3.5.5 \(\int \frac {1}{x^3 \sqrt [3]{-a+b x}} \, dx\)

Optimal. Leaf size=136 \[ \frac {b^2 \log (x)}{9 a^{7/3}}-\frac {b^2 \log \left (\sqrt [3]{b x-a}+\sqrt [3]{a}\right )}{3 a^{7/3}}-\frac {2 b^2 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b x-a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3}}+\frac {2 b (b x-a)^{2/3}}{3 a^2 x}+\frac {(b x-a)^{2/3}}{2 a x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {51, 56, 617, 204, 31} \begin {gather*} \frac {b^2 \log (x)}{9 a^{7/3}}-\frac {b^2 \log \left (\sqrt [3]{b x-a}+\sqrt [3]{a}\right )}{3 a^{7/3}}-\frac {2 b^2 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b x-a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3}}+\frac {2 b (b x-a)^{2/3}}{3 a^2 x}+\frac {(b x-a)^{2/3}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(-a + b*x)^(1/3)),x]

[Out]

(-a + b*x)^(2/3)/(2*a*x^2) + (2*b*(-a + b*x)^(2/3))/(3*a^2*x) - (2*b^2*ArcTan[(a^(1/3) - 2*(-a + b*x)^(1/3))/(
Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(7/3)) + (b^2*Log[x])/(9*a^(7/3)) - (b^2*Log[a^(1/3) + (-a + b*x)^(1/3)])/(3*a
^(7/3))

Rule 31

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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 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 {1}{x^3 \sqrt [3]{-a+b x}} \, dx &=\frac {(-a+b x)^{2/3}}{2 a x^2}+\frac {(2 b) \int \frac {1}{x^2 \sqrt [3]{-a+b x}} \, dx}{3 a}\\ &=\frac {(-a+b x)^{2/3}}{2 a x^2}+\frac {2 b (-a+b x)^{2/3}}{3 a^2 x}+\frac {\left (2 b^2\right ) \int \frac {1}{x \sqrt [3]{-a+b x}} \, dx}{9 a^2}\\ &=\frac {(-a+b x)^{2/3}}{2 a x^2}+\frac {2 b (-a+b x)^{2/3}}{3 a^2 x}+\frac {b^2 \log (x)}{9 a^{7/3}}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+x} \, dx,x,\sqrt [3]{-a+b x}\right )}{3 a^{7/3}}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{a^{2/3}-\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{-a+b x}\right )}{3 a^2}\\ &=\frac {(-a+b x)^{2/3}}{2 a x^2}+\frac {2 b (-a+b x)^{2/3}}{3 a^2 x}+\frac {b^2 \log (x)}{9 a^{7/3}}-\frac {b^2 \log \left (\sqrt [3]{a}+\sqrt [3]{-a+b x}\right )}{3 a^{7/3}}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{-a+b x}}{\sqrt [3]{a}}\right )}{3 a^{7/3}}\\ &=\frac {(-a+b x)^{2/3}}{2 a x^2}+\frac {2 b (-a+b x)^{2/3}}{3 a^2 x}-\frac {2 b^2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{-a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{7/3}}+\frac {b^2 \log (x)}{9 a^{7/3}}-\frac {b^2 \log \left (\sqrt [3]{a}+\sqrt [3]{-a+b x}\right )}{3 a^{7/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 38, normalized size = 0.28 \begin {gather*} \frac {3 b^2 (b x-a)^{2/3} \, _2F_1\left (\frac {2}{3},3;\frac {5}{3};1-\frac {b x}{a}\right )}{2 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(-a + b*x)^(1/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.13, size = 160, normalized size = 1.18 \begin {gather*} -\frac {2 b^2 \log \left (\sqrt [3]{b x-a}+\sqrt [3]{a}\right )}{9 a^{7/3}}+\frac {b^2 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b x-a}+(b x-a)^{2/3}\right )}{9 a^{7/3}}-\frac {2 b^2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b x-a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3}}+\frac {(b x-a)^{2/3} (4 (b x-a)+7 a)}{6 a^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-a + b*x)^(2/3)*(7*a + 4*(-a + b*x)))/(6*a^2*x^2) - (2*b^2*ArcTan[1/Sqrt[3] - (2*(-a + b*x)^(1/3))/(Sqrt[3]*
a^(1/3))])/(3*Sqrt[3]*a^(7/3)) - (2*b^2*Log[a^(1/3) + (-a + b*x)^(1/3)])/(9*a^(7/3)) + (b^2*Log[a^(2/3) - a^(1
/3)*(-a + b*x)^(1/3) + (-a + b*x)^(2/3)])/(9*a^(7/3))

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fricas [A]  time = 0.66, size = 374, normalized size = 2.75 \begin {gather*} \left [\frac {6 \, \sqrt {\frac {1}{3}} a b^{2} x^{2} \sqrt {\frac {\left (-a\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b x + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x - a\right )}^{\frac {2}{3}} \left (-a\right )^{\frac {2}{3}} + {\left (b x - a\right )}^{\frac {1}{3}} a + \left (-a\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a\right )^{\frac {1}{3}}}{a}} - 3 \, {\left (b x - a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {2}{3}} - 3 \, a}{x}\right ) + 2 \, \left (-a\right )^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x - a\right )}^{\frac {2}{3}} + {\left (b x - a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + \left (-a\right )^{\frac {2}{3}}\right ) - 4 \, \left (-a\right )^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x - a\right )}^{\frac {1}{3}} - \left (-a\right )^{\frac {1}{3}}\right ) + 3 \, {\left (4 \, a b x + 3 \, a^{2}\right )} {\left (b x - a\right )}^{\frac {2}{3}}}{18 \, a^{3} x^{2}}, \frac {12 \, \sqrt {\frac {1}{3}} a b^{2} x^{2} \sqrt {-\frac {\left (-a\right )^{\frac {1}{3}}}{a}} \arctan \left (\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x - a\right )}^{\frac {1}{3}} + \left (-a\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a\right )^{\frac {1}{3}}}{a}}\right ) + 2 \, \left (-a\right )^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x - a\right )}^{\frac {2}{3}} + {\left (b x - a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + \left (-a\right )^{\frac {2}{3}}\right ) - 4 \, \left (-a\right )^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x - a\right )}^{\frac {1}{3}} - \left (-a\right )^{\frac {1}{3}}\right ) + 3 \, {\left (4 \, a b x + 3 \, a^{2}\right )} {\left (b x - a\right )}^{\frac {2}{3}}}{18 \, a^{3} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(6*sqrt(1/3)*a*b^2*x^2*sqrt((-a)^(1/3)/a)*log((2*b*x + 3*sqrt(1/3)*(2*(b*x - a)^(2/3)*(-a)^(2/3) + (b*x
- a)^(1/3)*a + (-a)^(1/3)*a)*sqrt((-a)^(1/3)/a) - 3*(b*x - a)^(1/3)*(-a)^(2/3) - 3*a)/x) + 2*(-a)^(2/3)*b^2*x^
2*log((b*x - a)^(2/3) + (b*x - a)^(1/3)*(-a)^(1/3) + (-a)^(2/3)) - 4*(-a)^(2/3)*b^2*x^2*log((b*x - a)^(1/3) -
(-a)^(1/3)) + 3*(4*a*b*x + 3*a^2)*(b*x - a)^(2/3))/(a^3*x^2), 1/18*(12*sqrt(1/3)*a*b^2*x^2*sqrt(-(-a)^(1/3)/a)
*arctan(sqrt(1/3)*(2*(b*x - a)^(1/3) + (-a)^(1/3))*sqrt(-(-a)^(1/3)/a)) + 2*(-a)^(2/3)*b^2*x^2*log((b*x - a)^(
2/3) + (b*x - a)^(1/3)*(-a)^(1/3) + (-a)^(2/3)) - 4*(-a)^(2/3)*b^2*x^2*log((b*x - a)^(1/3) - (-a)^(1/3)) + 3*(
4*a*b*x + 3*a^2)*(b*x - a)^(2/3))/(a^3*x^2)]

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giac [A]  time = 2.24, size = 167, normalized size = 1.23 \begin {gather*} \frac {\frac {4 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x - a\right )}^{\frac {1}{3}} + \left (-a\right )^{\frac {1}{3}}\right )}}{3 \, \left (-a\right )^{\frac {1}{3}}}\right )}{\left (-a\right )^{\frac {1}{3}} a^{2}} - \frac {2 \, b^{3} \log \left ({\left (b x - a\right )}^{\frac {2}{3}} + {\left (b x - a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + \left (-a\right )^{\frac {2}{3}}\right )}{\left (-a\right )^{\frac {1}{3}} a^{2}} - \frac {4 \, \left (-a\right )^{\frac {2}{3}} b^{3} \log \left ({\left | {\left (b x - a\right )}^{\frac {1}{3}} - \left (-a\right )^{\frac {1}{3}} \right |}\right )}{a^{3}} + \frac {3 \, {\left (4 \, {\left (b x - a\right )}^{\frac {5}{3}} b^{3} + 7 \, {\left (b x - a\right )}^{\frac {2}{3}} a b^{3}\right )}}{a^{2} b^{2} x^{2}}}{18 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(4*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x - a)^(1/3) + (-a)^(1/3))/(-a)^(1/3))/((-a)^(1/3)*a^2) - 2*b^3*l
og((b*x - a)^(2/3) + (b*x - a)^(1/3)*(-a)^(1/3) + (-a)^(2/3))/((-a)^(1/3)*a^2) - 4*(-a)^(2/3)*b^3*log(abs((b*x
- a)^(1/3) - (-a)^(1/3)))/a^3 + 3*(4*(b*x - a)^(5/3)*b^3 + 7*(b*x - a)^(2/3)*a*b^3)/(a^2*b^2*x^2))/b

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maple [A]  time = 0.01, size = 128, normalized size = 0.94 \begin {gather*} \frac {2 \sqrt {3}\, b^{2} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x -a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}-1\right )}{3}\right )}{9 a^{\frac {7}{3}}}-\frac {2 b^{2} \ln \left (a^{\frac {1}{3}}+\left (b x -a \right )^{\frac {1}{3}}\right )}{9 a^{\frac {7}{3}}}+\frac {b^{2} \ln \left (a^{\frac {2}{3}}-\left (b x -a \right )^{\frac {1}{3}} a^{\frac {1}{3}}+\left (b x -a \right )^{\frac {2}{3}}\right )}{9 a^{\frac {7}{3}}}+\frac {2 \left (b x -a \right )^{\frac {2}{3}} b}{3 a^{2} x}+\frac {\left (b x -a \right )^{\frac {2}{3}}}{2 a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x-a)^(2/3)/a/x^2+2/3*b*(b*x-a)^(2/3)/a^2/x-2/9*b^2*ln(a^(1/3)+(b*x-a)^(1/3))/a^(7/3)+1/9*b^2/a^(7/3)*ln
(a^(2/3)-(b*x-a)^(1/3)*a^(1/3)+(b*x-a)^(2/3))+2/9*b^2/a^(7/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2*(b*x-a)^(1/3)/a^(1
/3)-1))

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maxima [A]  time = 3.09, size = 159, normalized size = 1.17 \begin {gather*} \frac {2 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x - a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {7}{3}}} + \frac {b^{2} \log \left ({\left (b x - a\right )}^{\frac {2}{3}} - {\left (b x - a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{9 \, a^{\frac {7}{3}}} - \frac {2 \, b^{2} \log \left ({\left (b x - a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}{9 \, a^{\frac {7}{3}}} + \frac {4 \, {\left (b x - a\right )}^{\frac {5}{3}} b^{2} + 7 \, {\left (b x - a\right )}^{\frac {2}{3}} a b^{2}}{6 \, {\left ({\left (b x - a\right )}^{2} a^{2} + 2 \, {\left (b x - a\right )} a^{3} + a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x - a)^(1/3) - a^(1/3))/a^(1/3))/a^(7/3) + 1/9*b^2*log((b*x - a)^(2/3
) - (b*x - a)^(1/3)*a^(1/3) + a^(2/3))/a^(7/3) - 2/9*b^2*log((b*x - a)^(1/3) + a^(1/3))/a^(7/3) + 1/6*(4*(b*x
- a)^(5/3)*b^2 + 7*(b*x - a)^(2/3)*a*b^2)/((b*x - a)^2*a^2 + 2*(b*x - a)*a^3 + a^4)

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mupad [B]  time = 0.22, size = 216, normalized size = 1.59 \begin {gather*} \frac {\frac {7\,b^2\,{\left (b\,x-a\right )}^{2/3}}{6\,a}+\frac {2\,b^2\,{\left (b\,x-a\right )}^{5/3}}{3\,a^2}}{{\left (a-b\,x\right )}^2-2\,a\,\left (a-b\,x\right )+a^2}-\frac {\ln \left (\frac {4\,b^4\,{\left (b\,x-a\right )}^{1/3}}{9\,a^4}-\frac {{\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}^2}{9\,{\left (-a\right )}^{11/3}}\right )\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{9\,{\left (-a\right )}^{7/3}}+\frac {2\,b^2\,\ln \left (\frac {4\,b^4\,{\left (b\,x-a\right )}^{1/3}}{9\,a^4}-\frac {4\,b^4}{9\,{\left (-a\right )}^{11/3}}\right )}{9\,{\left (-a\right )}^{7/3}}+\frac {b^2\,\ln \left (\frac {4\,b^4\,{\left (b\,x-a\right )}^{1/3}}{9\,a^4}-\frac {9\,b^4\,{\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}^2}{{\left (-a\right )}^{11/3}}\right )\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}{{\left (-a\right )}^{7/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(b*x - a)^(1/3)),x)

[Out]

((7*b^2*(b*x - a)^(2/3))/(6*a) + (2*b^2*(b*x - a)^(5/3))/(3*a^2))/((a - b*x)^2 - 2*a*(a - b*x) + a^2) - (log((
4*b^4*(b*x - a)^(1/3))/(9*a^4) - (3^(1/2)*b^2*1i + b^2)^2/(9*(-a)^(11/3)))*(3^(1/2)*b^2*1i + b^2))/(9*(-a)^(7/
3)) + (2*b^2*log((4*b^4*(b*x - a)^(1/3))/(9*a^4) - (4*b^4)/(9*(-a)^(11/3))))/(9*(-a)^(7/3)) + (b^2*log((4*b^4*
(b*x - a)^(1/3))/(9*a^4) - (9*b^4*((3^(1/2)*1i)/9 - 1/9)^2)/(-a)^(11/3))*((3^(1/2)*1i)/9 - 1/9))/(-a)^(7/3)

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sympy [C]  time = 2.62, size = 2744, normalized size = 20.18

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x-a)**(1/3),x)

[Out]

-4*a**(14/3)*b**(10/3)*(-a/b + x)**(4/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(I*pi/3)/a**(1/3))*gamma(
2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I
*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b +
x)**(13/3)*exp(2*I*pi/3)*gamma(5/3)) - 4*a**(14/3)*b**(10/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*
(-a/b + x)**(1/3)*exp_polar(I*pi)/a**(1/3))*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma
(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp
(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3)) - 4*a**(14/3)*b**(10/3)
*(-a/b + x)**(4/3)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(5*I*pi/3)/a**(1/3))*gamma(2/3)/
(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3
)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(
13/3)*exp(2*I*pi/3)*gamma(5/3)) - 12*a**(11/3)*b**(13/3)*(-a/b + x)**(7/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*
exp_polar(I*pi/3)/a**(1/3))*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*
b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma
(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3)) - 12*a**(11/3)*b**(13/3)*(-a/b + x)**(7
/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(I*pi)/a**(1/3))*gamma(2/3)/(27*a**7*b**(4/3)*(-
a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a*
*5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*
gamma(5/3)) - 12*a**(11/3)*b**(13/3)*(-a/b + x)**(7/3)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_p
olar(5*I*pi/3)/a**(1/3))*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**
(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/
3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3)) - 12*a**(8/3)*b**(16/3)*(-a/b + x)**(10/3)
*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(I*pi/3)/a**(1/3))*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)
*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-
a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3)) - 1
2*a**(8/3)*b**(16/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(I*pi)/a**(1
/3))*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7
/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/
3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3)) - 12*a**(8/3)*b**(16/3)*(-a/b + x)**(10/3)*exp(-2*I*pi/3)*log(
1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(5*I*pi/3)/a**(1/3))*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*ex
p(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b
+ x)**(10/3)*exp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3)) - 4*a*
*(5/3)*b**(19/3)*(-a/b + x)**(13/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(I*pi/3)/a**(1/3))*gamma(2/3)/
(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3
)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(
13/3)*exp(2*I*pi/3)*gamma(5/3)) - 4*a**(5/3)*b**(19/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(-a/b
+ x)**(1/3)*exp_polar(I*pi)/a**(1/3))*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3)
+ 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*
pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3)) - 4*a**(5/3)*b**(19/3)*(-a/b
+ x)**(13/3)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(5*I*pi/3)/a**(1/3))*gamma(2/3)/(27*a
**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gam
ma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)
*exp(2*I*pi/3)*gamma(5/3)) + 21*a**4*b**4*(-a/b + x)**2*exp(2*I*pi/3)*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)*
*(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10
/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3
)) + 33*a**3*b**5*(-a/b + x)**3*exp(2*I*pi/3)*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gam
ma(5/3) + 81*a**6*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*e
xp(2*I*pi/3)*gamma(5/3) + 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3)) + 12*a**2*b**6*(-a/b
+ x)**4*exp(2*I*pi/3)*gamma(2/3)/(27*a**7*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**6*b**(7/
3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3) + 81*a**5*b**(10/3)*(-a/b + x)**(10/3)*exp(2*I*pi/3)*gamma(5/3)
+ 27*a**4*b**(13/3)*(-a/b + x)**(13/3)*exp(2*I*pi/3)*gamma(5/3))

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