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

Optimal. Leaf size=130 \[ -\frac {b^2 \log (x)}{9 a^{7/3}}+\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{7/3}}+\frac {2 b^2 \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3}}+\frac {2 b (a+b x)^{2/3}}{3 a^2 x}-\frac {(a+b x)^{2/3}}{2 a x^2} \]

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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, 55, 617, 204, 31} \begin {gather*} -\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 {2 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^{7/3}}+\frac {2 b (a+b x)^{2/3}}{3 a^2 x}-\frac {(a+b x)^{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))/(Sq
rt[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 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[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] && 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 \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 = 35, normalized size = 0.27 \begin {gather*} -\frac {3 b^2 (a+b x)^{2/3} \, _2F_1\left (\frac {2}{3},3;\frac {5}{3};\frac {b x}{a}+1\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 = 0.15, size = 149, normalized size = 1.15 \begin {gather*} \frac {2 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{9 a^{7/3}}-\frac {b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{9 a^{7/3}}+\frac {2 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^{7/3}}-\frac {(a+b x)^{2/3} (7 a-4 (a+b x))}{6 a^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((a + b*x)^(2/3)*(7*a - 4*(a + b*x)))/(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 = 0.96, size = 296, normalized size = 2.28 \begin {gather*} \left [\frac {6 \, \sqrt {\frac {1}{3}} a b^{2} x^{2} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} \log \left (\frac {2 \, b x + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {2}{3}} a^{\frac {2}{3}} - {\left (b x + a\right )}^{\frac {1}{3}} a - a^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} - 3 \, {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + 3 \, a}{x}\right ) - 2 \, a^{\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}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 4 \, a^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\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^{\frac {2}{3}} b^{2} x^{2} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{a^{\frac {1}{3}}}\right ) - 2 \, a^{\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}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 4 \, a^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\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(-1/a^(2/3))*log((2*b*x + 3*sqrt(1/3)*(2*(b*x + a)^(2/3)*a^(2/3) - (b*x + a)^
(1/3)*a - a^(4/3))*sqrt(-1/a^(2/3)) - 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^(2/3)*b^2*x^2*arctan(sqrt(1/3)*(2*(b*x + a)^(1/3) + a^(1/
3))/a^(1/3)) - 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 = 130, normalized size = 1.00 \begin {gather*} \frac {\frac {4 \, \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 {7}{3}}} - \frac {2 \, 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 {7}{3}}} + \frac {4 \, b^{3} \log \left ({\left | {\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {7}{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^(7/3) - 2*b^3*log((b*x + a)^(2
/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^(7/3) + 4*b^3*log(abs((b*x + a)^(1/3) - a^(1/3)))/a^(7/3) + 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 = 117, normalized size = 0.90 \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/a^(7/3)*ln(-a^(1/3)+(b*x+a)^(1/3))-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 = 3.03, size = 142, normalized size = 1.09 \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 = 0.23, size = 182, normalized size = 1.40 \begin {gather*} \frac {2\,b^2\,\ln \left ({\left (a+b\,x\right )}^{1/3}-a^{1/3}\right )}{9\,a^{7/3}}-\frac {\frac {7\,b^2\,{\left (a+b\,x\right )}^{2/3}}{6\,a}-\frac {2\,b^2\,{\left (a+b\,x\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 (a+b\,x\right )}^{1/3}}{9\,a^4}-\frac {{\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}^2}{9\,a^{11/3}}\right )\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{9\,a^{7/3}}+\frac {b^2\,\ln \left (\frac {4\,b^4\,{\left (a+b\,x\right )}^{1/3}}{9\,a^4}-\frac {9\,b^4\,{\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}^2}{a^{11/3}}\right )\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}{a^{7/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x)^(1/3)),x)

[Out]

(2*b^2*log((a + b*x)^(1/3) - a^(1/3)))/(9*a^(7/3)) - ((7*b^2*(a + b*x)^(2/3))/(6*a) - (2*b^2*(a + b*x)^(5/3))/
(3*a^2))/((a + b*x)^2 - 2*a*(a + b*x) + a^2) - (log((4*b^4*(a + b*x)^(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)) + (b^2*log((4*b^4*(a + b*x)^(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 = 2.59, size = 2730, normalized size = 21.00

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)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(2/3)/(2
7*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)*ga
mma(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(2*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) - 8
1*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)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*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)/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(2*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(4*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)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/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)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b
+ x)**(1/3)*exp_polar(2*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(4*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)/a**(1/3))*gamm
a(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(2*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)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*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)*gamm
a(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)*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**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)**(1
3/3)*exp(2*I*pi/3)*gamma(5/3))

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