3.5.12 \(\int \frac {1}{x^3 (a+b x)^{2/3}} \, dx\)

Optimal. Leaf size=130 \[ -\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{8/3}}-\frac {5 b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3}}+\frac {5 b \sqrt [3]{a+b x}}{6 a^2 x}-\frac {\sqrt [3]{a+b x}}{2 a x^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {51, 57, 617, 204, 31} \begin {gather*} -\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{8/3}}-\frac {5 b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3}}+\frac {5 b \sqrt [3]{a+b x}}{6 a^2 x}-\frac {\sqrt [3]{a+b x}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x)^(2/3)),x]

[Out]

-(a + b*x)^(1/3)/(2*a*x^2) + (5*b*(a + b*x)^(1/3))/(6*a^2*x) - (5*b^2*ArcTan[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sq
rt[3]*a^(1/3))])/(3*Sqrt[3]*a^(8/3)) - (5*b^2*Log[x])/(18*a^(8/3)) + (5*b^2*Log[a^(1/3) - (a + b*x)^(1/3)])/(6
*a^(8/3))

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 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[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]
&& PosQ[(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 (a+b x)^{2/3}} \, dx &=-\frac {\sqrt [3]{a+b x}}{2 a x^2}-\frac {(5 b) \int \frac {1}{x^2 (a+b x)^{2/3}} \, dx}{6 a}\\ &=-\frac {\sqrt [3]{a+b x}}{2 a x^2}+\frac {5 b \sqrt [3]{a+b x}}{6 a^2 x}+\frac {\left (5 b^2\right ) \int \frac {1}{x (a+b x)^{2/3}} \, dx}{9 a^2}\\ &=-\frac {\sqrt [3]{a+b x}}{2 a x^2}+\frac {5 b \sqrt [3]{a+b x}}{6 a^2 x}-\frac {5 b^2 \log (x)}{18 a^{8/3}}-\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x}\right )}{6 a^{8/3}}-\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x}\right )}{6 a^{7/3}}\\ &=-\frac {\sqrt [3]{a+b x}}{2 a x^2}+\frac {5 b \sqrt [3]{a+b x}}{6 a^2 x}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{8/3}}+\frac {\left (5 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^{8/3}}\\ &=-\frac {\sqrt [3]{a+b x}}{2 a x^2}+\frac {5 b \sqrt [3]{a+b x}}{6 a^2 x}-\frac {5 b^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{8/3}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{8/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 33, normalized size = 0.25 \begin {gather*} -\frac {3 b^2 \sqrt [3]{a+b x} \, _2F_1\left (\frac {1}{3},3;\frac {4}{3};\frac {b x}{a}+1\right )}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x)^(2/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 149, normalized size = 1.15 \begin {gather*} \frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{9 a^{8/3}}-\frac {5 b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{18 a^{8/3}}-\frac {5 b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} a^{8/3}}-\frac {\sqrt [3]{a+b x} (8 a-5 (a+b x))}{6 a^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((a + b*x)^(1/3)*(8*a - 5*(a + b*x)))/(a^2*x^2) - (5*b^2*ArcTan[1/Sqrt[3] + (2*(a + b*x)^(1/3))/(Sqrt[3]*
a^(1/3))])/(3*Sqrt[3]*a^(8/3)) + (5*b^2*Log[a^(1/3) - (a + b*x)^(1/3)])/(9*a^(8/3)) - (5*b^2*Log[a^(2/3) + a^(
1/3)*(a + b*x)^(1/3) + (a + b*x)^(2/3)])/(18*a^(8/3))

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fricas [A]  time = 0.82, size = 162, normalized size = 1.25 \begin {gather*} -\frac {10 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} a b^{2} x^{2} \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}}\right )}}{3 \, a^{2}}\right ) + 5 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (a^{2}\right )}^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}}\right ) - 10 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 3 \, {\left (5 \, a^{2} b x - 3 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {1}{3}}}{18 \, a^{4} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(10*sqrt(3)*(a^2)^(1/6)*a*b^2*x^2*arctan(1/3*(a^2)^(1/6)*(sqrt(3)*(a^2)^(1/3)*a + 2*sqrt(3)*(a^2)^(2/3)*
(b*x + a)^(1/3))/a^2) + 5*(a^2)^(2/3)*b^2*x^2*log((b*x + a)^(2/3)*a + (a^2)^(1/3)*a + (a^2)^(2/3)*(b*x + a)^(1
/3)) - 10*(a^2)^(2/3)*b^2*x^2*log((b*x + a)^(1/3)*a - (a^2)^(2/3)) - 3*(5*a^2*b*x - 3*a^3)*(b*x + a)^(1/3))/(a
^4*x^2)

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giac [A]  time = 2.05, size = 130, normalized size = 1.00 \begin {gather*} -\frac {\frac {10 \, \sqrt {3} b^{3} \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 )}{a^{\frac {8}{3}}} + \frac {5 \, b^{3} \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 )}{a^{\frac {8}{3}}} - \frac {10 \, b^{3} \log \left ({\left | {\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {8}{3}}} - \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {4}{3}} b^{3} - 8 \, {\left (b x + a\right )}^{\frac {1}{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)^(2/3),x, algorithm="giac")

[Out]

-1/18*(10*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3))/a^(8/3) + 5*b^3*log((b*x + a)^
(2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^(8/3) - 10*b^3*log(abs((b*x + a)^(1/3) - a^(1/3)))/a^(8/3) - 3*(5
*(b*x + a)^(4/3)*b^3 - 8*(b*x + a)^(1/3)*a*b^3)/(a^2*b^2*x^2))/b

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maple [A]  time = 0.01, size = 117, normalized size = 0.90 \begin {gather*} -\frac {5 \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 {8}{3}}}+\frac {5 b^{2} \ln \left (-a^{\frac {1}{3}}+\left (b x +a \right )^{\frac {1}{3}}\right )}{9 a^{\frac {8}{3}}}-\frac {5 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 )}{18 a^{\frac {8}{3}}}+\frac {5 \left (b x +a \right )^{\frac {1}{3}} b}{6 a^{2} x}-\frac {\left (b x +a \right )^{\frac {1}{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)^(2/3),x)

[Out]

-1/2*(b*x+a)^(1/3)/a/x^2+5/6*b*(b*x+a)^(1/3)/a^2/x+5/9*b^2/a^(8/3)*ln(-a^(1/3)+(b*x+a)^(1/3))-5/18*b^2/a^(8/3)
*ln(a^(2/3)+(b*x+a)^(1/3)*a^(1/3)+(b*x+a)^(2/3))-5/9*b^2/a^(8/3)*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 = 3.04, size = 142, normalized size = 1.09 \begin {gather*} -\frac {5 \, \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 {8}{3}}} - \frac {5 \, 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 )}{18 \, a^{\frac {8}{3}}} + \frac {5 \, b^{2} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {8}{3}}} + \frac {5 \, {\left (b x + a\right )}^{\frac {4}{3}} b^{2} - 8 \, {\left (b x + a\right )}^{\frac {1}{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)^(2/3),x, algorithm="maxima")

[Out]

-5/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3))/a^(8/3) - 5/18*b^2*log((b*x + a)^(2
/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^(8/3) + 5/9*b^2*log((b*x + a)^(1/3) - a^(1/3))/a^(8/3) + 1/6*(5*(b*
x + a)^(4/3)*b^2 - 8*(b*x + a)^(1/3)*a*b^2)/((b*x + a)^2*a^2 - 2*(b*x + a)*a^3 + a^4)

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mupad [B]  time = 0.13, size = 175, normalized size = 1.35 \begin {gather*} \frac {5\,b^2\,\ln \left ({\left (a+b\,x\right )}^{1/3}-a^{1/3}\right )}{9\,a^{8/3}}-\frac {\frac {4\,b^2\,{\left (a+b\,x\right )}^{1/3}}{3\,a}-\frac {5\,b^2\,{\left (a+b\,x\right )}^{4/3}}{6\,a^2}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2}+\frac {5\,b^2\,\ln \left (\frac {5\,b^2\,{\left (a+b\,x\right )}^{1/3}}{a^2}-\frac {5\,b^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{5/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}}-\frac {5\,b^2\,\ln \left (\frac {5\,b^2\,{\left (a+b\,x\right )}^{1/3}}{a^2}+\frac {5\,b^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{5/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x)^(2/3)),x)

[Out]

(5*b^2*log((a + b*x)^(1/3) - a^(1/3)))/(9*a^(8/3)) - ((4*b^2*(a + b*x)^(1/3))/(3*a) - (5*b^2*(a + b*x)^(4/3))/
(6*a^2))/((a + b*x)^2 - 2*a*(a + b*x) + a^2) + (5*b^2*log((5*b^2*(a + b*x)^(1/3))/a^2 - (5*b^2*((3^(1/2)*1i)/2
- 1/2))/a^(5/3))*((3^(1/2)*1i)/2 - 1/2))/(9*a^(8/3)) - (5*b^2*log((5*b^2*(a + b*x)^(1/3))/a^2 + (5*b^2*((3^(1
/2)*1i)/2 + 1/2))/a^(5/3))*((3^(1/2)*1i)/2 + 1/2))/(9*a^(8/3))

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sympy [C]  time = 2.73, size = 2728, normalized size = 20.98

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*a**(13/3)*b**(8/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(1/3)/(5
4*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*g
amma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3)*
exp(2*I*pi/3)*gamma(4/3)) + 10*a**(13/3)*b**(8/3)*(a/b + x)**(2/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar
(2*I*pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3
)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54
*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) + 10*a**(13/3)*b**(8/3)*(a/b + x)**(2/3)*exp(-2*I*
pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**
(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b**(8/
3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) -
30*a**(10/3)*b**(11/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(1/3)/
(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)
*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3
)*exp(2*I*pi/3)*gamma(4/3)) - 30*a**(10/3)*b**(11/3)*(a/b + x)**(5/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_po
lar(2*I*pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(
5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) -
54*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) - 30*a**(10/3)*b**(11/3)*(a/b + x)**(5/3)*exp(-
2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b +
x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b*
*(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3
)) + 30*a**(7/3)*b**(14/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(1/
3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi
/3)*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(1
1/3)*exp(2*I*pi/3)*gamma(4/3)) + 30*a**(7/3)*b**(14/3)*(a/b + x)**(8/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_
polar(2*I*pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b*
*(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3)
- 54*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) + 30*a**(7/3)*b**(14/3)*(a/b + x)**(8/3)*exp(
-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b +
x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b
**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/
3)) - 10*a**(4/3)*b**(17/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(
1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*
pi/3)*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**
(11/3)*exp(2*I*pi/3)*gamma(4/3)) - 10*a**(4/3)*b**(17/3)*(a/b + x)**(11/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*e
xp_polar(2*I*pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6
*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4
/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) - 10*a**(4/3)*b**(17/3)*(a/b + x)**(11/3)*
exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a
/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a*
*5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamm
a(4/3)) - 24*a**4*b**3*(a/b + x)*exp(2*I*pi/3)*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gam
ma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp
(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) + 39*a**3*b**4*(a/b + x)
**2*exp(2*I*pi/3)*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(
a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a*
*4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) - 15*a**2*b**5*(a/b + x)**3*exp(2*I*pi/3)*gamma(1/3)/
(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)
*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3
)*exp(2*I*pi/3)*gamma(4/3))

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