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

Optimal. Leaf size=223 \[ -\frac {\left (a+b x^3\right )^{5/3} (a d+b c)}{5 b^2 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^2 d}-\frac {c^2 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{11/3}}+\frac {c^2 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3}}+\frac {c^2 (b c-a d)^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{11/3}}+\frac {c^2 \left (a+b x^3\right )^{2/3}}{2 d^3} \]

[Out]

1/2*c^2*(b*x^3+a)^(2/3)/d^3-1/5*(a*d+b*c)*(b*x^3+a)^(5/3)/b^2/d^2+1/8*(b*x^3+a)^(8/3)/b^2/d-1/6*c^2*(-a*d+b*c)
^(2/3)*ln(d*x^3+c)/d^(11/3)+1/2*c^2*(-a*d+b*c)^(2/3)*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(11/3)+1/3
*c^2*(-a*d+b*c)^(2/3)*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a)^(1/3)/(-a*d+b*c)^(1/3))*3^(1/2))/d^(11/3)*3^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {446, 88, 50, 56, 617, 204, 31} \[ -\frac {\left (a+b x^3\right )^{5/3} (a d+b c)}{5 b^2 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^2 d}+\frac {c^2 \left (a+b x^3\right )^{2/3}}{2 d^3}-\frac {c^2 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{11/3}}+\frac {c^2 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3}}+\frac {c^2 (b c-a d)^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{11/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(a + b*x^3)^(2/3))/(2*d^3) - ((b*c + a*d)*(a + b*x^3)^(5/3))/(5*b^2*d^2) + (a + b*x^3)^(8/3)/(8*b^2*d) +
(c^2*(b*c - a*d)^(2/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(11/3
)) - (c^2*(b*c - a*d)^(2/3)*Log[c + d*x^3])/(6*d^(11/3)) + (c^2*(b*c - a*d)^(2/3)*Log[(b*c - a*d)^(1/3) + d^(1
/3)*(a + b*x^3)^(1/3)])/(2*d^(11/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 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 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}}{c+d x^3} \, 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) (a+b x)^{2/3}}{b d^2}+\frac {(a+b x)^{5/3}}{b d}+\frac {c^2 (a+b x)^{2/3}}{d^2 (c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {(b c+a d) \left (a+b x^3\right )^{5/3}}{5 b^2 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^2 d}+\frac {c^2 \operatorname {Subst}\left (\int \frac {(a+b x)^{2/3}}{c+d x} \, dx,x,x^3\right )}{3 d^2}\\ &=\frac {c^2 \left (a+b x^3\right )^{2/3}}{2 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{5/3}}{5 b^2 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^2 d}-\frac {\left (c^2 (b c-a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )}{3 d^3}\\ &=\frac {c^2 \left (a+b x^3\right )^{2/3}}{2 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{5/3}}{5 b^2 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^2 d}-\frac {c^2 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{11/3}}+\frac {\left (c^2 (b c-a d)^{2/3}\right ) \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^{11/3}}-\frac {\left (c^2 (b c-a d)\right ) \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^4}\\ &=\frac {c^2 \left (a+b x^3\right )^{2/3}}{2 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{5/3}}{5 b^2 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^2 d}-\frac {c^2 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{11/3}}+\frac {c^2 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3}}-\frac {\left (c^2 (b c-a d)^{2/3}\right ) \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^{11/3}}\\ &=\frac {c^2 \left (a+b x^3\right )^{2/3}}{2 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{5/3}}{5 b^2 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^2 d}+\frac {c^2 (b c-a d)^{2/3} \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 )}{\sqrt {3} d^{11/3}}-\frac {c^2 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{11/3}}+\frac {c^2 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3}}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 104, normalized size = 0.47 \[ \frac {\left (a+b x^3\right )^{2/3} \left (-3 a^2 d^2-20 b^2 c^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )+2 a b d \left (d x^3-4 c\right )+b^2 \left (20 c^2-8 c d x^3+5 d^2 x^6\right )\right )}{40 b^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^(2/3)*(-3*a^2*d^2 + 2*a*b*d*(-4*c + d*x^3) + b^2*(20*c^2 - 8*c*d*x^3 + 5*d^2*x^6) - 20*b^2*c^2*Hy
pergeometric2F1[2/3, 1, 5/3, (d*(a + b*x^3))/(-(b*c) + a*d)]))/(40*b^2*d^3)

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fricas [B]  time = 1.10, size = 398, normalized size = 1.78 \[ \frac {40 \, \sqrt {3} b^{2} c^{2} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) - 20 \, b^{2} c^{2} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )} - {\left (b c - a d\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}}\right ) + 40 \, b^{2} c^{2} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left (-d \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right ) + 3 \, {\left (5 \, b^{2} d^{2} x^{6} + 20 \, b^{2} c^{2} - 8 \, a b c d - 3 \, a^{2} d^{2} - 2 \, {\left (4 \, b^{2} c d - a b d^{2}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{120 \, b^{2} d^{3}} \]

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]

1/120*(40*sqrt(3)*b^2*c^2*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)
*d*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3) - sqrt(3)*(b*c - a*d))/(b*c - a*d)) - 20*b^2*c^2*((b^2*c^2 - 2*
a*b*c*d + a^2*d^2)/d^2)^(1/3)*log((b*x^3 + a)^(1/3)*d*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b*x^3 + a
)^(2/3)*(b*c - a*d) - (b*c - a*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)) + 40*b^2*c^2*((b^2*c^2 - 2*a*b*
c*d + a^2*d^2)/d^2)^(1/3)*log(-d*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b*x^3 + a)^(1/3)*(b*c - a*d))
+ 3*(5*b^2*d^2*x^6 + 20*b^2*c^2 - 8*a*b*c*d - 3*a^2*d^2 - 2*(4*b^2*c*d - a*b*d^2)*x^3)*(b*x^3 + a)^(2/3))/(b^2
*d^3)

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giac [A]  time = 0.27, size = 350, normalized size = 1.57 \[ \frac {{\left (b^{19} c^{3} d^{5} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} - a b^{18} c^{2} d^{6} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{19} c d^{8} - a b^{18} d^{9}\right )}} + \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{5}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{5}} + \frac {20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{16} c^{2} d^{5} - 8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{15} c d^{6} + 5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{14} d^{7} - 8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a b^{14} d^{7}}{40 \, b^{16} d^{8}} \]

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^19*c^3*d^5*(-(b*c - a*d)/d)^(1/3) - a*b^18*c^2*d^6*(-(b*c - a*d)/d)^(1/3))*(-(b*c - a*d)/d)^(1/3)*log(a
bs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^19*c*d^8 - a*b^18*d^9) + 1/3*sqrt(3)*(-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))/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))/d^5 + 1/40*(20*(b*x^3 + a)^(2/3)*b^16*c^2*d^5 - 8*(b*x^3 + a)^(5/3)*b^15*c*d^6 + 5*(b*x^3 + a)^(8/3)*b^
14*d^7 - 8*(b*x^3 + a)^(5/3)*a*b^14*d^7)/(b^16*d^8)

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maple [F]  time = 0.71, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{8}}{d \,x^{3}+c}\, dx \]

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 \[ \text {Exception raised: ValueError} \]

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 = 5.11, size = 385, normalized size = 1.73 \[ \left (\frac {a^2}{2\,b^2\,d}+\frac {\left (\frac {2\,a}{b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{b^4\,d^2}\right )\,\left (b^3\,c-a\,b^2\,d\right )}{2\,b^2\,d}\right )\,{\left (b\,x^3+a\right )}^{2/3}-\left (\frac {2\,a}{5\,b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{5\,b^4\,d^2}\right )\,{\left (b\,x^3+a\right )}^{5/3}+\frac {{\left (b\,x^3+a\right )}^{8/3}}{8\,b^2\,d}+\frac {c^2\,\ln \left (\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c^4\,d^2-2\,a\,b\,c^5\,d+b^2\,c^6\right )}{d^5}-\frac {c^4\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{9\,d^{22/3}}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{11/3}}-\frac {c^2\,\ln \left (\frac {c^4\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^5}-\frac {c^4\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{16/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{11/3}}+\frac {c^2\,\ln \left (\frac {c^4\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^5}-\frac {c^4\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{7/3}}{4\,d^{16/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{d^{11/3}} \]

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^2/(2*b^2*d) + (((2*a)/(b^2*d) + (b^3*c - a*b^2*d)/(b^4*d^2))*(b^3*c - a*b^2*d))/(2*b^2*d))*(a + b*x^3)^(2/3
) - ((2*a)/(5*b^2*d) + (b^3*c - a*b^2*d)/(5*b^4*d^2))*(a + b*x^3)^(5/3) + (a + b*x^3)^(8/3)/(8*b^2*d) + (c^2*l
og(((a + b*x^3)^(1/3)*(b^2*c^6 + a^2*c^4*d^2 - 2*a*b*c^5*d))/d^5 - (c^4*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2
))/(9*d^(22/3)))*(a*d - b*c)^(2/3))/(3*d^(11/3)) - (c^2*log((c^4*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^5 - (c^4*(
(3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(7/3))/d^(16/3))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(2/3))/(3*d^(11/3)) + (c
^2*log((c^4*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^5 - (c^4*(3^(1/2)*1i - 1)^2*(a*d - b*c)^(7/3))/(4*d^(16/3)))*((
3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(2/3))/d^(11/3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8} \left (a + b x^{3}\right )^{\frac {2}{3}}}{c + d x^{3}}\, dx \]

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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