3.5.69 \(\int \frac {x^8}{(a+b x^3)^{2/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=201 \[ -\frac {\sqrt [3]{a+b x^3} (a d+b c)}{b^2 d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{7/3} (b c-a d)^{2/3}}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3} (b c-a d)^{2/3}}-\frac {c^2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{7/3} (b c-a d)^{2/3}} \]

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Rubi [A]  time = 0.21, antiderivative size = 201, 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, 88, 58, 617, 204, 31} \begin {gather*} -\frac {\sqrt [3]{a+b x^3} (a d+b c)}{b^2 d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{7/3} (b c-a d)^{2/3}}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3} (b c-a d)^{2/3}}-\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^{7/3} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

-(((b*c + a*d)*(a + b*x^3)^(1/3))/(b^2*d^2)) + (a + b*x^3)^(4/3)/(4*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^(7/3)*(b*c - a*d)^(2/3)) - (c^2*Log[c + d*x^3])/(6*d^(7/
3)*(b*c - a*d)^(2/3)) + (c^2*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(7/3)*(b*c - a*d)^(2/3))

Rule 31

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

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[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
] && NegQ[(b*c - a*d)/b]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 )^{2/3} \left (c+d x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {-b c-a d}{b d^2 (a+b x)^{2/3}}+\frac {\sqrt [3]{a+b x}}{b d}+\frac {c^2}{d^2 (a+b x)^{2/3} (c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {(b c+a d) \sqrt [3]{a+b x^3}}{b^2 d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}+\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 d^2}\\ &=-\frac {(b c+a d) \sqrt [3]{a+b x^3}}{b^2 d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{7/3} (b c-a d)^{2/3}}+\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^{7/3} (b c-a d)^{2/3}}+\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^{8/3} \sqrt [3]{b c-a d}}\\ &=-\frac {(b c+a d) \sqrt [3]{a+b x^3}}{b^2 d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{7/3} (b c-a d)^{2/3}}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3} (b c-a d)^{2/3}}+\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^{7/3} (b c-a d)^{2/3}}\\ &=-\frac {(b c+a d) \sqrt [3]{a+b x^3}}{b^2 d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 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^{7/3} (b c-a d)^{2/3}}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{7/3} (b c-a d)^{2/3}}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3} (b c-a d)^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 211, normalized size = 1.05 \begin {gather*} \frac {-\frac {12 \sqrt [3]{a+b x^3} (a d+b c)}{b^2}+\frac {3 d \left (a+b x^3\right )^{4/3}}{b^2}-\frac {2 c^2 \left (\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 )-2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+2 \sqrt {3} \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 )\right )}{\sqrt [3]{d} (b c-a d)^{2/3}}}{12 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/((a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

((-12*(b*c + a*d)*(a + b*x^3)^(1/3))/b^2 + (3*d*(a + b*x^3)^(4/3))/b^2 - (2*c^2*(2*Sqrt[3]*ArcTan[(1 - (2*d^(1
/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - 2*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + Lo
g[(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)]))/(d^(1/3)*(b*c
- a*d)^(2/3)))/(12*d^2)

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IntegrateAlgebraic [A]  time = 0.42, size = 246, normalized size = 1.22 \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (-3 a d-4 b c+b d x^3\right )}{4 b^2 d^2}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{7/3} (b c-a d)^{2/3}}-\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^{7/3} (b c-a d)^{2/3}}-\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^{7/3} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8/((a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

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

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fricas [B]  time = 0.47, size = 1156, normalized size = 5.75

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[-1/12*(2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(2/3)*b^2*c^2*log(-(b*x^3 + a)^(2/3)*(b*c*d - a*d^2) - (b^2*c^2*
d - 2*a*b*c*d^2 + a^2*d^3)^(1/3)*(b*c - a*d) + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(2/3)*(b*x^3 + a)^(1/3)) -
4*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(2/3)*b^2*c^2*log(-(b*x^3 + a)^(1/3)*(b*c*d - a*d^2) - (b^2*c^2*d - 2*a*
b*c*d^2 + a^2*d^3)^(2/3)) + 6*sqrt(1/3)*(b^3*c^3*d - a*b^2*c^2*d^2)*sqrt(-(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^
(1/3)/d)*log((b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2 - 2*(b^2*c*d - a*b*d^2)*x^3 + 3*sqrt(1/3)*(2*(b*x^3 + a)^(2/3)*(
b*c*d - a*d^2) - (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(1/3)*(b*c - a*d) + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(
2/3)*(b*x^3 + a)^(1/3))*sqrt(-(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(1/3)/d) + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*
d^3)^(1/3)*(b*x^3 + a)^(1/3)*(b*c - a*d))/(d*x^3 + c)) + 3*(4*b^3*c^3*d - 5*a*b^2*c^2*d^2 - 2*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)^(1/3))/(b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2
*b^2*d^5), -1/12*(2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(2/3)*b^2*c^2*log(-(b*x^3 + a)^(2/3)*(b*c*d - a*d^2) -
(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(1/3)*(b*c - a*d) + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(2/3)*(b*x^3 + a)
^(1/3)) - 4*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(2/3)*b^2*c^2*log(-(b*x^3 + a)^(1/3)*(b*c*d - a*d^2) - (b^2*c^
2*d - 2*a*b*c*d^2 + a^2*d^3)^(2/3)) - 12*sqrt(1/3)*(b^3*c^3*d - a*b^2*c^2*d^2)*sqrt((b^2*c^2*d - 2*a*b*c*d^2 +
a^2*d^3)^(1/3)/d)*arctan(-sqrt(1/3)*((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(1/3)*(b*c - a*d) - 2*(b^2*c^2*d - 2
*a*b*c*d^2 + a^2*d^3)^(2/3)*(b*x^3 + a)^(1/3))*sqrt((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(1/3)/d)/(b^2*c^2 - 2*
a*b*c*d + a^2*d^2)) + 3*(4*b^3*c^3*d - 5*a*b^2*c^2*d^2 - 2*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)^(1/3))/(b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)]

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giac [A]  time = 0.30, size = 312, normalized size = 1.55 \begin {gather*} -\frac {b^{10} c^{2} d^{2} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{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^{11} c d^{4} - a b^{10} d^{5}\right )}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{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 c d^{3} - \sqrt {3} a d^{4}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{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 c d^{3} - a d^{4}\right )}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{7} c d^{2} - {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{6} d^{3} + 4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a b^{6} d^{3}}{4 \, b^{8} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-1/3*b^10*c^2*d^2*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^11*c*d^4 - a*
b^10*d^5) + (-b*c*d^2 + a*d^3)^(1/3)*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*c*d^3 - sqrt(3)*a*d^4) + 1/6*(-b*c*d^2 + a*d^3)^(1/3)*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*c*d^3 - a*d^4) - 1/4*(4*(b*x^3 + a)^
(1/3)*b^7*c*d^2 - (b*x^3 + a)^(4/3)*b^6*d^3 + 4*(b*x^3 + a)^(1/3)*a*b^6*d^3)/(b^8*d^4)

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\left (b \,x^{3}+a \right )^{\frac {2}{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)^(2/3)/(d*x^3+c),x)

[Out]

int(x^8/(b*x^3+a)^(2/3)/(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)^(2/3)/(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 = 4.69, size = 292, normalized size = 1.45 \begin {gather*} \frac {{\left (b\,x^3+a\right )}^{4/3}}{4\,b^2\,d}-\left (\frac {2\,a}{b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{b^4\,d^2}\right )\,{\left (b\,x^3+a\right )}^{1/3}-\frac {\ln \left (3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}+\frac {\left (c^2+\sqrt {3}\,c^2\,1{}\mathrm {i}\right )\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{6\,d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )\,\left (c^2+\sqrt {3}\,c^2\,1{}\mathrm {i}\right )}{6\,d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}+\frac {c^2\,\ln \left (3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}-\frac {c^2\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )}{3\,d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}+\frac {c^2\,\ln \left (3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}-\frac {c^2\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((a + b*x^3)^(2/3)*(c + d*x^3)),x)

[Out]

(a + b*x^3)^(4/3)/(4*b^2*d) - ((2*a)/(b^2*d) + (b^3*c - a*b^2*d)/(b^4*d^2))*(a + b*x^3)^(1/3) - (log(3*c^2*(a
+ b*x^3)^(1/3) + ((3^(1/2)*c^2*1i + c^2)*(9*a*d^3 - 9*b*c*d^2))/(6*d^(7/3)*(a*d - b*c)^(2/3)))*(3^(1/2)*c^2*1i
+ c^2))/(6*d^(7/3)*(a*d - b*c)^(2/3)) + (c^2*log(3*c^2*(a + b*x^3)^(1/3) - (c^2*(9*a*d^3 - 9*b*c*d^2))/(3*d^(
7/3)*(a*d - b*c)^(2/3))))/(3*d^(7/3)*(a*d - b*c)^(2/3)) + (c^2*log(3*c^2*(a + b*x^3)^(1/3) - (c^2*((3^(1/2)*1i
)/6 - 1/6)*(9*a*d^3 - 9*b*c*d^2))/(d^(7/3)*(a*d - b*c)^(2/3)))*((3^(1/2)*1i)/6 - 1/6))/(d^(7/3)*(a*d - b*c)^(2
/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 {2}{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)**(2/3)/(d*x**3+c),x)

[Out]

Integral(x**8/((a + b*x**3)**(2/3)*(c + d*x**3)), x)

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