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

Optimal. Leaf size=220 \[ -\frac {\left (a+b x^3\right )^{4/3} (a d+b c)}{4 b^2 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^2 d}+\frac {c^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c^2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}+\frac {c^2 \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{10/3}}+\frac {c^2 \sqrt [3]{a+b x^3}}{d^3} \]

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Rubi [A]  time = 0.26, antiderivative size = 220, 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, 58, 617, 204, 31} \begin {gather*} -\frac {\left (a+b x^3\right )^{4/3} (a d+b c)}{4 b^2 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^2 d}+\frac {c^2 \sqrt [3]{a+b x^3}}{d^3}+\frac {c^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c^2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}+\frac {c^2 \sqrt [3]{b c-a d} \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^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(a + b*x^3)^(1/3))/d^3 - ((b*c + a*d)*(a + b*x^3)^(4/3))/(4*b^2*d^2) + (a + b*x^3)^(7/3)/(7*b^2*d) + (c^2
*(b*c - a*d)^(1/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(10/3)) +
(c^2*(b*c - a*d)^(1/3)*Log[c + d*x^3])/(6*d^(10/3)) - (c^2*(b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*
(a + b*x^3)^(1/3)])/(2*d^(10/3))

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 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 \sqrt [3]{a+b x^3}}{c+d x^3} \, 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) \sqrt [3]{a+b x}}{b d^2}+\frac {(a+b x)^{4/3}}{b d}+\frac {c^2 \sqrt [3]{a+b x}}{d^2 (c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {(b c+a d) \left (a+b x^3\right )^{4/3}}{4 b^2 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^2 d}+\frac {c^2 \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{c+d x} \, dx,x,x^3\right )}{3 d^2}\\ &=\frac {c^2 \sqrt [3]{a+b x^3}}{d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{4/3}}{4 b^2 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^2 d}-\frac {\left (c^2 (b c-a d)\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 d^3}\\ &=\frac {c^2 \sqrt [3]{a+b x^3}}{d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{4/3}}{4 b^2 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^2 d}+\frac {c^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {\left (c^2 \sqrt [3]{b c-a d}\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^{10/3}}-\frac {\left (c^2 (b c-a d)^{2/3}\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^{11/3}}\\ &=\frac {c^2 \sqrt [3]{a+b x^3}}{d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{4/3}}{4 b^2 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^2 d}+\frac {c^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c^2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}-\frac {\left (c^2 \sqrt [3]{b c-a d}\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^{10/3}}\\ &=\frac {c^2 \sqrt [3]{a+b x^3}}{d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{4/3}}{4 b^2 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^2 d}+\frac {c^2 \sqrt [3]{b c-a d} \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^{10/3}}+\frac {c^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c^2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 230, normalized size = 1.05 \begin {gather*} \frac {-\frac {21 d \left (a+b x^3\right )^{4/3} (a d+b c)}{b^2}+\frac {12 d^2 \left (a+b x^3\right )^{7/3}}{b^2}+\frac {14 c^2 \sqrt [3]{b c-a d} \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}}+84 c^2 \sqrt [3]{a+b x^3}}{84 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(84*c^2*(a + b*x^3)^(1/3) - (21*d*(b*c + a*d)*(a + b*x^3)^(4/3))/b^2 + (12*d^2*(a + b*x^3)^(7/3))/b^2 + (14*c^
2*(b*c - a*d)^(1/3)*(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)] + 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)]))/d^(1/3))/(84*d^3)

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IntegrateAlgebraic [A]  time = 0.37, size = 283, normalized size = 1.29 \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (-3 a^2 d^2-7 a b c d+a b d^2 x^3+28 b^2 c^2-7 b^2 c d x^3+4 b^2 d^2 x^6\right )}{28 b^2 d^3}-\frac {c^2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{10/3}}+\frac {c^2 \sqrt [3]{b c-a d} \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^{10/3}}+\frac {c^2 \sqrt [3]{b c-a d} \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^{10/3}} \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)^(1/3)*(28*b^2*c^2 - 7*a*b*c*d - 3*a^2*d^2 - 7*b^2*c*d*x^3 + a*b*d^2*x^3 + 4*b^2*d^2*x^6))/(28*b^2
*d^3) + (c^2*(b*c - a*d)^(1/3)*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^(10/3)) - (c^2*(b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*d^(10/3)) +
(c^2*(b*c - a*d)^(1/3)*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^(10/3))

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fricas [A]  time = 0.46, size = 282, normalized size = 1.28 \begin {gather*} -\frac {28 \, \sqrt {3} b^{2} c^{2} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 14 \, b^{2} c^{2} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \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 ) - 28 \, b^{2} c^{2} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) - 3 \, {\left (4 \, b^{2} d^{2} x^{6} + 28 \, b^{2} c^{2} - 7 \, a b c d - 3 \, a^{2} d^{2} - {\left (7 \, b^{2} c d - a b d^{2}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{84 \, b^{2} d^{3}} \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]

-1/84*(28*sqrt(3)*b^2*c^2*(-(b*c - a*d)/d)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*d*(-(b*c - a*d)/d)^(
2/3) - sqrt(3)*(b*c - a*d))/(b*c - a*d)) + 14*b^2*c^2*(-(b*c - a*d)/d)^(1/3)*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)) - 28*b^2*c^2*(-(b*c - a*d)/d)^(1/3)*log((b*x^3 + a)^
(1/3) - (-(b*c - a*d)/d)^(1/3)) - 3*(4*b^2*d^2*x^6 + 28*b^2*c^2 - 7*a*b*c*d - 3*a^2*d^2 - (7*b^2*c*d - a*b*d^2
)*x^3)*(b*x^3 + a)^(1/3))/(b^2*d^3)

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giac [A]  time = 0.25, size = 320, normalized size = 1.45 \begin {gather*} \frac {{\left (b^{17} c^{3} d^{4} - a b^{16} c^{2} d^{5}\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^{17} c d^{7} - a b^{16} d^{8}\right )}} - \frac {\sqrt {3} {\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 )}{3 \, 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 \, d^{4}} + \frac {28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{14} c^{2} d^{4} - 7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{13} c d^{5} + 4 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{12} d^{6} - 7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a b^{12} d^{6}}{28 \, b^{14} d^{7}} \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^17*c^3*d^4 - a*b^16*c^2*d^5)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3))
)/(b^17*c*d^7 - a*b^16*d^8) - 1/3*sqrt(3)*(-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))/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))/d^4 + 1/28*(28*(b*x^3 + a)^(1/3)*b^14*c
^2*d^4 - 7*(b*x^3 + a)^(4/3)*b^13*c*d^5 + 4*(b*x^3 + a)^(7/3)*b^12*d^6 - 7*(b*x^3 + a)^(4/3)*a*b^12*d^6)/(b^14
*d^7)

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maple [F]  time = 0.63, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{8}}{d \,x^{3}+c}\, 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 = 4.93, size = 336, normalized size = 1.53 \begin {gather*} \left (\frac {a^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 )}{b^2\,d}\right )\,{\left (b\,x^3+a\right )}^{1/3}-\left (\frac {a}{2\,b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{4\,b^4\,d^2}\right )\,{\left (b\,x^3+a\right )}^{4/3}+\frac {{\left (b\,x^3+a\right )}^{7/3}}{7\,b^2\,d}+\frac {c^2\,\ln \left ({\left (a\,d-b\,c\right )}^{1/3}-d^{1/3}\,{\left (b\,x^3+a\right )}^{1/3}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{10/3}}-\frac {c^2\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (b\,c^3-a\,c^2\,d\right )}{d}-\frac {3\,c^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{10/3}}+\frac {c^2\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (b\,c^3-a\,c^2\,d\right )}{d}+\frac {9\,c^2\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{4/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{10/3}} \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^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))/(b^2*d))*(a + b*x^3)^(1/3) -
(a/(2*b^2*d) + (b^3*c - a*b^2*d)/(4*b^4*d^2))*(a + b*x^3)^(4/3) + (a + b*x^3)^(7/3)/(7*b^2*d) + (c^2*log((a*d
- b*c)^(1/3) - d^(1/3)*(a + b*x^3)^(1/3))*(a*d - b*c)^(1/3))/(3*d^(10/3)) - (c^2*log((3*(a + b*x^3)^(1/3)*(b*c
^3 - a*c^2*d))/d - (3*c^2*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(4/3))/d^(4/3))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c
)^(1/3))/(3*d^(10/3)) + (c^2*log((3*(a + b*x^3)^(1/3)*(b*c^3 - a*c^2*d))/d + (9*c^2*((3^(1/2)*1i)/6 - 1/6)*(a*
d - b*c)^(4/3))/d^(4/3))*((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(1/3))/d^(10/3)

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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}}}{c + d x^{3}}\, 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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