3.4.90 \(\int \frac {(a+b x)^{4/3}}{x^2} \, dx\)

Optimal. Leaf size=107 \[ -\frac {(a+b x)^{4/3}}{x}+4 b \sqrt [3]{a+b x}-\frac {2}{3} \sqrt [3]{a} b \log (x)+2 \sqrt [3]{a} b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )-\frac {4 \sqrt [3]{a} b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {47, 50, 57, 617, 204, 31} \begin {gather*} -\frac {(a+b x)^{4/3}}{x}+4 b \sqrt [3]{a+b x}-\frac {2}{3} \sqrt [3]{a} b \log (x)+2 \sqrt [3]{a} b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )-\frac {4 \sqrt [3]{a} b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(4/3)/x^2,x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 33, normalized size = 0.31 \begin {gather*} \frac {3 b (a+b x)^{7/3} \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {b x}{a}+1\right )}{7 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(4/3)/x^2,x]

[Out]

(3*b*(a + b*x)^(7/3)*Hypergeometric2F1[2, 7/3, 10/3, 1 + (b*x)/a])/(7*a^2)

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IntegrateAlgebraic [A]  time = 0.19, size = 135, normalized size = 1.26 \begin {gather*} -\frac {2}{3} \sqrt [3]{a} b \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )+\frac {\sqrt [3]{a+b x} (3 (a+b x)-4 a)}{x}+\frac {4}{3} \sqrt [3]{a} b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )-\frac {4 \sqrt [3]{a} b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(4/3)/x^2,x]

[Out]

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

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fricas [A]  time = 0.72, size = 111, normalized size = 1.04 \begin {gather*} -\frac {4 \, \sqrt {3} a^{\frac {1}{3}} b x \arctan \left (\frac {2 \, \sqrt {3} {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) + 2 \, a^{\frac {1}{3}} b x \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 {1}{3}} b x \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - 3 \, {\left (3 \, b x - a\right )} {\left (b x + a\right )}^{\frac {1}{3}}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(4/3)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.35, size = 119, normalized size = 1.11 \begin {gather*} -\frac {4 \, \sqrt {3} a^{\frac {1}{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 ) + 2 \, a^{\frac {1}{3}} 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 ) - 4 \, a^{\frac {1}{3}} b^{2} \log \left ({\left | {\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right ) - 9 \, {\left (b x + a\right )}^{\frac {1}{3}} b^{2} + \frac {3 \, {\left (b x + a\right )}^{\frac {1}{3}} a b}{x}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(4/3)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 103, normalized size = 0.96 \begin {gather*} -\frac {4 \sqrt {3}\, a^{\frac {1}{3}} b \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{3}+\frac {4 a^{\frac {1}{3}} b \ln \left (-a^{\frac {1}{3}}+\left (b x +a \right )^{\frac {1}{3}}\right )}{3}-\frac {2 a^{\frac {1}{3}} b \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 )}{3}+3 \left (b x +a \right )^{\frac {1}{3}} b -\frac {\left (b x +a \right )^{\frac {1}{3}} a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(4/3)/x^2,x)

[Out]

3*b*(b*x+a)^(1/3)-a*(b*x+a)^(1/3)/x+4/3*b*a^(1/3)*ln(-a^(1/3)+(b*x+a)^(1/3))-2/3*b*a^(1/3)*ln(a^(2/3)+(b*x+a)^
(1/3)*a^(1/3)+(b*x+a)^(2/3))-4/3*b*a^(1/3)*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 = 104, normalized size = 0.97 \begin {gather*} -\frac {4}{3} \, \sqrt {3} a^{\frac {1}{3}} b \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 ) - \frac {2}{3} \, a^{\frac {1}{3}} b \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 ) + \frac {4}{3} \, a^{\frac {1}{3}} b \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + 3 \, {\left (b x + a\right )}^{\frac {1}{3}} b - \frac {{\left (b x + a\right )}^{\frac {1}{3}} a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(4/3)/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 131, normalized size = 1.22 \begin {gather*} 3\,b\,{\left (a+b\,x\right )}^{1/3}+\frac {4\,a^{1/3}\,b\,\ln \left (12\,a^{4/3}\,b-12\,a\,b\,{\left (a+b\,x\right )}^{1/3}\right )}{3}-\frac {a\,{\left (a+b\,x\right )}^{1/3}}{x}+\frac {2\,a^{1/3}\,b\,\ln \left (12\,a\,b\,{\left (a+b\,x\right )}^{1/3}-6\,a^{4/3}\,b\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{3}-\frac {2\,a^{1/3}\,b\,\ln \left (12\,a\,b\,{\left (a+b\,x\right )}^{1/3}+6\,a^{4/3}\,b\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(4/3)/x^2,x)

[Out]

3*b*(a + b*x)^(1/3) + (4*a^(1/3)*b*log(12*a^(4/3)*b - 12*a*b*(a + b*x)^(1/3)))/3 - (a*(a + b*x)^(1/3))/x + (2*
a^(1/3)*b*log(12*a*b*(a + b*x)^(1/3) - 6*a^(4/3)*b*(3^(1/2)*1i - 1))*(3^(1/2)*1i - 1))/3 - (2*a^(1/3)*b*log(12
*a*b*(a + b*x)^(1/3) + 6*a^(4/3)*b*(3^(1/2)*1i + 1))*(3^(1/2)*1i + 1))/3

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sympy [C]  time = 2.58, size = 719, normalized size = 6.72 \begin {gather*} \frac {28 a^{\frac {10}{3}} b e^{\frac {2 i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} \sqrt [3]{\frac {a}{b} + x}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac {7}{3}\right )}{9 a^{3} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right ) - 9 a^{2} b \left (\frac {a}{b} + x\right ) e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right )} + \frac {28 a^{\frac {10}{3}} b \log {\left (1 - \frac {\sqrt [3]{b} \sqrt [3]{\frac {a}{b} + x} e^{\frac {2 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac {7}{3}\right )}{9 a^{3} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right ) - 9 a^{2} b \left (\frac {a}{b} + x\right ) e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right )} + \frac {28 a^{\frac {10}{3}} b e^{- \frac {2 i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} \sqrt [3]{\frac {a}{b} + x} e^{\frac {4 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac {7}{3}\right )}{9 a^{3} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right ) - 9 a^{2} b \left (\frac {a}{b} + x\right ) e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right )} - \frac {28 a^{\frac {7}{3}} b^{2} \left (\frac {a}{b} + x\right ) e^{\frac {2 i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} \sqrt [3]{\frac {a}{b} + x}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac {7}{3}\right )}{9 a^{3} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right ) - 9 a^{2} b \left (\frac {a}{b} + x\right ) e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right )} - \frac {28 a^{\frac {7}{3}} b^{2} \left (\frac {a}{b} + x\right ) \log {\left (1 - \frac {\sqrt [3]{b} \sqrt [3]{\frac {a}{b} + x} e^{\frac {2 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac {7}{3}\right )}{9 a^{3} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right ) - 9 a^{2} b \left (\frac {a}{b} + x\right ) e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right )} - \frac {28 a^{\frac {7}{3}} b^{2} \left (\frac {a}{b} + x\right ) e^{- \frac {2 i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} \sqrt [3]{\frac {a}{b} + x} e^{\frac {4 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac {7}{3}\right )}{9 a^{3} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right ) - 9 a^{2} b \left (\frac {a}{b} + x\right ) e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right )} + \frac {84 a^{3} b^{\frac {4}{3}} \sqrt [3]{\frac {a}{b} + x} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {7}{3}\right )}{9 a^{3} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right ) - 9 a^{2} b \left (\frac {a}{b} + x\right ) e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right )} - \frac {63 a^{2} b^{\frac {7}{3}} \left (\frac {a}{b} + x\right )^{\frac {4}{3}} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {7}{3}\right )}{9 a^{3} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right ) - 9 a^{2} b \left (\frac {a}{b} + x\right ) e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {10}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(4/3)/x**2,x)

[Out]

28*a**(10/3)*b*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(7/3)/(9*a**3*exp(2*I*pi/3)*gamm
a(10/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3)) + 28*a**(10/3)*b*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp
_polar(2*I*pi/3)/a**(1/3))*gamma(7/3)/(9*a**3*exp(2*I*pi/3)*gamma(10/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gam
ma(10/3)) + 28*a**(10/3)*b*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamm
a(7/3)/(9*a**3*exp(2*I*pi/3)*gamma(10/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3)) - 28*a**(7/3)*b**2*(a
/b + x)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(7/3)/(9*a**3*exp(2*I*pi/3)*gamma(10/3)
- 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3)) - 28*a**(7/3)*b**2*(a/b + x)*log(1 - b**(1/3)*(a/b + x)**(1/3
)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(7/3)/(9*a**3*exp(2*I*pi/3)*gamma(10/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3
)*gamma(10/3)) - 28*a**(7/3)*b**2*(a/b + x)*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(7/3)/(9*a**3*exp(2*I*pi/3)*gamma(10/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3)) + 84
*a**3*b**(4/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(7/3)/(9*a**3*exp(2*I*pi/3)*gamma(10/3) - 9*a**2*b*(a/b + x
)*exp(2*I*pi/3)*gamma(10/3)) - 63*a**2*b**(7/3)*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(7/3)/(9*a**3*exp(2*I*pi/3
)*gamma(10/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(10/3))

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