3.20.93 \(\int \frac {\sqrt [3]{-6-2 x+x^2}}{-1+x} \, dx\)

Optimal. Leaf size=141 \[ \frac {3}{2} \sqrt [3]{x^2-2 x-6}-\frac {1}{2} \sqrt [3]{7} \log \left (7^{2/3} \sqrt [3]{x^2-2 x-6}+7\right )+\frac {1}{4} \sqrt [3]{7} \log \left (-\sqrt [3]{7} \left (x^2-2 x-6\right )^{2/3}+7^{2/3} \sqrt [3]{x^2-2 x-6}-7\right )+\frac {1}{2} \sqrt {3} \sqrt [3]{7} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x^2-2 x-6}}{\sqrt {3} \sqrt [3]{7}}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 103, normalized size of antiderivative = 0.73, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {694, 266, 50, 58, 617, 204, 31} \begin {gather*} \frac {3}{2} \sqrt [3]{(x-1)^2-7}-\frac {3}{4} \sqrt [3]{7} \log \left (\sqrt [3]{(x-1)^2-7}+\sqrt [3]{7}\right )+\frac {1}{2} \sqrt [3]{7} \log (1-x)+\frac {1}{2} \sqrt {3} \sqrt [3]{7} \tan ^{-1}\left (\frac {\sqrt [3]{7}-2 \sqrt [3]{(x-1)^2-7}}{\sqrt {3} \sqrt [3]{7}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6 - 2*x + x^2)^(1/3)/(-1 + x),x]

[Out]

(3*(-7 + (-1 + x)^2)^(1/3))/2 + (Sqrt[3]*7^(1/3)*ArcTan[(7^(1/3) - 2*(-7 + (-1 + x)^2)^(1/3))/(Sqrt[3]*7^(1/3)
)])/2 - (3*7^(1/3)*Log[7^(1/3) + (-7 + (-1 + x)^2)^(1/3)])/4 + (7^(1/3)*Log[1 - x])/2

Rule 31

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d
*x)^(1/3)], x] + Dist[3/(2*b*q^2), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x
] && NegQ[(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 {\sqrt [3]{-6-2 x+x^2}}{-1+x} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt [3]{-7+x^2}}{x} \, dx,x,-1+x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt [3]{-7+x}}{x} \, dx,x,(-1+x)^2\right )\\ &=\frac {3}{2} \sqrt [3]{-7+(-1+x)^2}-\frac {7}{2} \operatorname {Subst}\left (\int \frac {1}{(-7+x)^{2/3} x} \, dx,x,(-1+x)^2\right )\\ &=\frac {3}{2} \sqrt [3]{-7+(-1+x)^2}+\frac {1}{2} \sqrt [3]{7} \log (1-x)-\frac {1}{4} \left (3 \sqrt [3]{7}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{7}+x} \, dx,x,\sqrt [3]{-7+(-1+x)^2}\right )-\frac {1}{4} \left (3\ 7^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{7^{2/3}-\sqrt [3]{7} x+x^2} \, dx,x,\sqrt [3]{-7+(-1+x)^2}\right )\\ &=\frac {3}{2} \sqrt [3]{-7+(-1+x)^2}-\frac {3}{4} \sqrt [3]{7} \log \left (\sqrt [3]{7}+\sqrt [3]{-7+(-1+x)^2}\right )+\frac {1}{2} \sqrt [3]{7} \log (1-x)-\frac {1}{2} \left (3 \sqrt [3]{7}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{-7+(-1+x)^2}}{\sqrt [3]{7}}\right )\\ &=\frac {3}{2} \sqrt [3]{-7+(-1+x)^2}+\frac {1}{2} \sqrt {3} \sqrt [3]{7} \tan ^{-1}\left (\frac {7-2\ 7^{2/3} \sqrt [3]{-7+(-1+x)^2}}{7 \sqrt {3}}\right )-\frac {3}{4} \sqrt [3]{7} \log \left (\sqrt [3]{7}+\sqrt [3]{-7+(-1+x)^2}\right )+\frac {1}{2} \sqrt [3]{7} \log (1-x)\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 129, normalized size = 0.91 \begin {gather*} \frac {1}{4} \left (6 \sqrt [3]{x^2-2 x-6}+2 \sqrt {3} \sqrt [3]{7} \tan ^{-1}\left (\frac {7-2\ 7^{2/3} \sqrt [3]{x^2-2 x-6}}{7 \sqrt {3}}\right )-2 \sqrt [3]{7} \log \left (\sqrt [3]{(x-1)^2-7}+\sqrt [3]{7}\right )+\sqrt [3]{7} \log \left (\left ((x-1)^2-7\right )^{2/3}-\sqrt [3]{7} \sqrt [3]{(x-1)^2-7}+7^{2/3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 - 2*x + x^2)^(1/3)/(-1 + x),x]

[Out]

(6*(-6 - 2*x + x^2)^(1/3) + 2*Sqrt[3]*7^(1/3)*ArcTan[(7 - 2*7^(2/3)*(-6 - 2*x + x^2)^(1/3))/(7*Sqrt[3])] - 2*7
^(1/3)*Log[7^(1/3) + (-7 + (-1 + x)^2)^(1/3)] + 7^(1/3)*Log[7^(2/3) - 7^(1/3)*(-7 + (-1 + x)^2)^(1/3) + (-7 +
(-1 + x)^2)^(2/3)])/4

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IntegrateAlgebraic [A]  time = 0.25, size = 141, normalized size = 1.00 \begin {gather*} \frac {3}{2} \sqrt [3]{-6-2 x+x^2}+\frac {1}{2} \sqrt {3} \sqrt [3]{7} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-6-2 x+x^2}}{\sqrt {3} \sqrt [3]{7}}\right )-\frac {1}{2} \sqrt [3]{7} \log \left (7+7^{2/3} \sqrt [3]{-6-2 x+x^2}\right )+\frac {1}{4} \sqrt [3]{7} \log \left (-7+7^{2/3} \sqrt [3]{-6-2 x+x^2}-\sqrt [3]{7} \left (-6-2 x+x^2\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-6 - 2*x + x^2)^(1/3)/(-1 + x),x]

[Out]

(3*(-6 - 2*x + x^2)^(1/3))/2 + (Sqrt[3]*7^(1/3)*ArcTan[1/Sqrt[3] - (2*(-6 - 2*x + x^2)^(1/3))/(Sqrt[3]*7^(1/3)
)])/2 - (7^(1/3)*Log[7 + 7^(2/3)*(-6 - 2*x + x^2)^(1/3)])/2 + (7^(1/3)*Log[-7 + 7^(2/3)*(-6 - 2*x + x^2)^(1/3)
- 7^(1/3)*(-6 - 2*x + x^2)^(2/3)])/4

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fricas [A]  time = 1.24, size = 102, normalized size = 0.72 \begin {gather*} \frac {1}{2} \, \sqrt {3} \left (-7\right )^{\frac {1}{3}} \arctan \left (\frac {2}{21} \, \sqrt {3} \left (-7\right )^{\frac {2}{3}} {\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{4} \, \left (-7\right )^{\frac {1}{3}} \log \left (\left (-7\right )^{\frac {2}{3}} + \left (-7\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 6\right )}^{\frac {2}{3}}\right ) + \frac {1}{2} \, \left (-7\right )^{\frac {1}{3}} \log \left (-\left (-7\right )^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-2*x-6)^(1/3)/(-1+x),x, algorithm="fricas")

[Out]

1/2*sqrt(3)*(-7)^(1/3)*arctan(2/21*sqrt(3)*(-7)^(2/3)*(x^2 - 2*x - 6)^(1/3) - 1/3*sqrt(3)) - 1/4*(-7)^(1/3)*lo
g((-7)^(2/3) + (-7)^(1/3)*(x^2 - 2*x - 6)^(1/3) + (x^2 - 2*x - 6)^(2/3)) + 1/2*(-7)^(1/3)*log(-(-7)^(1/3) + (x
^2 - 2*x - 6)^(1/3)) + 3/2*(x^2 - 2*x - 6)^(1/3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}}}{x - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-2*x-6)^(1/3)/(-1+x),x, algorithm="giac")

[Out]

integrate((x^2 - 2*x - 6)^(1/3)/(x - 1), x)

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maple [C]  time = 9.40, size = 1819, normalized size = 12.90

method	result	size
risch	\(\text {Expression too large to display}\)	\(1819\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-2*x-6)^(1/3)/(-1+x),x,method=_RETURNVERBOSE)

[Out]

3/2*(x^2-2*x-6)^(1/3)+(7/12*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*ln((108*RootOf(_Z^3+7)*x^
4-432*RootOf(_Z^3+7)*x^3-2502*RootOf(_Z^3+7)*x^2+5868*RootOf(_Z^3+7)*x+13716*RootOf(_Z^3+7)+21*RootOf(36*RootO
f(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*RootOf(_Z^3+7)^3*x^4-84*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3
+7)+49*_Z^2)*RootOf(_Z^3+7)^3*x^3-42*RootOf(_Z^3+7)^3*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)
*x^2-980*RootOf(_Z^3+7)^2*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)^2*x^2+252*RootOf(_Z^3+7)^3*
RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*x+5880*RootOf(_Z^3+7)^2*RootOf(36*RootOf(_Z^3+7)^2+42
*_Z*RootOf(_Z^3+7)+49*_Z^2)^2*x+320040*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)-58380*RootOf(3
6*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*x^2+136920*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49
*_Z^2)*x+2520*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*x^4-10080*RootOf(36*RootOf(_Z^3+7)^2+42
*_Z*RootOf(_Z^3+7)+49*_Z^2)*x^3-819*(x^4-4*x^3-8*x^2+24*x+36)^(1/3)*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z
^3+7)+49*_Z^2)*RootOf(_Z^3+7)*x^2+50274*(x^4-4*x^3-8*x^2+24*x+36)^(2/3)+490*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*R
ootOf(_Z^3+7)+49*_Z^2)^2*RootOf(_Z^3+7)^2*x^4-1960*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)^2*
RootOf(_Z^3+7)^2*x^3+1638*(x^4-4*x^3-8*x^2+24*x+36)^(1/3)*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_
Z^2)*RootOf(_Z^3+7)*x-7560*(x^4-4*x^3-8*x^2+24*x+36)^(2/3)*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*
_Z^2)*RootOf(_Z^3+7)^2+6480*RootOf(_Z^3+7)^2*(x^4-4*x^3-8*x^2+24*x+36)^(1/3)*x^2-12960*RootOf(_Z^3+7)^2*(x^4-4
*x^3-8*x^2+24*x+36)^(1/3)*x+4914*(x^4-4*x^3-8*x^2+24*x+36)^(1/3)*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+
7)+49*_Z^2)*RootOf(_Z^3+7)-38880*RootOf(_Z^3+7)^2*(x^4-4*x^3-8*x^2+24*x+36)^(1/3))/(x^2-2*x-6)/(-1+x)^2)+1/2*R
ootOf(_Z^3+7)*ln(-(-9360*RootOf(_Z^3+7)*x^4+37440*RootOf(_Z^3+7)*x^3+110160*RootOf(_Z^3+7)*x^2-295200*RootOf(_
Z^3+7)*x-548640*RootOf(_Z^3+7)+840*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*RootOf(_Z^3+7)^3*x
^4-3360*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*RootOf(_Z^3+7)^3*x^3-1680*RootOf(_Z^3+7)^3*Ro
otOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*x^2-98*RootOf(_Z^3+7)^2*RootOf(36*RootOf(_Z^3+7)^2+42*_
Z*RootOf(_Z^3+7)+49*_Z^2)^2*x^2+10080*RootOf(_Z^3+7)^3*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2
)*x+588*RootOf(_Z^3+7)^2*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)^2*x-32004*RootOf(36*RootOf(_
Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)+6426*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*x^2-17220
*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*x-546*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7
)+49*_Z^2)*x^4+2184*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*x^3+16758*(x^4-4*x^3-8*x^2+24*x+3
6)^(1/3)*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*RootOf(_Z^3+7)*x^2-9828*(x^4-4*x^3-8*x^2+24*
x+36)^(2/3)+49*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)^2*RootOf(_Z^3+7)^2*x^4-196*RootOf(36*R
ootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)^2*RootOf(_Z^3+7)^2*x^3-33516*(x^4-4*x^3-8*x^2+24*x+36)^(1/3)*Roo
tOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*RootOf(_Z^3+7)*x-15120*(x^4-4*x^3-8*x^2+24*x+36)^(2/3)*R
ootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*RootOf(_Z^3+7)^2+12960*RootOf(_Z^3+7)^2*(x^4-4*x^3-8*x
^2+24*x+36)^(1/3)*x^2-25920*RootOf(_Z^3+7)^2*(x^4-4*x^3-8*x^2+24*x+36)^(1/3)*x-100548*(x^4-4*x^3-8*x^2+24*x+36
)^(1/3)*RootOf(36*RootOf(_Z^3+7)^2+42*_Z*RootOf(_Z^3+7)+49*_Z^2)*RootOf(_Z^3+7)-77760*RootOf(_Z^3+7)^2*(x^4-4*
x^3-8*x^2+24*x+36)^(1/3))/(x^2-2*x-6)/(-1+x)^2))/(x^2-2*x-6)^(2/3)*((x^2-2*x-6)^2)^(1/3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}}}{x - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-2*x-6)^(1/3)/(-1+x),x, algorithm="maxima")

[Out]

integrate((x^2 - 2*x - 6)^(1/3)/(x - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^2-2\,x-6\right )}^{1/3}}{x-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 2*x - 6)^(1/3)/(x - 1),x)

[Out]

int((x^2 - 2*x - 6)^(1/3)/(x - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{x^{2} - 2 x - 6}}{x - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-2*x-6)**(1/3)/(-1+x),x)

[Out]

Integral((x**2 - 2*x - 6)**(1/3)/(x - 1), x)

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