3.685 \(\int \frac {\sqrt [3]{x}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=117 \[ \frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{2/3} b^{4/3}}-\frac {\log (a+b x)}{6 a^{2/3} b^{4/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{4/3}}-\frac {\sqrt [3]{x}}{b (a+b x)} \]

[Out]

-x^(1/3)/b/(b*x+a)+1/2*ln(a^(1/3)+b^(1/3)*x^(1/3))/a^(2/3)/b^(4/3)-1/6*ln(b*x+a)/a^(2/3)/b^(4/3)-1/3*arctan(1/
3*(a^(1/3)-2*b^(1/3)*x^(1/3))/a^(1/3)*3^(1/2))/a^(2/3)/b^(4/3)*3^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {47, 58, 617, 204, 31} \[ \frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{2/3} b^{4/3}}-\frac {\log (a+b x)}{6 a^{2/3} b^{4/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{4/3}}-\frac {\sqrt [3]{x}}{b (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[x^(1/3)/(a + b*x)^2,x]

[Out]

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

Rule 31

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

Rule 47

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

Rubi steps

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

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.23 \[ \frac {3 x^{4/3} \, _2F_1\left (\frac {4}{3},2;\frac {7}{3};-\frac {b x}{a}\right )}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(1/3)/(a + b*x)^2,x]

[Out]

(3*x^(4/3)*Hypergeometric2F1[4/3, 2, 7/3, -((b*x)/a)])/(4*a^2)

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fricas [B]  time = 0.50, size = 389, normalized size = 3.32 \[ \left [-\frac {6 \, a^{2} b x^{\frac {1}{3}} - 3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x + a^{2} b\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{\frac {2}{3}} - \left (a^{2} b\right )^{\frac {1}{3}} a + \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x^{\frac {1}{3}}}{b x + a}\right ) + \left (a^{2} b\right )^{\frac {2}{3}} {\left (b x + a\right )} \log \left (a b x^{\frac {2}{3}} + \left (a^{2} b\right )^{\frac {1}{3}} a - \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right ) - 2 \, \left (a^{2} b\right )^{\frac {2}{3}} {\left (b x + a\right )} \log \left (a b x^{\frac {1}{3}} + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{6 \, {\left (a^{2} b^{3} x + a^{3} b^{2}\right )}}, -\frac {6 \, a^{2} b x^{\frac {1}{3}} - 6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x + a^{2} b\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (a^{2} b\right )^{\frac {1}{3}} a - 2 \, \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) + \left (a^{2} b\right )^{\frac {2}{3}} {\left (b x + a\right )} \log \left (a b x^{\frac {2}{3}} + \left (a^{2} b\right )^{\frac {1}{3}} a - \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right ) - 2 \, \left (a^{2} b\right )^{\frac {2}{3}} {\left (b x + a\right )} \log \left (a b x^{\frac {1}{3}} + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{6 \, {\left (a^{2} b^{3} x + a^{3} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(6*a^2*b*x^(1/3) - 3*sqrt(1/3)*(a*b^2*x + a^2*b)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x - a^2 + 3*sqrt(1/3)
*(2*a*b*x^(2/3) - (a^2*b)^(1/3)*a + (a^2*b)^(2/3)*x^(1/3))*sqrt(-(a^2*b)^(1/3)/b) - 3*(a^2*b)^(1/3)*a*x^(1/3))
/(b*x + a)) + (a^2*b)^(2/3)*(b*x + a)*log(a*b*x^(2/3) + (a^2*b)^(1/3)*a - (a^2*b)^(2/3)*x^(1/3)) - 2*(a^2*b)^(
2/3)*(b*x + a)*log(a*b*x^(1/3) + (a^2*b)^(2/3)))/(a^2*b^3*x + a^3*b^2), -1/6*(6*a^2*b*x^(1/3) - 6*sqrt(1/3)*(a
*b^2*x + a^2*b)*sqrt((a^2*b)^(1/3)/b)*arctan(-sqrt(1/3)*((a^2*b)^(1/3)*a - 2*(a^2*b)^(2/3)*x^(1/3))*sqrt((a^2*
b)^(1/3)/b)/a^2) + (a^2*b)^(2/3)*(b*x + a)*log(a*b*x^(2/3) + (a^2*b)^(1/3)*a - (a^2*b)^(2/3)*x^(1/3)) - 2*(a^2
*b)^(2/3)*(b*x + a)*log(a*b*x^(1/3) + (a^2*b)^(2/3)))/(a^2*b^3*x + a^3*b^2)]

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giac [A]  time = 1.13, size = 136, normalized size = 1.16 \[ -\frac {\left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x^{\frac {1}{3}} - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a b^{2}} - \frac {x^{\frac {1}{3}}}{{\left (b x + a\right )} b} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{\frac {2}{3}} + x^{\frac {1}{3}} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-a/b)^(1/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/(a*b) + 1/3*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*(2*x^
(1/3) + (-a/b)^(1/3))/(-a/b)^(1/3))/(a*b^2) - x^(1/3)/((b*x + a)*b) + 1/6*(-a*b^2)^(1/3)*log(x^(2/3) + x^(1/3)
*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b^2)

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maple [A]  time = 0.01, size = 112, normalized size = 0.96 \[ -\frac {x^{\frac {1}{3}}}{\left (b x +a \right ) b}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}+\frac {\ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}-\frac {\ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/3)/(b*x+a)^2,x)

[Out]

-x^(1/3)/b/(b*x+a)+1/3/b^2/(a/b)^(2/3)*ln(x^(1/3)+(a/b)^(1/3))-1/6/b^2/(a/b)^(2/3)*ln(x^(2/3)-(a/b)^(1/3)*x^(1
/3)+(a/b)^(2/3))+1/3/b^2/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x^(1/3)-1))

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maxima [A]  time = 3.03, size = 120, normalized size = 1.03 \[ -\frac {x^{\frac {1}{3}}}{b^{2} x + a b} + \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{\frac {2}{3}} - x^{\frac {1}{3}} \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {\log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^(1/3)/(b^2*x + a*b) + 1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^(1/3) - (a/b)^(1/3))/(a/b)^(1/3))/(b^2*(a/b)^(2/3
)) - 1/6*log(x^(2/3) - x^(1/3)*(a/b)^(1/3) + (a/b)^(2/3))/(b^2*(a/b)^(2/3)) + 1/3*log(x^(1/3) + (a/b)^(1/3))/(
b^2*(a/b)^(2/3))

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mupad [B]  time = 0.06, size = 120, normalized size = 1.03 \[ \frac {\ln \left (3\,b\,x^{1/3}+3\,a^{1/3}\,b^{2/3}\right )}{3\,a^{2/3}\,b^{4/3}}-\frac {x^{1/3}}{b\,\left (a+b\,x\right )}+\frac {\ln \left (3\,b\,x^{1/3}+\frac {3\,a^{1/3}\,b^{2/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^{2/3}\,b^{4/3}}-\frac {\ln \left (3\,b\,x^{1/3}-\frac {3\,a^{1/3}\,b^{2/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^{2/3}\,b^{4/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/3)/(a + b*x)^2,x)

[Out]

log(3*b*x^(1/3) + 3*a^(1/3)*b^(2/3))/(3*a^(2/3)*b^(4/3)) - x^(1/3)/(b*(a + b*x)) + (log(3*b*x^(1/3) + (3*a^(1/
3)*b^(2/3)*(3^(1/2)*1i - 1))/2)*(3^(1/2)*1i - 1))/(6*a^(2/3)*b^(4/3)) - (log(3*b*x^(1/3) - (3*a^(1/3)*b^(2/3)*
(3^(1/2)*1i + 1))/2)*(3^(1/2)*1i + 1))/(6*a^(2/3)*b^(4/3))

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sympy [A]  time = 71.72, size = 607, normalized size = 5.19 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {2}{3}}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {3 x^{\frac {4}{3}}}{4 a^{2}} & \text {for}\: b = 0 \\- \frac {3}{2 b^{2} x^{\frac {2}{3}}} & \text {for}\: a = 0 \\- \frac {2 \sqrt [3]{-1} a^{\frac {4}{3}} \sqrt [3]{\frac {1}{b}} \log {\left (- \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac {1}{b}} + \sqrt [3]{x} \right )}}{6 a^{2} b + 6 a b^{2} x} + \frac {\sqrt [3]{-1} a^{\frac {4}{3}} \sqrt [3]{\frac {1}{b}} \log {\left (4 \left (-1\right )^{\frac {2}{3}} a^{\frac {2}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}} + 4 \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{x} \sqrt [3]{\frac {1}{b}} + 4 x^{\frac {2}{3}} \right )}}{6 a^{2} b + 6 a b^{2} x} + \frac {2 \sqrt [3]{-1} \sqrt {3} a^{\frac {4}{3}} \sqrt [3]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\sqrt {3}}{3} - \frac {2 \left (-1\right )^{\frac {2}{3}} \sqrt {3} \sqrt [3]{x}}{3 \sqrt [3]{a} \sqrt [3]{\frac {1}{b}}} \right )}}{6 a^{2} b + 6 a b^{2} x} - \frac {2 \sqrt [3]{-1} a^{\frac {4}{3}} \sqrt [3]{\frac {1}{b}} \log {\relax (2 )}}{6 a^{2} b + 6 a b^{2} x} - \frac {2 \sqrt [3]{-1} \sqrt [3]{a} b x \sqrt [3]{\frac {1}{b}} \log {\left (- \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac {1}{b}} + \sqrt [3]{x} \right )}}{6 a^{2} b + 6 a b^{2} x} + \frac {\sqrt [3]{-1} \sqrt [3]{a} b x \sqrt [3]{\frac {1}{b}} \log {\left (4 \left (-1\right )^{\frac {2}{3}} a^{\frac {2}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}} + 4 \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{x} \sqrt [3]{\frac {1}{b}} + 4 x^{\frac {2}{3}} \right )}}{6 a^{2} b + 6 a b^{2} x} + \frac {2 \sqrt [3]{-1} \sqrt {3} \sqrt [3]{a} b x \sqrt [3]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\sqrt {3}}{3} - \frac {2 \left (-1\right )^{\frac {2}{3}} \sqrt {3} \sqrt [3]{x}}{3 \sqrt [3]{a} \sqrt [3]{\frac {1}{b}}} \right )}}{6 a^{2} b + 6 a b^{2} x} - \frac {2 \sqrt [3]{-1} \sqrt [3]{a} b x \sqrt [3]{\frac {1}{b}} \log {\relax (2 )}}{6 a^{2} b + 6 a b^{2} x} - \frac {6 a \sqrt [3]{x}}{6 a^{2} b + 6 a b^{2} x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/3)/(b*x+a)**2,x)

[Out]

Piecewise((zoo/x**(2/3), Eq(a, 0) & Eq(b, 0)), (3*x**(4/3)/(4*a**2), Eq(b, 0)), (-3/(2*b**2*x**(2/3)), Eq(a, 0
)), (-2*(-1)**(1/3)*a**(4/3)*(1/b)**(1/3)*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x**(1/3))/(6*a**2*b + 6*a*b
**2*x) + (-1)**(1/3)*a**(4/3)*(1/b)**(1/3)*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*
*(1/3)*(1/b)**(1/3) + 4*x**(2/3))/(6*a**2*b + 6*a*b**2*x) + 2*(-1)**(1/3)*sqrt(3)*a**(4/3)*(1/b)**(1/3)*atan(s
qrt(3)/3 - 2*(-1)**(2/3)*sqrt(3)*x**(1/3)/(3*a**(1/3)*(1/b)**(1/3)))/(6*a**2*b + 6*a*b**2*x) - 2*(-1)**(1/3)*a
**(4/3)*(1/b)**(1/3)*log(2)/(6*a**2*b + 6*a*b**2*x) - 2*(-1)**(1/3)*a**(1/3)*b*x*(1/b)**(1/3)*log(-(-1)**(1/3)
*a**(1/3)*(1/b)**(1/3) + x**(1/3))/(6*a**2*b + 6*a*b**2*x) + (-1)**(1/3)*a**(1/3)*b*x*(1/b)**(1/3)*log(4*(-1)*
*(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x**(1/3)*(1/b)**(1/3) + 4*x**(2/3))/(6*a**2*b + 6*a*b**2
*x) + 2*(-1)**(1/3)*sqrt(3)*a**(1/3)*b*x*(1/b)**(1/3)*atan(sqrt(3)/3 - 2*(-1)**(2/3)*sqrt(3)*x**(1/3)/(3*a**(1
/3)*(1/b)**(1/3)))/(6*a**2*b + 6*a*b**2*x) - 2*(-1)**(1/3)*a**(1/3)*b*x*(1/b)**(1/3)*log(2)/(6*a**2*b + 6*a*b*
*2*x) - 6*a*x**(1/3)/(6*a**2*b + 6*a*b**2*x), True))

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