf(_Z^3-9)^2+3*_Z*R
ootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^2+131994*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf
(_Z^3-9)^2*x^2+967776*(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+2
46360*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x+263988*RootOf(RootOf(_Z^3-9)^2+3*
_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x-1935552*(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+322592*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2*x^2+1579341*(x^2-x+1)^(1/3)*RootOf(RootO
f(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x^2-1290368*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2*x-6317364*
(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+349010*RootOf(_Z^3-9)*x^3
+373983*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^3+1290368*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2+63173
64*(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)-3756990*RootOf(_Z^3-9)*x
^2-4025817*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^2-1834695*(x^2-x+1)^(2/3)*x+5851050*RootOf(_Z
^3-9)*x+6269715*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x+3669390*(x^2-x+1)^(2/3)-4455010*RootOf(_
Z^3-9)-4773783*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2))/(1+x)^3)+1/3*RootOf(RootOf(_Z^3-9)^2+3*_Z*
RootOf(_Z^3-9)+9*_Z^2)*ln((3*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^3+234*Root
Of(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^3+6*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_
Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^2+468*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2
*x^2+1872*(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+12*RootOf(Roo
tOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x+936*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9
*_Z^2)^2*RootOf(_Z^3-9)^2*x-3744*(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+624*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2*x^2-153*(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^2-2496*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2*x+612*(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-8*RootOf(_Z^3-9)*x^3-624*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(
_Z^3-9)+9*_Z^2)*x^3+2496*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2-612*(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)+201*RootOf(_Z^3-9)*x^2+15678*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*
_Z^2)*x^2+6075*(x^2-x+1)^(2/3)*x-249*RootOf(_Z^3-9)*x-19422*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2
)*x-12150*(x^2-x+1)^(2/3)+217*RootOf(_Z^3-9)+16926*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2))/(1+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 + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

int(1/((x + 1)*(x^2 - x + 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 + 1\right ) \sqrt [3]{x^{2} - x + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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