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

Optimal. Leaf size=251 \[ -\frac {\left (a+b x^3\right )^{7/3} (a d+b c)}{7 b^2 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac {c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}-\frac {c^2 (b c-a d)^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{13/3}}-\frac {c^2 \sqrt [3]{a+b x^3} (b c-a d)}{d^4}+\frac {c^2 \left (a+b x^3\right )^{4/3}}{4 d^3} \]

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Rubi [A]  time = 0.36, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 9, 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 )^{7/3} (a d+b c)}{7 b^2 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^2 d}+\frac {c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac {c^2 \sqrt [3]{a+b x^3} (b c-a d)}{d^4}-\frac {c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}-\frac {c^2 (b c-a d)^{4/3} \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^{13/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^2*(b*c - a*d)*(a + b*x^3)^(1/3))/d^4) + (c^2*(a + b*x^3)^(4/3))/(4*d^3) - ((b*c + a*d)*(a + b*x^3)^(7/3))
/(7*b^2*d^2) + (a + b*x^3)^(10/3)/(10*b^2*d) - (c^2*(b*c - a*d)^(4/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3)
)/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(13/3)) - (c^2*(b*c - a*d)^(4/3)*Log[c + d*x^3])/(6*d^(13/3)) + (c^2
*(b*c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(13/3))

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 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, 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 {(-b c-a d) (a+b x)^{4/3}}{b d^2}+\frac {(a+b x)^{7/3}}{b d}+\frac {c^2 (a+b x)^{4/3}}{d^2 (c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^2 d}+\frac {c^2 \operatorname {Subst}\left (\int \frac {(a+b x)^{4/3}}{c+d x} \, dx,x,x^3\right )}{3 d^2}\\ &=\frac {c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac {\left (c^2 (b c-a d)\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{c+d x} \, dx,x,x^3\right )}{3 d^3}\\ &=-\frac {c^2 (b c-a d) \sqrt [3]{a+b x^3}}{d^4}+\frac {c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^2 d}+\frac {\left (c^2 (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 d^4}\\ &=-\frac {c^2 (b c-a d) \sqrt [3]{a+b x^3}}{d^4}+\frac {c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac {c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {\left (c^2 (b c-a d)^{4/3}\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^{13/3}}+\frac {\left (c^2 (b c-a d)^{5/3}\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^{14/3}}\\ &=-\frac {c^2 (b c-a d) \sqrt [3]{a+b x^3}}{d^4}+\frac {c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac {c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}+\frac {\left (c^2 (b c-a d)^{4/3}\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^{13/3}}\\ &=-\frac {c^2 (b c-a d) \sqrt [3]{a+b x^3}}{d^4}+\frac {c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac {c^2 (b c-a d)^{4/3} \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^{13/3}}-\frac {c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 258, normalized size = 1.03 \begin {gather*} \frac {-\frac {60 d \left (a+b x^3\right )^{7/3} (a d+b c)}{b^2}+\frac {42 d^2 \left (a+b x^3\right )^{10/3}}{b^2}-\frac {70 c^2 (b c-a d) \left (\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 )+6 \sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{d^{4/3}}+105 c^2 \left (a+b x^3\right )^{4/3}}{420 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(105*c^2*(a + b*x^3)^(4/3) - (60*d*(b*c + a*d)*(a + b*x^3)^(7/3))/b^2 + (42*d^2*(a + b*x^3)^(10/3))/b^2 - (70*
c^2*(b*c - a*d)*(6*d^(1/3)*(a + b*x^3)^(1/3) + (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^(4/3))/(420*d^3)

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IntegrateAlgebraic [A]  time = 0.46, size = 340, normalized size = 1.35 \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (-6 a^3 d^3-20 a^2 b c d^2+2 a^2 b d^3 x^3+175 a b^2 c^2 d-40 a b^2 c d^2 x^3+22 a b^2 d^3 x^6-140 b^3 c^3+35 b^3 c^2 d x^3-20 b^3 c d^2 x^6+14 b^3 d^3 x^9\right )}{140 b^2 d^4}+\frac {c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{13/3}}-\frac {c^2 (b c-a d)^{4/3} \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^{13/3}}-\frac {c^2 (b c-a d)^{4/3} \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^{13/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^(1/3)*(-140*b^3*c^3 + 175*a*b^2*c^2*d - 20*a^2*b*c*d^2 - 6*a^3*d^3 + 35*b^3*c^2*d*x^3 - 40*a*b^2*
c*d^2*x^3 + 2*a^2*b*d^3*x^3 - 20*b^3*c*d^2*x^6 + 22*a*b^2*d^3*x^6 + 14*b^3*d^3*x^9))/(140*b^2*d^4) - (c^2*(b*c
- a*d)^(4/3)*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^(13/3)
) + (c^2*(b*c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*d^(13/3)) - (c^2*(b*c - a*d)
^(4/3)*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^
(13/3))

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fricas [A]  time = 0.56, size = 369, normalized size = 1.47 \begin {gather*} \frac {140 \, \sqrt {3} {\left (b^{3} c^{3} - a b^{2} c^{2} d\right )} \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 ) + 70 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d\right )} \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 ) - 140 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d\right )} \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 (14 \, b^{3} d^{3} x^{9} - 2 \, {\left (10 \, b^{3} c d^{2} - 11 \, a b^{2} d^{3}\right )} x^{6} - 140 \, b^{3} c^{3} + 175 \, a b^{2} c^{2} d - 20 \, a^{2} b c d^{2} - 6 \, a^{3} d^{3} + {\left (35 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 2 \, a^{2} b d^{3}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{420 \, b^{2} d^{4}} \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="fricas")

[Out]

1/420*(140*sqrt(3)*(b^3*c^3 - a*b^2*c^2*d)*(-(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)) + 70*(b^3*c^3 - a*b^2*c^2*d)*(-(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)) - 140*(b^3*c^3 - a*
b^2*c^2*d)*(-(b*c - a*d)/d)^(1/3)*log((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)) + 3*(14*b^3*d^3*x^9 - 2*(10*
b^3*c*d^2 - 11*a*b^2*d^3)*x^6 - 140*b^3*c^3 + 175*a*b^2*c^2*d - 20*a^2*b*c*d^2 - 6*a^3*d^3 + (35*b^3*c^2*d - 4
0*a*b^2*c*d^2 + 2*a^2*b*d^3)*x^3)*(b*x^3 + a)^(1/3))/(b^2*d^4)

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giac [A]  time = 0.30, size = 394, normalized size = 1.57 \begin {gather*} -\frac {{\left (b^{24} c^{4} d^{6} - 2 \, a b^{23} c^{3} d^{7} + a^{2} b^{22} c^{2} d^{8}\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^{23} c d^{10} - a b^{22} d^{11}\right )}} + \frac {\sqrt {3} {\left (b c^{3} - a c^{2} d\right )} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} \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^{3} - a c^{2} d\right )} {\left (-b c d^{2} + a d^{3}\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 )}{6 \, d^{5}} - \frac {140 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{21} c^{3} d^{6} - 35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{20} c^{2} d^{7} - 140 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a b^{20} c^{2} d^{7} + 20 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{19} c d^{8} - 14 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} b^{18} d^{9} + 20 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a b^{18} d^{9}}{140 \, b^{20} d^{10}} \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]

-1/3*(b^24*c^4*d^6 - 2*a*b^23*c^3*d^7 + a^2*b^22*c^2*d^8)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (
-(b*c - a*d)/d)^(1/3)))/(b^23*c*d^10 - a*b^22*d^11) + 1/3*sqrt(3)*(b*c^3 - a*c^2*d)*(-b*c*d^2 + a*d^3)^(1/3)*a
rctan(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^3 - a*
c^2*d)*(-b*c*d^2 + a*d^3)^(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))/d^5 - 1/140*(140*(b*x^3 + a)^(1/3)*b^21*c^3*d^6 - 35*(b*x^3 + a)^(4/3)*b^20*c^2*d^7 - 140*(b*x^3
+ a)^(1/3)*a*b^20*c^2*d^7 + 20*(b*x^3 + a)^(7/3)*b^19*c*d^8 - 14*(b*x^3 + a)^(10/3)*b^18*d^9 + 20*(b*x^3 + a)^
(7/3)*a*b^18*d^9)/(b^20*d^10)

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{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)^(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 = 5.12, size = 477, normalized size = 1.90 \begin {gather*} \left (\frac {a^2}{4\,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 )}{4\,b^2\,d}\right )\,{\left (b\,x^3+a\right )}^{4/3}-\left (\frac {2\,a}{7\,b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{7\,b^4\,d^2}\right )\,{\left (b\,x^3+a\right )}^{7/3}+\frac {{\left (b\,x^3+a\right )}^{10/3}}{10\,b^2\,d}+\frac {c^2\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c^2\,d^2-2\,a\,b\,c^3\,d+b^2\,c^4\right )}{d^2}-\frac {c^2\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{13/3}}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{13/3}}-\frac {\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 (b^3\,c-a\,b^2\,d\right )}{b^2\,d}-\frac {c^2\,\ln \left (\frac {3\,c^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{7/3}}+\frac {3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{13/3}}+\frac {c^2\,\ln \left (\frac {3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^2}-\frac {9\,c^2\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{7/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{13/3}} \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^2/(4*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))/(4*b^2*d))*(a + b*x^3)^(4/3
) - ((2*a)/(7*b^2*d) + (b^3*c - a*b^2*d)/(7*b^4*d^2))*(a + b*x^3)^(7/3) + (a + b*x^3)^(10/3)/(10*b^2*d) + (c^2
*log((3*(a + b*x^3)^(1/3)*(b^2*c^4 + a^2*c^2*d^2 - 2*a*b*c^3*d))/d^2 - (c^2*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c
*d^2))/(3*d^(13/3)))*(a*d - b*c)^(4/3))/(3*d^(13/3)) - ((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)*(b^3*c - a*b^2*d))/(b^2*d) - (c^2*log((3*c^2*((3^(1/2)*1
i)/2 + 1/2)*(a*d - b*c)^(7/3))/d^(7/3) + (3*c^2*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^2)*((3^(1/2)*1i)/2 + 1/2)*(
a*d - b*c)^(4/3))/(3*d^(13/3)) + (c^2*log((3*c^2*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^2 - (9*c^2*((3^(1/2)*1i)/6
- 1/6)*(a*d - b*c)^(7/3))/d^(7/3))*((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(4/3))/d^(13/3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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]

Timed out

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