3.5.18 \(\int \frac {1}{x^2 (a+b x)^{4/3}} \, dx\)

Optimal. Leaf size=113 \[ \frac {2 b \log (x)}{3 a^{7/3}}-\frac {2 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{7/3}}-\frac {4 b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{7/3}}-\frac {4 b}{a^2 \sqrt [3]{a+b x}}-\frac {1}{a x \sqrt [3]{a+b x}} \]

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Rubi [A]  time = 0.04, antiderivative size = 115, normalized size of antiderivative = 1.02, 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 {4 (a+b x)^{2/3}}{a^2 x}+\frac {2 b \log (x)}{3 a^{7/3}}-\frac {2 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{7/3}}-\frac {4 b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{7/3}}+\frac {3}{a x \sqrt [3]{a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x)^(4/3)),x]

[Out]

3/(a*x*(a + b*x)^(1/3)) - (4*(a + b*x)^(2/3))/(a^2*x) - (4*b*ArcTan[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sqrt[3]*a^(
1/3))])/(Sqrt[3]*a^(7/3)) + (2*b*Log[x])/(3*a^(7/3)) - (2*b*Log[a^(1/3) - (a + b*x)^(1/3)])/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^2 (a+b x)^{4/3}} \, dx &=\frac {3}{a x \sqrt [3]{a+b x}}+\frac {4 \int \frac {1}{x^2 \sqrt [3]{a+b x}} \, dx}{a}\\ &=\frac {3}{a x \sqrt [3]{a+b x}}-\frac {4 (a+b x)^{2/3}}{a^2 x}-\frac {(4 b) \int \frac {1}{x \sqrt [3]{a+b x}} \, dx}{3 a^2}\\ &=\frac {3}{a x \sqrt [3]{a+b x}}-\frac {4 (a+b x)^{2/3}}{a^2 x}+\frac {2 b \log (x)}{3 a^{7/3}}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x}\right )}{a^{7/3}}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x}\right )}{a^2}\\ &=\frac {3}{a x \sqrt [3]{a+b x}}-\frac {4 (a+b x)^{2/3}}{a^2 x}+\frac {2 b \log (x)}{3 a^{7/3}}-\frac {2 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{7/3}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}\right )}{a^{7/3}}\\ &=\frac {3}{a x \sqrt [3]{a+b x}}-\frac {4 (a+b x)^{2/3}}{a^2 x}-\frac {4 b \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{7/3}}+\frac {2 b \log (x)}{3 a^{7/3}}-\frac {2 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{7/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 31, normalized size = 0.27 \begin {gather*} -\frac {3 b \, _2F_1\left (-\frac {1}{3},2;\frac {2}{3};\frac {b x}{a}+1\right )}{a^2 \sqrt [3]{a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x)^(4/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.17, size = 138, normalized size = 1.22 \begin {gather*} -\frac {4 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{7/3}}+\frac {2 b \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{3 a^{7/3}}-\frac {4 b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} a^{7/3}}+\frac {3 a-4 (a+b x)}{a^2 x \sqrt [3]{a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.83, size = 407, normalized size = 3.60 \begin {gather*} \left [\frac {6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{2} + a^{2} b x\right )} \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 (b^{2} x^{2} + a b x\right )} \left (-a\right )^{\frac {2}{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 ) - 4 \, {\left (b^{2} x^{2} + a b x\right )} \left (-a\right )^{\frac {2}{3}} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} + \left (-a\right )^{\frac {1}{3}}\right ) - 3 \, {\left (4 \, a b x + a^{2}\right )} {\left (b x + a\right )}^{\frac {2}{3}}}{3 \, {\left (a^{3} b x^{2} + a^{4} x\right )}}, -\frac {12 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{2} + a^{2} b x\right )} \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 (b^{2} x^{2} + a b x\right )} \left (-a\right )^{\frac {2}{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 ) + 4 \, {\left (b^{2} x^{2} + a b x\right )} \left (-a\right )^{\frac {2}{3}} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} + \left (-a\right )^{\frac {1}{3}}\right ) + 3 \, {\left (4 \, a b x + a^{2}\right )} {\left (b x + a\right )}^{\frac {2}{3}}}{3 \, {\left (a^{3} b x^{2} + a^{4} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x+a)^(4/3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.40, size = 120, normalized size = 1.06 \begin {gather*} -\frac {4 \, \sqrt {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 )}{3 \, a^{\frac {7}{3}}} + \frac {2 \, 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 )}{3 \, a^{\frac {7}{3}}} - \frac {4 \, b \log \left ({\left | {\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{3 \, a^{\frac {7}{3}}} - \frac {4 \, {\left (b x + a\right )} b - 3 \, a b}{{\left ({\left (b x + a\right )}^{\frac {4}{3}} - {\left (b x + a\right )}^{\frac {1}{3}} a\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x+a)^(4/3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 108, normalized size = 0.96 \begin {gather*} -\frac {4 \sqrt {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 a^{\frac {7}{3}}}-\frac {4 b \ln \left (-a^{\frac {1}{3}}+\left (b x +a \right )^{\frac {1}{3}}\right )}{3 a^{\frac {7}{3}}}+\frac {2 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 a^{\frac {7}{3}}}-\frac {3 b}{\left (b x +a \right )^{\frac {1}{3}} a^{2}}-\frac {\left (b x +a \right )^{\frac {2}{3}}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x+a)^(4/3),x)

[Out]

-3*b/a^2/(b*x+a)^(1/3)-1/a^2*(b*x+a)^(2/3)/x-4/3*b/a^(7/3)*ln(-a^(1/3)+(b*x+a)^(1/3))+2/3*b/a^(7/3)*ln(a^(2/3)
+(b*x+a)^(1/3)*a^(1/3)+(b*x+a)^(2/3))-4/3*b/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 = 2.99, size = 122, normalized size = 1.08 \begin {gather*} -\frac {4 \, \sqrt {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 )}{3 \, a^{\frac {7}{3}}} - \frac {4 \, {\left (b x + a\right )} b - 3 \, a b}{{\left (b x + a\right )}^{\frac {4}{3}} a^{2} - {\left (b x + a\right )}^{\frac {1}{3}} a^{3}} + \frac {2 \, 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 )}{3 \, a^{\frac {7}{3}}} - \frac {4 \, b \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{3 \, a^{\frac {7}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x+a)^(4/3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 173, normalized size = 1.53 \begin {gather*} -\frac {\frac {3\,b}{a}-\frac {4\,b\,\left (a+b\,x\right )}{a^2}}{a\,{\left (a+b\,x\right )}^{1/3}-{\left (a+b\,x\right )}^{4/3}}+\frac {\ln \left (a^{7/3}\,{\left (2\,b-\sqrt {3}\,b\,2{}\mathrm {i}\right )}^2-16\,a^2\,b^2\,{\left (a+b\,x\right )}^{1/3}\right )\,\left (2\,b-\sqrt {3}\,b\,2{}\mathrm {i}\right )}{3\,a^{7/3}}+\frac {\ln \left (a^{7/3}\,{\left (2\,b+\sqrt {3}\,b\,2{}\mathrm {i}\right )}^2-16\,a^2\,b^2\,{\left (a+b\,x\right )}^{1/3}\right )\,\left (2\,b+\sqrt {3}\,b\,2{}\mathrm {i}\right )}{3\,a^{7/3}}-\frac {4\,b\,\ln \left (16\,a^{7/3}\,b^2-16\,a^2\,b^2\,{\left (a+b\,x\right )}^{1/3}\right )}{3\,a^{7/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x)^(4/3)),x)

[Out]

(log(a^(7/3)*(2*b - 3^(1/2)*b*2i)^2 - 16*a^2*b^2*(a + b*x)^(1/3))*(2*b - 3^(1/2)*b*2i))/(3*a^(7/3)) - ((3*b)/a
- (4*b*(a + b*x))/a^2)/(a*(a + b*x)^(1/3) - (a + b*x)^(4/3)) + (log(a^(7/3)*(2*b + 3^(1/2)*b*2i)^2 - 16*a^2*b
^2*(a + b*x)^(1/3))*(2*b + 3^(1/2)*b*2i))/(3*a^(7/3)) - (4*b*log(16*a^(7/3)*b^2 - 16*a^2*b^2*(a + b*x)^(1/3)))
/(3*a^(7/3))

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sympy [C]  time = 2.52, size = 857, normalized size = 7.58

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x+a)**(4/3),x)

[Out]

-9*a**(4/3)*b**(2/3)*exp(2*I*pi/3)*gamma(-1/3)/(-9*a**(10/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 9*a**
(7/3)*b*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(2/3)) + 12*a**(1/3)*b**(5/3)*(a/b + x)*exp(2*I*pi/3)*gamma(-1/3)/
(-9*a**(10/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 9*a**(7/3)*b*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(2/
3)) - 4*a*b*(a/b + x)**(1/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(-1/3)/(-9*a**(10/
3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 9*a**(7/3)*b*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(2/3)) - 4*a*b
*(a/b + x)**(1/3)*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(-1/3)/(
-9*a**(10/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 9*a**(7/3)*b*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(2/3
)) - 4*a*b*(a/b + x)**(1/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(-1/3)/(-9*a*
*(10/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 9*a**(7/3)*b*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(2/3)) +
4*b**2*(a/b + x)**(4/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(-1/3)/(-9*a**(10/3)*(a
/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 9*a**(7/3)*b*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(2/3)) + 4*b**2*(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(-1/3)/(-9*a
**(10/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 9*a**(7/3)*b*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(2/3)) +
4*b**2*(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(-1/3)/(-9*a**(1
0/3)*(a/b + x)**(1/3)*exp(2*I*pi/3)*gamma(2/3) + 9*a**(7/3)*b*(a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(2/3))

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