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

Optimal. Leaf size=125 \[ -\frac {\log \left (\sqrt [3]{2} \sqrt [3]{x^2-2 x-3}+2\right )}{2\ 2^{2/3}}+\frac {\log \left (2^{2/3} \left (x^2-2 x-3\right )^{2/3}-2 \sqrt [3]{2} \sqrt [3]{x^2-2 x-3}+4\right )}{4\ 2^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{2} \sqrt [3]{x^2-2 x-3}}{\sqrt {3}}\right )}{2\ 2^{2/3}} \]

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Rubi [A]  time = 0.06, antiderivative size = 84, normalized size of antiderivative = 0.67, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {694, 266, 56, 617, 204, 31} \begin {gather*} -\frac {3 \log \left (\sqrt [3]{(x-1)^2-4}+2^{2/3}\right )}{4\ 2^{2/3}}+\frac {\log (1-x)}{2\ 2^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-\sqrt [3]{2} \sqrt [3]{(x-1)^2-4}}{\sqrt {3}}\right )}{2\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[3]*ArcTan[(1 - 2^(1/3)*(-4 + (-1 + x)^2)^(1/3))/Sqrt[3]])/2^(2/3) - (3*Log[2^(2/3) + (-4 + (-1 + x)
^2)^(1/3)])/(4*2^(2/3)) + Log[1 - x]/(2*2^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(
1/3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && Ne
gQ[(b*c - a*d)/b]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 36, normalized size = 0.29 \begin {gather*} \frac {3}{16} \left ((x-1)^2-4\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {1}{4} \left (4-(x-1)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-4 + (-1 + x)^2)^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, (4 - (-1 + x)^2)/4])/16

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IntegrateAlgebraic [A]  time = 0.23, size = 125, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{2} \sqrt [3]{-3-2 x+x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {\log \left (2+\sqrt [3]{2} \sqrt [3]{-3-2 x+x^2}\right )}{2\ 2^{2/3}}+\frac {\log \left (4-2 \sqrt [3]{2} \sqrt [3]{-3-2 x+x^2}+2^{2/3} \left (-3-2 x+x^2\right )^{2/3}\right )}{4\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[3]*ArcTan[1/Sqrt[3] - (2^(1/3)*(-3 - 2*x + x^2)^(1/3))/Sqrt[3]])/2^(2/3) - Log[2 + 2^(1/3)*(-3 - 2*
x + x^2)^(1/3)]/(2*2^(2/3)) + Log[4 - 2*2^(1/3)*(-3 - 2*x + x^2)^(1/3) + 2^(2/3)*(-3 - 2*x + x^2)^(2/3)]/(4*2^
(2/3))

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fricas [A]  time = 0.45, size = 113, normalized size = 0.90 \begin {gather*} \frac {1}{4} \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (2 \, \left (-1\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x - 3\right )}^{\frac {1}{3}} + 4^{\frac {1}{3}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{2} - 2 \, x - 3\right )}^{\frac {1}{3}} - 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 3\right )}^{\frac {2}{3}}\right ) + \frac {1}{8} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (x^{2} - 2 \, x - 3\right )}^{\frac {1}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*4^(1/6)*sqrt(3)*(-1)^(1/3)*arctan(1/6*4^(1/6)*sqrt(3)*(2*(-1)^(1/3)*(x^2 - 2*x - 3)^(1/3) + 4^(1/3))) - 1/
16*4^(2/3)*(-1)^(1/3)*log(-4^(1/3)*(-1)^(2/3)*(x^2 - 2*x - 3)^(1/3) - 4^(2/3)*(-1)^(1/3) + (x^2 - 2*x - 3)^(2/
3)) + 1/8*4^(2/3)*(-1)^(1/3)*log(4^(1/3)*(-1)^(2/3) + (x^2 - 2*x - 3)^(1/3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} - 2 \, x - 3\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^2-2*x-3)^(1/3),x, algorithm="giac")

[Out]

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

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maple [C]  time = 8.34, size = 806, normalized size = 6.45

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{3}+2\right ) \ln \left (-\frac {4 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}+2\right )^{2} x^{2}-20 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )^{3} x^{2}-8 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}+2\right )^{2} x +40 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )^{3} x +120 \left (x^{2}-2 x -3\right )^{\frac {2}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )^{2}-198 \left (x^{2}-2 x -3\right )^{\frac {1}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )+3 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) x^{2}-120 \left (x^{2}-2 x -3\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}+2\right )^{2}-15 \RootOf \left (\textit {\_Z}^{3}+2\right ) x^{2}-6 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) x +30 \RootOf \left (\textit {\_Z}^{3}+2\right ) x -21 \left (x^{2}-2 x -3\right )^{\frac {2}{3}}-37 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right )+185 \RootOf \left (\textit {\_Z}^{3}+2\right )}{\left (-1+x \right )^{2}}\right )}{4}+\frac {\RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) \ln \left (-\frac {80 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}+2\right )^{2} x^{2}-4 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )^{3} x^{2}-160 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}+2\right )^{2} x +8 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )^{3} x -240 \left (x^{2}-2 x -3\right )^{\frac {2}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+84 \left (x^{2}-2 x -3\right )^{\frac {1}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )-140 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) x^{2}+240 \left (x^{2}-2 x -3\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+7 \RootOf \left (\textit {\_Z}^{3}+2\right ) x^{2}+280 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right ) x -14 \RootOf \left (\textit {\_Z}^{3}+2\right ) x +198 \left (x^{2}-2 x -3\right )^{\frac {2}{3}}+740 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+4 \textit {\_Z}^{2}\right )-37 \RootOf \left (\textit {\_Z}^{3}+2\right )}{\left (-1+x \right )^{2}}\right )}{2}\)	\(806\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*RootOf(_Z^3+2)*ln(-(4*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^2*x^2-20*RootOf
(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^3*x^2-8*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+
2)+4*_Z^2)^2*RootOf(_Z^3+2)^2*x+40*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^3*x+120*
(x^2-2*x-3)^(2/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^2-198*(x^2-2*x-3)^(1/3)*R
ootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)+3*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2
)+4*_Z^2)*x^2-120*(x^2-2*x-3)^(1/3)*RootOf(_Z^3+2)^2-15*RootOf(_Z^3+2)*x^2-6*RootOf(RootOf(_Z^3+2)^2+2*_Z*Root
Of(_Z^3+2)+4*_Z^2)*x+30*RootOf(_Z^3+2)*x-21*(x^2-2*x-3)^(2/3)-37*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4
*_Z^2)+185*RootOf(_Z^3+2))/(-1+x)^2)+1/2*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*ln(-(80*RootOf(Ro
otOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^2*x^2-4*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2
)+4*_Z^2)*RootOf(_Z^3+2)^3*x^2-160*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^2*x+8*
RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^3*x-240*(x^2-2*x-3)^(2/3)*RootOf(RootOf(_Z^
3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^2+84*(x^2-2*x-3)^(1/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_
Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)-140*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*x^2+240*(x^2-2*x-3)^(1/3
)*RootOf(_Z^3+2)^2+7*RootOf(_Z^3+2)*x^2+280*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*x-14*RootOf(_Z
^3+2)*x+198*(x^2-2*x-3)^(2/3)+740*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)-37*RootOf(_Z^3+2))/(-1+x
)^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} - 2 \, x - 3\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^2-2*x-3)^(1/3),x, algorithm="maxima")

[Out]

integrate(1/((x^2 - 2*x - 3)^(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^2-2\,x-3\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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