3.24.33 \(\int \frac {-d+c x}{x^7 \sqrt [3]{-b+a x^3}} \, dx\)

Optimal. Leaf size=184 \[ \frac {2 a^2 d \log \left (\sqrt [3]{a x^3-b}+\sqrt [3]{b}\right )}{27 b^{7/3}}-\frac {a^2 d \log \left (-\sqrt [3]{b} \sqrt [3]{a x^3-b}+\left (a x^3-b\right )^{2/3}+b^{2/3}\right )}{27 b^{7/3}}+\frac {2 a^2 d \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{a x^3-b}}{\sqrt {3} \sqrt [3]{b}}\right )}{9 \sqrt {3} b^{7/3}}+\frac {\left (a x^3-b\right )^{2/3} \left (27 a c x^4-20 a d x^3+18 b c x-15 b d\right )}{90 b^2 x^6} \]

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Rubi [A]  time = 0.25, antiderivative size = 198, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1844, 266, 51, 56, 617, 204, 31, 271, 264} \begin {gather*} \frac {a^2 d \log \left (\sqrt [3]{a x^3-b}+\sqrt [3]{b}\right )}{9 b^{7/3}}+\frac {2 a^2 d \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a x^3-b}}{\sqrt {3} \sqrt [3]{b}}\right )}{9 \sqrt {3} b^{7/3}}-\frac {a^2 d \log (x)}{9 b^{7/3}}+\frac {3 a c \left (a x^3-b\right )^{2/3}}{10 b^2 x^2}-\frac {2 a d \left (a x^3-b\right )^{2/3}}{9 b^2 x^3}+\frac {c \left (a x^3-b\right )^{2/3}}{5 b x^5}-\frac {d \left (a x^3-b\right )^{2/3}}{6 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-d + c*x)/(x^7*(-b + a*x^3)^(1/3)),x]

[Out]

-1/6*(d*(-b + a*x^3)^(2/3))/(b*x^6) + (c*(-b + a*x^3)^(2/3))/(5*b*x^5) - (2*a*d*(-b + a*x^3)^(2/3))/(9*b^2*x^3
) + (3*a*c*(-b + a*x^3)^(2/3))/(10*b^2*x^2) + (2*a^2*d*ArcTan[(b^(1/3) - 2*(-b + a*x^3)^(1/3))/(Sqrt[3]*b^(1/3
))])/(9*Sqrt[3]*b^(7/3)) - (a^2*d*Log[x])/(9*b^(7/3)) + (a^2*d*Log[b^(1/3) + (-b + a*x^3)^(1/3)])/(9*b^(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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

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]

Rule 1844

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {-d+c x}{x^7 \sqrt [3]{-b+a x^3}} \, dx &=\int \left (-\frac {d}{x^7 \sqrt [3]{-b+a x^3}}+\frac {c}{x^6 \sqrt [3]{-b+a x^3}}\right ) \, dx\\ &=c \int \frac {1}{x^6 \sqrt [3]{-b+a x^3}} \, dx-d \int \frac {1}{x^7 \sqrt [3]{-b+a x^3}} \, dx\\ &=\frac {c \left (-b+a x^3\right )^{2/3}}{5 b x^5}+\frac {(3 a c) \int \frac {1}{x^3 \sqrt [3]{-b+a x^3}} \, dx}{5 b}-\frac {1}{3} d \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt [3]{-b+a x}} \, dx,x,x^3\right )\\ &=-\frac {d \left (-b+a x^3\right )^{2/3}}{6 b x^6}+\frac {c \left (-b+a x^3\right )^{2/3}}{5 b x^5}+\frac {3 a c \left (-b+a x^3\right )^{2/3}}{10 b^2 x^2}-\frac {(2 a d) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [3]{-b+a x}} \, dx,x,x^3\right )}{9 b}\\ &=-\frac {d \left (-b+a x^3\right )^{2/3}}{6 b x^6}+\frac {c \left (-b+a x^3\right )^{2/3}}{5 b x^5}-\frac {2 a d \left (-b+a x^3\right )^{2/3}}{9 b^2 x^3}+\frac {3 a c \left (-b+a x^3\right )^{2/3}}{10 b^2 x^2}-\frac {\left (2 a^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{-b+a x}} \, dx,x,x^3\right )}{27 b^2}\\ &=-\frac {d \left (-b+a x^3\right )^{2/3}}{6 b x^6}+\frac {c \left (-b+a x^3\right )^{2/3}}{5 b x^5}-\frac {2 a d \left (-b+a x^3\right )^{2/3}}{9 b^2 x^3}+\frac {3 a c \left (-b+a x^3\right )^{2/3}}{10 b^2 x^2}-\frac {a^2 d \log (x)}{9 b^{7/3}}+\frac {\left (a^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{b}+x} \, dx,x,\sqrt [3]{-b+a x^3}\right )}{9 b^{7/3}}-\frac {\left (a^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{b^{2/3}-\sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{-b+a x^3}\right )}{9 b^2}\\ &=-\frac {d \left (-b+a x^3\right )^{2/3}}{6 b x^6}+\frac {c \left (-b+a x^3\right )^{2/3}}{5 b x^5}-\frac {2 a d \left (-b+a x^3\right )^{2/3}}{9 b^2 x^3}+\frac {3 a c \left (-b+a x^3\right )^{2/3}}{10 b^2 x^2}-\frac {a^2 d \log (x)}{9 b^{7/3}}+\frac {a^2 d \log \left (\sqrt [3]{b}+\sqrt [3]{-b+a x^3}\right )}{9 b^{7/3}}-\frac {\left (2 a^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt [3]{b}}\right )}{9 b^{7/3}}\\ &=-\frac {d \left (-b+a x^3\right )^{2/3}}{6 b x^6}+\frac {c \left (-b+a x^3\right )^{2/3}}{5 b x^5}-\frac {2 a d \left (-b+a x^3\right )^{2/3}}{9 b^2 x^3}+\frac {3 a c \left (-b+a x^3\right )^{2/3}}{10 b^2 x^2}+\frac {2 a^2 d \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{7/3}}-\frac {a^2 d \log (x)}{9 b^{7/3}}+\frac {a^2 d \log \left (\sqrt [3]{b}+\sqrt [3]{-b+a x^3}\right )}{9 b^{7/3}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 65, normalized size = 0.35 \begin {gather*} \frac {\left (a x^3-b\right )^{2/3} \left (b c \left (3 a x^3+2 b\right )-5 a^2 d x^5 \, _2F_1\left (\frac {2}{3},3;\frac {5}{3};1-\frac {a x^3}{b}\right )\right )}{10 b^3 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-d + c*x)/(x^7*(-b + a*x^3)^(1/3)),x]

[Out]

((-b + a*x^3)^(2/3)*(b*c*(2*b + 3*a*x^3) - 5*a^2*d*x^5*Hypergeometric2F1[2/3, 3, 5/3, 1 - (a*x^3)/b]))/(10*b^3
*x^5)

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IntegrateAlgebraic [A]  time = 17.91, size = 184, normalized size = 1.00 \begin {gather*} \frac {\left (-b+a x^3\right )^{2/3} \left (-15 b d+18 b c x-20 a d x^3+27 a c x^4\right )}{90 b^2 x^6}+\frac {2 a^2 d \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt {3} \sqrt [3]{b}}\right )}{9 \sqrt {3} b^{7/3}}+\frac {2 a^2 d \log \left (\sqrt [3]{b}+\sqrt [3]{-b+a x^3}\right )}{27 b^{7/3}}-\frac {a^2 d \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{-b+a x^3}+\left (-b+a x^3\right )^{2/3}\right )}{27 b^{7/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-d + c*x)/(x^7*(-b + a*x^3)^(1/3)),x]

[Out]

((-b + a*x^3)^(2/3)*(-15*b*d + 18*b*c*x - 20*a*d*x^3 + 27*a*c*x^4))/(90*b^2*x^6) + (2*a^2*d*ArcTan[1/Sqrt[3] -
(2*(-b + a*x^3)^(1/3))/(Sqrt[3]*b^(1/3))])/(9*Sqrt[3]*b^(7/3)) + (2*a^2*d*Log[b^(1/3) + (-b + a*x^3)^(1/3)])/
(27*b^(7/3)) - (a^2*d*Log[b^(2/3) - b^(1/3)*(-b + a*x^3)^(1/3) + (-b + a*x^3)^(2/3)])/(27*b^(7/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x-d)/x^7/(a*x^3-b)^(1/3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {c x - d}{{\left (a x^{3} - b\right )}^{\frac {1}{3}} x^{7}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x-d)/x^7/(a*x^3-b)^(1/3),x, algorithm="giac")

[Out]

integrate((c*x - d)/((a*x^3 - b)^(1/3)*x^7), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {c x -d}{x^{7} \left (a \,x^{3}-b \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x-d)/x^7/(a*x^3-b)^(1/3),x)

[Out]

int((c*x-d)/x^7/(a*x^3-b)^(1/3),x)

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maxima [A]  time = 0.67, size = 219, normalized size = 1.19 \begin {gather*} -\frac {1}{54} \, {\left (\frac {4 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}} - b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {7}{3}}} + \frac {2 \, a^{2} \log \left ({\left (a x^{3} - b\right )}^{\frac {2}{3}} - {\left (a x^{3} - b\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {7}{3}}} - \frac {4 \, a^{2} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}{b^{\frac {7}{3}}} + \frac {3 \, {\left (4 \, {\left (a x^{3} - b\right )}^{\frac {5}{3}} a^{2} + 7 \, {\left (a x^{3} - b\right )}^{\frac {2}{3}} a^{2} b\right )}}{{\left (a x^{3} - b\right )}^{2} b^{2} + 2 \, {\left (a x^{3} - b\right )} b^{3} + b^{4}}\right )} d + \frac {c {\left (\frac {5 \, {\left (a x^{3} - b\right )}^{\frac {2}{3}} a}{x^{2}} - \frac {2 \, {\left (a x^{3} - b\right )}^{\frac {5}{3}}}{x^{5}}\right )}}{10 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x-d)/x^7/(a*x^3-b)^(1/3),x, algorithm="maxima")

[Out]

-1/54*(4*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(2*(a*x^3 - b)^(1/3) - b^(1/3))/b^(1/3))/b^(7/3) + 2*a^2*log((a*x^3 -
b)^(2/3) - (a*x^3 - b)^(1/3)*b^(1/3) + b^(2/3))/b^(7/3) - 4*a^2*log((a*x^3 - b)^(1/3) + b^(1/3))/b^(7/3) + 3*(
4*(a*x^3 - b)^(5/3)*a^2 + 7*(a*x^3 - b)^(2/3)*a^2*b)/((a*x^3 - b)^2*b^2 + 2*(a*x^3 - b)*b^3 + b^4))*d + 1/10*c
*(5*(a*x^3 - b)^(2/3)*a/x^2 - 2*(a*x^3 - b)^(5/3)/x^5)/b^2

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mupad [B]  time = 2.28, size = 250, normalized size = 1.36 \begin {gather*} \frac {2\,a^2\,d\,\ln \left ({\left (a\,x^3-b\right )}^{1/3}+b^{1/3}\right )}{27\,b^{7/3}}-\frac {\frac {7\,a^2\,d\,{\left (a\,x^3-b\right )}^{2/3}}{18\,b}+\frac {2\,a^2\,d\,{\left (a\,x^3-b\right )}^{5/3}}{9\,b^2}}{{\left (b-a\,x^3\right )}^2-2\,b\,\left (b-a\,x^3\right )+b^2}+\frac {2\,a^2\,d\,\ln \left (\frac {4\,a^4\,d^2\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{81\,b^{11/3}}+\frac {4\,a^4\,d^2\,{\left (a\,x^3-b\right )}^{1/3}}{81\,b^4}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{27\,b^{7/3}}-\frac {2\,a^2\,d\,\ln \left (\frac {4\,a^4\,d^2\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{81\,b^{11/3}}+\frac {4\,a^4\,d^2\,{\left (a\,x^3-b\right )}^{1/3}}{81\,b^4}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{27\,b^{7/3}}+\frac {c\,{\left (a\,x^3-b\right )}^{2/3}\,\left (3\,a\,x^3+2\,b\right )}{10\,b^2\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(d - c*x)/(x^7*(a*x^3 - b)^(1/3)),x)

[Out]

(2*a^2*d*log((a*x^3 - b)^(1/3) + b^(1/3)))/(27*b^(7/3)) - ((7*a^2*d*(a*x^3 - b)^(2/3))/(18*b) + (2*a^2*d*(a*x^
3 - b)^(5/3))/(9*b^2))/((b - a*x^3)^2 - 2*b*(b - a*x^3) + b^2) + (2*a^2*d*log((4*a^4*d^2*((3^(1/2)*1i)/2 - 1/2
)^2)/(81*b^(11/3)) + (4*a^4*d^2*(a*x^3 - b)^(1/3))/(81*b^4))*((3^(1/2)*1i)/2 - 1/2))/(27*b^(7/3)) - (2*a^2*d*l
og((4*a^4*d^2*((3^(1/2)*1i)/2 + 1/2)^2)/(81*b^(11/3)) + (4*a^4*d^2*(a*x^3 - b)^(1/3))/(81*b^4))*((3^(1/2)*1i)/
2 + 1/2))/(27*b^(7/3)) + (c*(a*x^3 - b)^(2/3)*(2*b + 3*a*x^3))/(10*b^2*x^5)

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sympy [C]  time = 2.96, size = 320, normalized size = 1.74 \begin {gather*} c \left (\begin {cases} - \frac {a^{\frac {5}{3}} \left (-1 + \frac {b}{a x^{3}}\right )^{\frac {2}{3}} e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {5}{3}\right )}{3 b^{2} \Gamma \left (\frac {1}{3}\right )} - \frac {2 a^{\frac {2}{3}} \left (-1 + \frac {b}{a x^{3}}\right )^{\frac {2}{3}} e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {5}{3}\right )}{9 b x^{3} \Gamma \left (\frac {1}{3}\right )} & \text {for}\: \left |{\frac {b}{a x^{3}}}\right | > 1 \\\frac {3 a^{\frac {11}{3}} x^{6} \left (1 - \frac {b}{a x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{9 a^{2} b^{2} x^{6} \Gamma \left (\frac {1}{3}\right ) - 9 a b^{3} x^{3} \Gamma \left (\frac {1}{3}\right )} - \frac {a^{\frac {8}{3}} b x^{3} \left (1 - \frac {b}{a x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{9 a^{2} b^{2} x^{6} \Gamma \left (\frac {1}{3}\right ) - 9 a b^{3} x^{3} \Gamma \left (\frac {1}{3}\right )} - \frac {2 a^{\frac {5}{3}} b^{2} \left (1 - \frac {b}{a x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{9 a^{2} b^{2} x^{6} \Gamma \left (\frac {1}{3}\right ) - 9 a b^{3} x^{3} \Gamma \left (\frac {1}{3}\right )} & \text {otherwise} \end {cases}\right ) + \frac {d \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 \sqrt [3]{a} x^{7} \Gamma \left (\frac {10}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x-d)/x**7/(a*x**3-b)**(1/3),x)

[Out]

c*Piecewise((-a**(5/3)*(-1 + b/(a*x**3))**(2/3)*exp(-I*pi/3)*gamma(-5/3)/(3*b**2*gamma(1/3)) - 2*a**(2/3)*(-1
+ b/(a*x**3))**(2/3)*exp(-I*pi/3)*gamma(-5/3)/(9*b*x**3*gamma(1/3)), Abs(b/(a*x**3)) > 1), (3*a**(11/3)*x**6*(
1 - b/(a*x**3))**(2/3)*gamma(-5/3)/(9*a**2*b**2*x**6*gamma(1/3) - 9*a*b**3*x**3*gamma(1/3)) - a**(8/3)*b*x**3*
(1 - b/(a*x**3))**(2/3)*gamma(-5/3)/(9*a**2*b**2*x**6*gamma(1/3) - 9*a*b**3*x**3*gamma(1/3)) - 2*a**(5/3)*b**2
*(1 - b/(a*x**3))**(2/3)*gamma(-5/3)/(9*a**2*b**2*x**6*gamma(1/3) - 9*a*b**3*x**3*gamma(1/3)), True)) + d*gamm
a(7/3)*hyper((1/3, 7/3), (10/3,), b*exp_polar(2*I*pi)/(a*x**3))/(3*a**(1/3)*x**7*gamma(10/3))

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