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

Optimal. Leaf size=174 \[ \frac {1}{4} \sqrt [3]{6 x^2-2 x-1}-\frac {1}{12} \sqrt [3]{\frac {7}{6}} \log \left (\sqrt [3]{6} 7^{2/3} \sqrt [3]{6 x^2-2 x-1}+7\right )+\frac {1}{24} \sqrt [3]{\frac {7}{6}} \log \left (6^{2/3} \sqrt [3]{7} \left (6 x^2-2 x-1\right )^{2/3}-\sqrt [3]{6} 7^{2/3} \sqrt [3]{6 x^2-2 x-1}+7\right )+\frac {\sqrt [3]{\frac {7}{2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {2}{7}} \sqrt [3]{6 x^2-2 x-1}}{\sqrt [6]{3}}\right )}{4\ 3^{5/6}} \]

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Rubi [A]  time = 0.16, antiderivative size = 119, normalized size of antiderivative = 0.68, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {694, 266, 50, 58, 617, 204, 31} \begin {gather*} \frac {\sqrt [3]{(6 x-1)^2-7}}{4 \sqrt [3]{6}}+\frac {1}{12} \sqrt [3]{\frac {7}{6}} \log (1-6 x)-\frac {1}{8} \sqrt [3]{\frac {7}{6}} \log \left (\sqrt [3]{(6 x-1)^2-7}+\sqrt [3]{7}\right )+\frac {\sqrt [3]{\frac {7}{2}} \tan ^{-1}\left (\frac {7-2\ 7^{2/3} \sqrt [3]{(6 x-1)^2-7}}{7 \sqrt {3}}\right )}{4\ 3^{5/6}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7 + (-1 + 6*x)^2)^(1/3)/(4*6^(1/3)) + ((7/2)^(1/3)*ArcTan[(7 - 2*7^(2/3)*(-7 + (-1 + 6*x)^2)^(1/3))/(7*Sqrt[
3])])/(4*3^(5/6)) + ((7/6)^(1/3)*Log[1 - 6*x])/12 - ((7/6)^(1/3)*Log[7^(1/3) + (-7 + (-1 + 6*x)^2)^(1/3)])/8

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]{-1-2 x+6 x^2}}{-1+6 x} \, dx &=\frac {1}{6} \operatorname {Subst}\left (\int \frac {\sqrt [3]{-\frac {7}{6}+\frac {x^2}{6}}}{x} \, dx,x,-1+6 x\right )\\ &=\frac {1}{12} \operatorname {Subst}\left (\int \frac {\sqrt [3]{-\frac {7}{6}+\frac {x}{6}}}{x} \, dx,x,(-1+6 x)^2\right )\\ &=\frac {\sqrt [3]{-7+(-1+6 x)^2}}{4 \sqrt [3]{6}}-\frac {7}{72} \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {7}{6}+\frac {x}{6}\right )^{2/3} x} \, dx,x,(-1+6 x)^2\right )\\ &=\frac {\sqrt [3]{-7+(-1+6 x)^2}}{4 \sqrt [3]{6}}+\frac {1}{12} \sqrt [3]{\frac {7}{6}} \log (1-6 x)-\frac {1}{8} \sqrt [3]{\frac {7}{6}} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {7}{6}}+x} \, dx,x,\sqrt [3]{-1-2 x+6 x^2}\right )-\frac {1}{8} \left (\frac {7}{6}\right )^{2/3} \operatorname {Subst}\left (\int \frac {1}{\left (\frac {7}{6}\right )^{2/3}-\sqrt [3]{\frac {7}{6}} x+x^2} \, dx,x,\sqrt [3]{-1-2 x+6 x^2}\right )\\ &=\frac {\sqrt [3]{-7+(-1+6 x)^2}}{4 \sqrt [3]{6}}+\frac {1}{12} \sqrt [3]{\frac {7}{6}} \log (1-6 x)-\frac {1}{8} \sqrt [3]{\frac {7}{6}} \log \left (\sqrt [3]{7}+\sqrt [3]{6} \sqrt [3]{-1-2 x+6 x^2}\right )-\frac {1}{4} \sqrt [3]{\frac {7}{6}} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 \sqrt [3]{\frac {6}{7}} \sqrt [3]{-1-2 x+6 x^2}\right )\\ &=\frac {\sqrt [3]{-7+(-1+6 x)^2}}{4 \sqrt [3]{6}}+\frac {\sqrt [3]{\frac {7}{2}} \tan ^{-1}\left (\frac {7-2 \sqrt [3]{6} 7^{2/3} \sqrt [3]{-1-2 x+6 x^2}}{7 \sqrt {3}}\right )}{4\ 3^{5/6}}+\frac {1}{12} \sqrt [3]{\frac {7}{6}} \log (1-6 x)-\frac {1}{8} \sqrt [3]{\frac {7}{6}} \log \left (\sqrt [3]{7}+\sqrt [3]{6} \sqrt [3]{-1-2 x+6 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 152, normalized size = 0.87 \begin {gather*} \frac {1}{4} \sqrt [3]{6 x^2-2 x-1}+\frac {\sqrt [3]{\frac {7}{2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {2}{7}} \sqrt [3]{6 x^2-2 x-1}}{\sqrt [6]{3}}\right )}{4\ 3^{5/6}}-\frac {1}{12} \sqrt [3]{\frac {7}{6}} \log \left (\sqrt [3]{(1-6 x)^2-7}+\sqrt [3]{7}\right )+\frac {1}{24} \sqrt [3]{\frac {7}{6}} \log \left (\left ((1-6 x)^2-7\right )^{2/3}-\sqrt [3]{7} \sqrt [3]{(1-6 x)^2-7}+7^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 - 2*x + 6*x^2)^(1/3)/4 + ((7/2)^(1/3)*ArcTan[1/Sqrt[3] - (2*(2/7)^(1/3)*(-1 - 2*x + 6*x^2)^(1/3))/3^(1/6)]
)/(4*3^(5/6)) - ((7/6)^(1/3)*Log[7^(1/3) + (-7 + (1 - 6*x)^2)^(1/3)])/12 + ((7/6)^(1/3)*Log[7^(2/3) - 7^(1/3)*
(-7 + (1 - 6*x)^2)^(1/3) + (-7 + (1 - 6*x)^2)^(2/3)])/24

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IntegrateAlgebraic [A]  time = 0.43, size = 174, normalized size = 1.00 \begin {gather*} \frac {1}{4} \sqrt [3]{-1-2 x+6 x^2}+\frac {\sqrt [3]{\frac {7}{2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {2}{7}} \sqrt [3]{-1-2 x+6 x^2}}{\sqrt [6]{3}}\right )}{4\ 3^{5/6}}-\frac {1}{12} \sqrt [3]{\frac {7}{6}} \log \left (7+\sqrt [3]{6} 7^{2/3} \sqrt [3]{-1-2 x+6 x^2}\right )+\frac {1}{24} \sqrt [3]{\frac {7}{6}} \log \left (7-\sqrt [3]{6} 7^{2/3} \sqrt [3]{-1-2 x+6 x^2}+6^{2/3} \sqrt [3]{7} \left (-1-2 x+6 x^2\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 - 2*x + 6*x^2)^(1/3)/4 + ((7/2)^(1/3)*ArcTan[1/Sqrt[3] - (2*(2/7)^(1/3)*(-1 - 2*x + 6*x^2)^(1/3))/3^(1/6)]
)/(4*3^(5/6)) - ((7/6)^(1/3)*Log[7 + 6^(1/3)*7^(2/3)*(-1 - 2*x + 6*x^2)^(1/3)])/12 + ((7/6)^(1/3)*Log[7 - 6^(1
/3)*7^(2/3)*(-1 - 2*x + 6*x^2)^(1/3) + 6^(2/3)*7^(1/3)*(-1 - 2*x + 6*x^2)^(2/3)])/24

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fricas [A]  time = 0.65, size = 146, normalized size = 0.84 \begin {gather*} \frac {1}{24} \cdot 6^{\frac {1}{6}} \sqrt {2} \left (-7\right )^{\frac {1}{3}} \arctan \left (\frac {1}{42} \cdot 6^{\frac {1}{6}} {\left (2 \cdot 6^{\frac {2}{3}} \sqrt {2} \left (-7\right )^{\frac {2}{3}} {\left (6 \, x^{2} - 2 \, x - 1\right )}^{\frac {1}{3}} - 7 \cdot 6^{\frac {1}{3}} \sqrt {2}\right )}\right ) - \frac {1}{144} \cdot 6^{\frac {2}{3}} \left (-7\right )^{\frac {1}{3}} \log \left (6^{\frac {2}{3}} \left (-7\right )^{\frac {1}{3}} {\left (6 \, x^{2} - 2 \, x - 1\right )}^{\frac {1}{3}} + 6^{\frac {1}{3}} \left (-7\right )^{\frac {2}{3}} + 6 \, {\left (6 \, x^{2} - 2 \, x - 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{72} \cdot 6^{\frac {2}{3}} \left (-7\right )^{\frac {1}{3}} \log \left (-6^{\frac {2}{3}} \left (-7\right )^{\frac {1}{3}} + 6 \, {\left (6 \, x^{2} - 2 \, x - 1\right )}^{\frac {1}{3}}\right ) + \frac {1}{4} \, {\left (6 \, x^{2} - 2 \, x - 1\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*6^(1/6)*sqrt(2)*(-7)^(1/3)*arctan(1/42*6^(1/6)*(2*6^(2/3)*sqrt(2)*(-7)^(2/3)*(6*x^2 - 2*x - 1)^(1/3) - 7*
6^(1/3)*sqrt(2))) - 1/144*6^(2/3)*(-7)^(1/3)*log(6^(2/3)*(-7)^(1/3)*(6*x^2 - 2*x - 1)^(1/3) + 6^(1/3)*(-7)^(2/
3) + 6*(6*x^2 - 2*x - 1)^(2/3)) + 1/72*6^(2/3)*(-7)^(1/3)*log(-6^(2/3)*(-7)^(1/3) + 6*(6*x^2 - 2*x - 1)^(1/3))
+ 1/4*(6*x^2 - 2*x - 1)^(1/3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 16.08, size = 1308, normalized size = 7.52

method	result	size
trager	\(\text {Expression too large to display}\)	\(1308\)
risch	\(\text {Expression too large to display}\)	\(2660\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(6*x^2-2*x-1)^(1/3)+7/4*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)*ln((-2394*RootOf(Roo
tOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)*RootOf(_Z^3+252)^4*x^2-317520*RootOf(RootOf(_Z^3+252)^2+12
6*_Z*RootOf(_Z^3+252)+15876*_Z^2)^2*RootOf(_Z^3+252)^3*x^2+798*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+25
2)+15876*_Z^2)*RootOf(_Z^3+252)^4*x+105840*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)^2*Roo
tOf(_Z^3+252)^3*x+50274*(6*x^2-2*x-1)^(2/3)*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)*Root
Of(_Z^3+252)^2+8892*RootOf(_Z^3+252)^2*x^2+1179360*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^
2)*RootOf(_Z^3+252)*x^2-2964*RootOf(_Z^3+252)^2*x-393120*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+158
76*_Z^2)*RootOf(_Z^3+252)*x-90720*(6*x^2-2*x-1)^(2/3)-15120*(6*x^2-2*x-1)^(1/3)*RootOf(_Z^3+252)-2111508*(6*x^
2-2*x-1)^(1/3)*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)-2413*RootOf(_Z^3+252)^2-320040*Ro
otOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)*RootOf(_Z^3+252))/(-1+6*x)^2)-1/72*ln(-(126*RootOf
(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)*RootOf(_Z^3+252)^4*x^2+317520*RootOf(RootOf(_Z^3+252)^
2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)^2*RootOf(_Z^3+252)^3*x^2-42*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3
+252)+15876*_Z^2)*RootOf(_Z^3+252)^4*x-105840*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)^2*
RootOf(_Z^3+252)^3*x+50274*(6*x^2-2*x-1)^(2/3)*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)*R
ootOf(_Z^3+252)^2+216*RootOf(_Z^3+252)^2*x^2+544320*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z
^2)*RootOf(_Z^3+252)*x^2-72*RootOf(_Z^3+252)^2*x-181440*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+1587
6*_Z^2)*RootOf(_Z^3+252)*x-9828*(6*x^2-2*x-1)^(2/3)-1638*(6*x^2-2*x-1)^(1/3)*RootOf(_Z^3+252)-2111508*(6*x^2-2
*x-1)^(1/3)*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)-127*RootOf(_Z^3+252)^2-320040*RootOf
(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)*RootOf(_Z^3+252))/(-1+6*x)^2)*RootOf(_Z^3+252)-7/4*ln(
-(126*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)*RootOf(_Z^3+252)^4*x^2+317520*RootOf(RootO
f(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)^2*RootOf(_Z^3+252)^3*x^2-42*RootOf(RootOf(_Z^3+252)^2+126*_Z
*RootOf(_Z^3+252)+15876*_Z^2)*RootOf(_Z^3+252)^4*x-105840*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15
876*_Z^2)^2*RootOf(_Z^3+252)^3*x+50274*(6*x^2-2*x-1)^(2/3)*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+1
5876*_Z^2)*RootOf(_Z^3+252)^2+216*RootOf(_Z^3+252)^2*x^2+544320*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+2
52)+15876*_Z^2)*RootOf(_Z^3+252)*x^2-72*RootOf(_Z^3+252)^2*x-181440*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z
^3+252)+15876*_Z^2)*RootOf(_Z^3+252)*x-9828*(6*x^2-2*x-1)^(2/3)-1638*(6*x^2-2*x-1)^(1/3)*RootOf(_Z^3+252)-2111
508*(6*x^2-2*x-1)^(1/3)*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)-127*RootOf(_Z^3+252)^2-3
20040*RootOf(RootOf(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)*RootOf(_Z^3+252))/(-1+6*x)^2)*RootOf(RootO
f(_Z^3+252)^2+126*_Z*RootOf(_Z^3+252)+15876*_Z^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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