3.5.81 \(\int \frac {x^8}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=203 \[ -\frac {a^2}{b^2 \sqrt [3]{a+b x^3} (b c-a d)}+\frac {\left (a+b x^3\right )^{2/3}}{2 b^2 d}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{5/3} (b c-a d)^{4/3}}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3} (b c-a d)^{4/3}}+\frac {c^2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{5/3} (b c-a d)^{4/3}} \]

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Rubi [A]  time = 0.24, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 87, 56, 617, 204, 31} \begin {gather*} -\frac {a^2}{b^2 \sqrt [3]{a+b x^3} (b c-a d)}+\frac {\left (a+b x^3\right )^{2/3}}{2 b^2 d}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{5/3} (b c-a d)^{4/3}}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3} (b c-a d)^{4/3}}+\frac {c^2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{5/3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

-(a^2/(b^2*(b*c - a*d)*(a + b*x^3)^(1/3))) + (a + b*x^3)^(2/3)/(2*b^2*d) + (c^2*ArcTan[(1 - (2*d^(1/3)*(a + b*
x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(5/3)*(b*c - a*d)^(4/3)) - (c^2*Log[c + d*x^3])/(6*d^(5/3)
*(b*c - a*d)^(4/3)) + (c^2*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(5/3)*(b*c - a*d)^(4/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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 {x^8}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{(a+b x)^{4/3} (c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {a^2}{b (b c-a d) (a+b x)^{4/3}}+\frac {1}{b d \sqrt [3]{a+b x}}+\frac {c^2}{d (-b c+a d) \sqrt [3]{a+b x} (c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {a^2}{b^2 (b c-a d) \sqrt [3]{a+b x^3}}+\frac {\left (a+b x^3\right )^{2/3}}{2 b^2 d}-\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )}{3 d (b c-a d)}\\ &=-\frac {a^2}{b^2 (b c-a d) \sqrt [3]{a+b x^3}}+\frac {\left (a+b x^3\right )^{2/3}}{2 b^2 d}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{5/3} (b c-a d)^{4/3}}+\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 d^{5/3} (b c-a d)^{4/3}}-\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{d}}+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 d^2 (b c-a d)}\\ &=-\frac {a^2}{b^2 (b c-a d) \sqrt [3]{a+b x^3}}+\frac {\left (a+b x^3\right )^{2/3}}{2 b^2 d}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{5/3} (b c-a d)^{4/3}}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3} (b c-a d)^{4/3}}-\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}\right )}{d^{5/3} (b c-a d)^{4/3}}\\ &=-\frac {a^2}{b^2 (b c-a d) \sqrt [3]{a+b x^3}}+\frac {\left (a+b x^3\right )^{2/3}}{2 b^2 d}+\frac {c^2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{5/3} (b c-a d)^{4/3}}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{5/3} (b c-a d)^{4/3}}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3} (b c-a d)^{4/3}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 101, normalized size = 0.50 \begin {gather*} \frac {-3 a^2 d^2-2 b^2 c^2 \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )+a b d \left (c-d x^3\right )+b^2 c \left (2 c+d x^3\right )}{2 b^2 d^2 \sqrt [3]{a+b x^3} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.46, size = 268, normalized size = 1.32 \begin {gather*} -\frac {3 a^2 d-a b c+a b d x^3-b^2 c x^3}{2 b^2 d \sqrt [3]{a+b x^3} (b c-a d)}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{5/3} (b c-a d)^{4/3}}-\frac {c^2 \log \left (-\sqrt [3]{d} \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{6 d^{5/3} (b c-a d)^{4/3}}+\frac {c^2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{b c-a d}}\right )}{\sqrt {3} d^{5/3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

-1/2*(-(a*b*c) + 3*a^2*d - b^2*c*x^3 + a*b*d*x^3)/(b^2*d*(b*c - a*d)*(a + b*x^3)^(1/3)) + (c^2*ArcTan[1/Sqrt[3
] - (2*d^(1/3)*(a + b*x^3)^(1/3))/(Sqrt[3]*(b*c - a*d)^(1/3))])/(Sqrt[3]*d^(5/3)*(b*c - a*d)^(4/3)) + (c^2*Log
[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*d^(5/3)*(b*c - a*d)^(4/3)) - (c^2*Log[(b*c - a*d)^(2/3) -
d^(1/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3) + d^(2/3)*(a + b*x^3)^(2/3)])/(6*d^(5/3)*(b*c - a*d)^(4/3))

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fricas [B]  time = 0.71, size = 1004, normalized size = 4.95

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[-1/6*(3*sqrt(1/3)*(a*b^3*c^3*d - a^2*b^2*c^2*d^2 + (b^4*c^3*d - a*b^3*c^2*d^2)*x^3)*sqrt((-b*c*d^2 + a*d^3)^(
1/3)/(b*c - a*d))*log((2*b*d^2*x^3 - b*c*d + 3*a*d^2 + 3*sqrt(1/3)*(2*(-b*c*d^2 + a*d^3)^(2/3)*(b*x^3 + a)^(2/
3) + (b*x^3 + a)^(1/3)*(b*c*d - a*d^2) + (-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d))*sqrt((-b*c*d^2 + a*d^3)^(1/3)/(
b*c - a*d)) - 3*(-b*c*d^2 + a*d^3)^(2/3)*(b*x^3 + a)^(1/3))/(d*x^3 + c)) + (b^3*c^2*x^3 + a*b^2*c^2)*(-b*c*d^2
+ a*d^3)^(2/3)*log((b*x^3 + a)^(2/3)*d^2 + (-b*c*d^2 + a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^
(2/3)) - 2*(b^3*c^2*x^3 + a*b^2*c^2)*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/
3)) - 3*(a*b^2*c^2*d^2 - 4*a^2*b*c*d^3 + 3*a^3*d^4 + (b^3*c^2*d^2 - 2*a*b^2*c*d^3 + a^2*b*d^4)*x^3)*(b*x^3 + a
)^(2/3))/(a*b^4*c^2*d^3 - 2*a^2*b^3*c*d^4 + a^3*b^2*d^5 + (b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*x^3), -1
/6*(6*sqrt(1/3)*(a*b^3*c^3*d - a^2*b^2*c^2*d^2 + (b^4*c^3*d - a*b^3*c^2*d^2)*x^3)*sqrt(-(-b*c*d^2 + a*d^3)^(1/
3)/(b*c - a*d))*arctan(sqrt(1/3)*(2*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(1/3))*sqrt(-(-b*c*d^2 + a*d^3)^(
1/3)/(b*c - a*d))/d) + (b^3*c^2*x^3 + a*b^2*c^2)*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(2/3)*d^2 + (-b*c*d^
2 + a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(2/3)) - 2*(b^3*c^2*x^3 + a*b^2*c^2)*(-b*c*d^2 + a*d
^3)^(2/3)*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/3)) - 3*(a*b^2*c^2*d^2 - 4*a^2*b*c*d^3 + 3*a^3*d^4 +
(b^3*c^2*d^2 - 2*a*b^2*c*d^3 + a^2*b*d^4)*x^3)*(b*x^3 + a)^(2/3))/(a*b^4*c^2*d^3 - 2*a^2*b^3*c*d^4 + a^3*b^2*
d^5 + (b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*x^3)]

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giac [A]  time = 0.25, size = 325, normalized size = 1.60 \begin {gather*} \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b^{2} c^{2} d^{3} - 2 \, \sqrt {3} a b c d^{4} + \sqrt {3} a^{2} d^{5}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )}} + \frac {c^{2} \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )}} - \frac {a^{2}}{{\left (b^{3} c - a b^{2} d\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{2 \, b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

(-b*c*d^2 + a*d^3)^(2/3)*c^2*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d
)^(1/3))/(sqrt(3)*b^2*c^2*d^3 - 2*sqrt(3)*a*b*c*d^4 + sqrt(3)*a^2*d^5) - 1/6*(-b*c*d^2 + a*d^3)^(2/3)*c^2*log(
(b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))/(b^2*c^2*d^3 - 2*a*b*c*
d^4 + a^2*d^5) + 1/3*c^2*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^2*c^2*
d - 2*a*b*c*d^2 + a^2*d^3) - a^2/((b^3*c - a*b^2*d)*(b*x^3 + a)^(1/3)) + 1/2*(b*x^3 + a)^(2/3)/(b^2*d)

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(x^8/(b*x^3+a)^(4/3)/(d*x^3+c),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

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mupad [B]  time = 4.99, size = 449, normalized size = 2.21 \begin {gather*} \frac {{\left (b\,x^3+a\right )}^{2/3}}{2\,b^2\,d}+\frac {a^2}{b^2\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}+\frac {c^2\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^4\,d^3-b\,c^5\,d^2\right )-\frac {c^4\,\left (9\,a^4\,d^9-36\,a^3\,b\,c\,d^8+54\,a^2\,b^2\,c^2\,d^7-36\,a\,b^3\,c^3\,d^6+9\,b^4\,c^4\,d^5\right )}{9\,d^{10/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )}{3\,d^{5/3}\,{\left (a\,d-b\,c\right )}^{4/3}}-\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^4\,d^3-b\,c^5\,d^2\right )-\frac {{\left (c^2+\sqrt {3}\,c^2\,1{}\mathrm {i}\right )}^2\,\left (9\,a^4\,d^9-36\,a^3\,b\,c\,d^8+54\,a^2\,b^2\,c^2\,d^7-36\,a\,b^3\,c^3\,d^6+9\,b^4\,c^4\,d^5\right )}{36\,d^{10/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (c^2+\sqrt {3}\,c^2\,1{}\mathrm {i}\right )}{6\,d^{5/3}\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {c^2\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^4\,d^3-b\,c^5\,d^2\right )-\frac {c^4\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\,\left (9\,a^4\,d^9-36\,a^3\,b\,c\,d^8+54\,a^2\,b^2\,c^2\,d^7-36\,a\,b^3\,c^3\,d^6+9\,b^4\,c^4\,d^5\right )}{d^{10/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{d^{5/3}\,{\left (a\,d-b\,c\right )}^{4/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((a + b*x^3)^(4/3)*(c + d*x^3)),x)

[Out]

(a + b*x^3)^(2/3)/(2*b^2*d) + a^2/(b^2*(a + b*x^3)^(1/3)*(a*d - b*c)) + (c^2*log((a + b*x^3)^(1/3)*(a*c^4*d^3
- b*c^5*d^2) - (c^4*(9*a^4*d^9 + 9*b^4*c^4*d^5 - 36*a*b^3*c^3*d^6 + 54*a^2*b^2*c^2*d^7 - 36*a^3*b*c*d^8))/(9*d
^(10/3)*(a*d - b*c)^(8/3))))/(3*d^(5/3)*(a*d - b*c)^(4/3)) - (log((a + b*x^3)^(1/3)*(a*c^4*d^3 - b*c^5*d^2) -
((3^(1/2)*c^2*1i + c^2)^2*(9*a^4*d^9 + 9*b^4*c^4*d^5 - 36*a*b^3*c^3*d^6 + 54*a^2*b^2*c^2*d^7 - 36*a^3*b*c*d^8)
)/(36*d^(10/3)*(a*d - b*c)^(8/3)))*(3^(1/2)*c^2*1i + c^2))/(6*d^(5/3)*(a*d - b*c)^(4/3)) + (c^2*log((a + b*x^3
)^(1/3)*(a*c^4*d^3 - b*c^5*d^2) - (c^4*((3^(1/2)*1i)/6 - 1/6)^2*(9*a^4*d^9 + 9*b^4*c^4*d^5 - 36*a*b^3*c^3*d^6
+ 54*a^2*b^2*c^2*d^7 - 36*a^3*b*c*d^8))/(d^(10/3)*(a*d - b*c)^(8/3)))*((3^(1/2)*1i)/6 - 1/6))/(d^(5/3)*(a*d -
b*c)^(4/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**3+a)**(4/3)/(d*x**3+c),x)

[Out]

Integral(x**8/((a + b*x**3)**(4/3)*(c + d*x**3)), x)

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