3.5.57 \(\int \frac {x^8}{\sqrt [3]{a+b x^3} (c+d x^3)} \, dx\)

Optimal. Leaf size=203 \[ -\frac {\left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^2 d^2}+\frac {\left (a+b x^3\right )^{5/3}}{5 b^2 d}+\frac {c^2 \log \left (c+d x^3\right )}{6 d^{8/3} \sqrt [3]{b c-a d}}-\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} \sqrt [3]{b c-a d}}-\frac {c^2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{8/3} \sqrt [3]{b c-a d}} \]

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Rubi [A]  time = 0.21, 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, 88, 56, 617, 204, 31} \begin {gather*} -\frac {\left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^2 d^2}+\frac {\left (a+b x^3\right )^{5/3}}{5 b^2 d}+\frac {c^2 \log \left (c+d x^3\right )}{6 d^{8/3} \sqrt [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 )}{2 d^{8/3} \sqrt [3]{b c-a 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^{8/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^3)^(1/3)*(c + d*x^3)),x]

[Out]

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

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

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Mathematica [C]  time = 0.09, size = 103, normalized size = 0.51 \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (3 a^2 d^2+5 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 (c-d x^3\right )+b^2 c \left (2 d x^3-5 c\right )\right )}{10 b^2 d^2 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/((a + b*x^3)^(1/3)*(c + d*x^3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.40, size = 247, normalized size = 1.22 \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (-3 a d-5 b c+2 b d x^3\right )}{10 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^{8/3} \sqrt [3]{b c-a d}}+\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^{8/3} \sqrt [3]{b c-a d}}-\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^{8/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8/((a + b*x^3)^(1/3)*(c + d*x^3)),x]

[Out]

((a + b*x^3)^(2/3)*(-5*b*c - 3*a*d + 2*b*d*x^3))/(10*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^(8/3)*(b*c - a*d)^(1/3)) - (c^2*Log[(b*c - a*d)^(1/3) + d^(1/
3)*(a + b*x^3)^(1/3)])/(3*d^(8/3)*(b*c - a*d)^(1/3)) + (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^(8/3)*(b*c - a*d)^(1/3))

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fricas [B]  time = 0.79, size = 768, normalized size = 3.78 \begin {gather*} \left [\frac {5 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{2} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}}\right ) - 10 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{2} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d - {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}\right ) + 15 \, \sqrt {\frac {1}{3}} {\left (b^{3} c^{3} d - a b^{2} c^{2} d^{2}\right )} \sqrt {\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \log \left (\frac {2 \, b d^{2} x^{3} - b c d + 3 \, a d^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c d - a d^{2}\right )} + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right )} \sqrt {\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} - 3 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{d x^{3} + c}\right ) - 3 \, {\left (5 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} - 3 \, a^{2} d^{4} - 2 \, {\left (b^{2} c d^{3} - a b d^{4}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{30 \, {\left (b^{3} c d^{4} - a b^{2} d^{5}\right )}}, \frac {5 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{2} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}}\right ) - 10 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{2} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d - {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}\right ) + 30 \, \sqrt {\frac {1}{3}} {\left (b^{3} c^{3} d - a b^{2} c^{2} d^{2}\right )} \sqrt {-\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}\right )} \sqrt {-\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}}}{d}\right ) - 3 \, {\left (5 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} - 3 \, a^{2} d^{4} - 2 \, {\left (b^{2} c d^{3} - a b d^{4}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{30 \, {\left (b^{3} c d^{4} - a b^{2} d^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[1/30*(5*(-b*c*d^2 + a*d^3)^(2/3)*b^2*c^2*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)) - 10*(-b*c*d^2 + a*d^3)^(2/3)*b^2*c^2*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a
*d^3)^(1/3)) + 15*sqrt(1/3)*(b^3*c^3*d - a*b^2*c^2*d^2)*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)) - 3*(5*b^2*c^2*d^2 - 2*a*b*c*d^3 - 3*a^2*d^4 - 2*(b^2*c*d^3 - a*b*
d^4)*x^3)*(b*x^3 + a)^(2/3))/(b^3*c*d^4 - a*b^2*d^5), 1/30*(5*(-b*c*d^2 + a*d^3)^(2/3)*b^2*c^2*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)) - 10*(-b*c*d^2 + a*d^3)^
(2/3)*b^2*c^2*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/3)) + 30*sqrt(1/3)*(b^3*c^3*d - a*b^2*c^2*d^2)*s
qrt(-(-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) - 3*(5*b^2*c^2*d^2 - 2*a*b*c*d^3 - 3*a^2*d^4 - 2*(b^2*c*d^3 -
a*b*d^4)*x^3)*(b*x^3 + a)^(2/3))/(b^3*c*d^4 - a*b^2*d^5)]

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giac [A]  time = 0.26, size = 313, normalized size = 1.54 \begin {gather*} -\frac {b^{12} c^{2} d^{3} \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^{13} c d^{5} - a b^{12} d^{6}\right )}} - \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 c d^{4} - \sqrt {3} a 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 c d^{4} - a d^{5}\right )}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{9} c d^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{8} d^{4} + 5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b^{8} d^{4}}{10 \, b^{10} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-1/3*b^12*c^2*d^3*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^13*c*d^5 - a*
b^12*d^6) - (-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*c*d^4 - sqrt(3)*a*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*c*d^4 - a*d^5) - 1/10*(5*(b*x^3 + a)
^(2/3)*b^9*c*d^3 - 2*(b*x^3 + a)^(5/3)*b^8*d^4 + 5*(b*x^3 + a)^(2/3)*a*b^8*d^4)/(b^10*d^5)

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maple [F]  time = 0.59, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\left (b \,x^{3}+a \right )^{\frac {1}{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)^(1/3)/(d*x^3+c),x)

[Out]

int(x^8/(b*x^3+a)^(1/3)/(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)^(1/3)/(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 = 267, normalized size = 1.32 \begin {gather*} \frac {{\left (b\,x^3+a\right )}^{5/3}}{5\,b^2\,d}-\left (\frac {a}{b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{2\,b^4\,d^2}\right )\,{\left (b\,x^3+a\right )}^{2/3}+\frac {c^2\,\ln \left (\frac {c^4\,{\left (b\,x^3+a\right )}^{1/3}}{d^3}+\frac {b\,c^5-a\,c^4\,d}{d^{10/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )}{3\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{1/3}}-\frac {\ln \left (\frac {c^4\,{\left (b\,x^3+a\right )}^{1/3}}{d^3}-\frac {c^4\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{10/3}}\right )\,\left (c^2+\sqrt {3}\,c^2\,1{}\mathrm {i}\right )}{6\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{1/3}}+\frac {c^2\,\ln \left (\frac {c^4\,{\left (b\,x^3+a\right )}^{1/3}}{d^3}-\frac {c^4\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{10/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{d^{8/3}\,{\left (a\,d-b\,c\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((a + b*x^3)^(1/3)*(c + d*x^3)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\sqrt [3]{a + b x^{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)**(1/3)/(d*x**3+c),x)

[Out]

Integral(x**8/((a + b*x**3)**(1/3)*(c + d*x**3)), x)

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