3.18.99 \(\int \frac {1}{\sqrt [3]{-1-x+x^2+x^3}} \, dx\) [1799]

Optimal. Leaf size=122 \[ -\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{-1-x+x^2+x^3}}{2+2 x+\sqrt [3]{-1-x+x^2+x^3}}\right )-\log \left (-1-x+\sqrt [3]{-1-x+x^2+x^3}\right )+\frac {1}{2} \log \left (1+2 x+x^2+(1+x) \sqrt [3]{-1-x+x^2+x^3}+\left (-1-x+x^2+x^3\right )^{2/3}\right ) \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 167, normalized size of antiderivative = 1.37, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2092, 2089, 62} \begin {gather*} -\frac {\sqrt {3} (-x-1)^{2/3} \sqrt [3]{x-1} \text {ArcTan}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x-1}}{\sqrt {3} \sqrt [3]{-x-1}}\right )}{\sqrt [3]{x^3+x^2-x-1}}-\frac {3 (-x-1)^{2/3} \sqrt [3]{x-1} \log \left (\frac {\sqrt [3]{x-1}}{\sqrt [3]{-x-1}}+1\right )}{2 \sqrt [3]{x^3+x^2-x-1}}-\frac {(-x-1)^{2/3} \sqrt [3]{x-1} \log \left (-\frac {8}{3} (x+1)\right )}{2 \sqrt [3]{x^3+x^2-x-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - x + x^2 + x^3)^(-1/3),x]

[Out]

-((Sqrt[3]*(-1 - x)^(2/3)*(-1 + x)^(1/3)*ArcTan[1/Sqrt[3] - (2*(-1 + x)^(1/3))/(Sqrt[3]*(-1 - x)^(1/3))])/(-1
- x + x^2 + x^3)^(1/3)) - (3*(-1 - x)^(2/3)*(-1 + x)^(1/3)*Log[1 + (-1 + x)^(1/3)/(-1 - x)^(1/3)])/(2*(-1 - x
+ x^2 + x^3)^(1/3)) - ((-1 - x)^(2/3)*(-1 + x)^(1/3)*Log[(-8*(1 + x))/3])/(2*(-1 - x + x^2 + x^3)^(1/3))

Rule 62

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

Rule 2089

Int[((a_.) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Dist[(a + b*x + d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*
x)^(2*p)), Int[(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b, d, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0]
&&  !IntegerQ[p]

Rule 2092

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - (c^2 - 3*b*d)*(x/(3*d)) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [3]{-1-x+x^2+x^3}} \, dx &=\text {Subst}\left (\int \frac {1}{\sqrt [3]{-\frac {16}{27}-\frac {4 x}{3}+x^3}} \, dx,x,\frac {1}{3}+x\right )\\ &=\frac {\left (4\ 2^{2/3} (-1-x)^{2/3} \sqrt [3]{-1+x}\right ) \text {Subst}\left (\int \frac {1}{\left (-\frac {16}{9}-\frac {8 x}{3}\right )^{2/3} \sqrt [3]{-\frac {16}{9}+\frac {4 x}{3}}} \, dx,x,\frac {1}{3}+x\right )}{3 \sqrt [3]{-1-x+x^2+x^3}}\\ &=-\frac {\sqrt {3} (-1-x)^{2/3} \sqrt [3]{-1+x} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-1+x}}{\sqrt {3} \sqrt [3]{-1-x}}\right )}{\sqrt [3]{-1-x+x^2+x^3}}-\frac {3 (-1-x)^{2/3} \sqrt [3]{-1+x} \log \left (1+\frac {\sqrt [3]{-1+x}}{\sqrt [3]{-1-x}}\right )}{2 \sqrt [3]{-1-x+x^2+x^3}}-\frac {(-1-x)^{2/3} \sqrt [3]{-1+x} \log (1+x)}{2 \sqrt [3]{-1-x+x^2+x^3}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 110, normalized size = 0.90 \begin {gather*} \frac {\sqrt [3]{-1+x} (1+x)^{2/3} \left (2 \sqrt {3} \text {ArcTan}\left (\frac {1+\frac {2}{\sqrt [3]{\frac {-1+x}{1+x}}}}{\sqrt {3}}\right )-2 \log \left (-1+\frac {1}{\sqrt [3]{\frac {-1+x}{1+x}}}\right )+\log \left (1+\frac {1}{\left (\frac {-1+x}{1+x}\right )^{2/3}}+\frac {1}{\sqrt [3]{\frac {-1+x}{1+x}}}\right )\right )}{2 \sqrt [3]{(-1+x) (1+x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.26, size = 352, normalized size = 2.89

method	result	size
trager	\(-\ln \left (-\frac {4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}} x -4 x^{2} \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}-2 x \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+3 x \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}+x^{2}+2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+3 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}-1}{1+x}\right )+\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}-5 x^{2} \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-6 x \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-3 \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}+3 x \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}+2 x^{2}-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+3 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}+4 x +2}{1+x}\right )\)	\(352\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(-(4*RootOf(_Z^2-_Z+1)^2*x^2+4*RootOf(_Z^2-_Z+1)^2*x+3*RootOf(_Z^2-_Z+1)*(x^3+x^2-x-1)^(2/3)-3*RootOf(_Z^2-
_Z+1)*(x^3+x^2-x-1)^(1/3)*x-4*x^2*RootOf(_Z^2-_Z+1)-3*RootOf(_Z^2-_Z+1)*(x^3+x^2-x-1)^(1/3)-2*x*RootOf(_Z^2-_Z
+1)+3*x*(x^3+x^2-x-1)^(1/3)+x^2+2*RootOf(_Z^2-_Z+1)+3*(x^3+x^2-x-1)^(1/3)-1)/(1+x))+RootOf(_Z^2-_Z+1)*ln((2*Ro
otOf(_Z^2-_Z+1)^2*x^2+2*RootOf(_Z^2-_Z+1)^2*x+3*RootOf(_Z^2-_Z+1)*(x^3+x^2-x-1)^(2/3)-5*x^2*RootOf(_Z^2-_Z+1)-
6*x*RootOf(_Z^2-_Z+1)-3*(x^3+x^2-x-1)^(2/3)+3*x*(x^3+x^2-x-1)^(1/3)+2*x^2-RootOf(_Z^2-_Z+1)+3*(x^3+x^2-x-1)^(1
/3)+4*x+2)/(1+x))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.41, size = 120, normalized size = 0.98 \begin {gather*} -\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x + 1\right )} + 2 \, \sqrt {3} {\left (x^{3} + x^{2} - x - 1\right )}^{\frac {1}{3}}}{3 \, {\left (x + 1\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} + x^{2} - x - 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + 2 \, x + {\left (x^{3} + x^{2} - x - 1\right )}^{\frac {2}{3}} + 1}{x^{2} + 2 \, x + 1}\right ) - \log \left (-\frac {x - {\left (x^{3} + x^{2} - x - 1\right )}^{\frac {1}{3}} + 1}{x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{x^{3} + x^{2} - x - 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

int(1/(x^2 - x + x^3 - 1)^(1/3), x)

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