3.5.14 \(\int \frac {x^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\)

Optimal. Leaf size=264 \[ \frac {\left (a+b x^3\right )^{4/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{4 b^3 d^3}-\frac {\left (a+b x^3\right )^{7/3} (2 a d+b c)}{7 b^3 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^3 d}-\frac {c^3 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^3 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}-\frac {c^3 \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{13/3}}-\frac {c^3 \sqrt [3]{a+b x^3}}{d^4} \]

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Rubi [A]  time = 0.39, antiderivative size = 264, 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} \left (a^2 d^2+a b c d+b^2 c^2\right )}{4 b^3 d^3}-\frac {\left (a+b x^3\right )^{7/3} (2 a d+b c)}{7 b^3 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^3 d}-\frac {c^3 \sqrt [3]{a+b x^3}}{d^4}-\frac {c^3 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^3 \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^{13/3}}-\frac {c^3 \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^{13/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^11*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

-((c^3*(a + b*x^3)^(1/3))/d^4) + ((b^2*c^2 + a*b*c*d + a^2*d^2)*(a + b*x^3)^(4/3))/(4*b^3*d^3) - ((b*c + 2*a*d
)*(a + b*x^3)^(7/3))/(7*b^3*d^2) + (a + b*x^3)^(10/3)/(10*b^3*d) - (c^3*(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^(13/3)) - (c^3*(b*c - a*d)^(1/3)*Log[c + d*x^3])
/(6*d^(13/3)) + (c^3*(b*c - a*d)^(1/3)*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^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{a+b x}}{c+d x} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \sqrt [3]{a+b x}}{b^2 d^3}+\frac {(-b c-2 a d) (a+b x)^{4/3}}{b^2 d^2}+\frac {(a+b x)^{7/3}}{b^2 d}-\frac {c^3 \sqrt [3]{a+b x}}{d^3 (c+d x)}\right ) \, dx,x,x^3\right )\\ &=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{4/3}}{4 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{7/3}}{7 b^3 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^3 d}-\frac {c^3 \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{c+d x} \, dx,x,x^3\right )}{3 d^3}\\ &=-\frac {c^3 \sqrt [3]{a+b x^3}}{d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{4/3}}{4 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{7/3}}{7 b^3 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^3 d}+\frac {\left (c^3 (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^4}\\ &=-\frac {c^3 \sqrt [3]{a+b x^3}}{d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{4/3}}{4 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{7/3}}{7 b^3 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^3 d}-\frac {c^3 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {\left (c^3 \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^{13/3}}+\frac {\left (c^3 (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^{14/3}}\\ &=-\frac {c^3 \sqrt [3]{a+b x^3}}{d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{4/3}}{4 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{7/3}}{7 b^3 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^3 d}-\frac {c^3 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^3 \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^{13/3}}+\frac {\left (c^3 \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^{13/3}}\\ &=-\frac {c^3 \sqrt [3]{a+b x^3}}{d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{4/3}}{4 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{7/3}}{7 b^3 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^3 d}-\frac {c^3 \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^{13/3}}-\frac {c^3 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^3 \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^{13/3}}\\ \end {align*}

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Mathematica [A]  time = 0.68, size = 270, normalized size = 1.02 \begin {gather*} \frac {\frac {105 d \left (a+b x^3\right )^{4/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{b^3}-\frac {60 d^2 \left (a+b x^3\right )^{7/3} (2 a d+b c)}{b^3}+\frac {42 d^3 \left (a+b x^3\right )^{10/3}}{b^3}-\frac {70 c^3 \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}}-420 c^3 \sqrt [3]{a+b x^3}}{420 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^11*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.50, size = 340, normalized size = 1.29 \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (9 a^3 d^3+15 a^2 b c d^2-3 a^2 b d^3 x^3+35 a b^2 c^2 d-5 a b^2 c d^2 x^3+2 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^3 d^4}+\frac {c^3 \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^{13/3}}-\frac {c^3 \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^{13/3}}-\frac {c^3 \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^{13/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^11*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

((a + b*x^3)^(1/3)*(-140*b^3*c^3 + 35*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 9*a^3*d^3 + 35*b^3*c^2*d*x^3 - 5*a*b^2*c*
d^2*x^3 - 3*a^2*b*d^3*x^3 - 20*b^3*c*d^2*x^6 + 2*a*b^2*d^3*x^6 + 14*b^3*d^3*x^9))/(140*b^3*d^4) - (c^3*(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^(13/3)) +
(c^3*(b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*d^(13/3)) - (c^3*(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^(13
/3))

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fricas [A]  time = 0.45, size = 325, normalized size = 1.23 \begin {gather*} -\frac {140 \, \sqrt {3} b^{3} c^{3} \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 \, b^{3} c^{3} \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 \, b^{3} c^{3} \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} - a b^{2} d^{3}\right )} x^{6} - 140 \, b^{3} c^{3} + 35 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} + 9 \, a^{3} d^{3} + {\left (35 \, b^{3} c^{2} d - 5 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{420 \, b^{3} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

-1/420*(140*sqrt(3)*b^3*c^3*((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*((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*((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 - a*b^2*d^3)*x^6 - 140*b^3*c^3 + 35*a*b^2*c^
2*d + 15*a^2*b*c*d^2 + 9*a^3*d^3 + (35*b^3*c^2*d - 5*a*b^2*c*d^2 - 3*a^2*b*d^3)*x^3)*(b*x^3 + a)^(1/3))/(b^3*d
^4)

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giac [A]  time = 0.29, size = 379, normalized size = 1.44 \begin {gather*} -\frac {{\left (b^{34} c^{4} d^{6} - a b^{33} c^{3} d^{7}\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^{34} c d^{10} - a b^{33} d^{11}\right )}} + \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} c^{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 d^{2} + a d^{3}\right )}^{\frac {1}{3}} c^{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^{30} c^{3} d^{6} - 35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{29} c^{2} d^{7} + 20 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{28} c d^{8} - 35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a b^{28} c d^{8} - 14 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} b^{27} d^{9} + 40 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a b^{27} d^{9} - 35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2} b^{27} d^{9}}{140 \, b^{30} d^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-1/3*(b^34*c^4*d^6 - a*b^33*c^3*d^7)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)
))/(b^34*c*d^10 - a*b^33*d^11) + 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(1/3)*c^3*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)^(1/3)*c^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
^30*c^3*d^6 - 35*(b*x^3 + a)^(4/3)*b^29*c^2*d^7 + 20*(b*x^3 + a)^(7/3)*b^28*c*d^8 - 35*(b*x^3 + a)^(4/3)*a*b^2
8*c*d^8 - 14*(b*x^3 + a)^(10/3)*b^27*d^9 + 40*(b*x^3 + a)^(7/3)*a*b^27*d^9 - 35*(b*x^3 + a)^(4/3)*a^2*b^27*d^9
)/(b^30*d^10)

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maple [F]  time = 0.65, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{11}}{d \,x^{3}+c}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(b*x^3+a)^(1/3)/(d*x^3+c),x)

[Out]

int(x^11*(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^11*(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 = 4.98, size = 442, normalized size = 1.67 \begin {gather*} \left (\frac {3\,a^2}{4\,b^3\,d}+\frac {\left (\frac {3\,a}{b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{b^6\,d^2}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{4\,b^3\,d}\right )\,{\left (b\,x^3+a\right )}^{4/3}-\left (\frac {3\,a}{7\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{7\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{7/3}-{\left (b\,x^3+a\right )}^{1/3}\,\left (\frac {a^3}{b^3\,d}+\frac {\left (\frac {3\,a^2}{b^3\,d}+\frac {\left (\frac {3\,a}{b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{b^6\,d^2}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{b^3\,d}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{b^3\,d}\right )+\frac {{\left (b\,x^3+a\right )}^{10/3}}{10\,b^3\,d}-\frac {c^3\,\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^{13/3}}-\frac {c^3\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (b\,c^4-a\,c^3\,d\right )}{d^2}+\frac {3\,c^3\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{7/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{13/3}}+\frac {c^3\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (b\,c^4-a\,c^3\,d\right )}{d^2}-\frac {9\,c^3\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{7/3}}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{13/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^11*(a + b*x^3)^(1/3))/(c + d*x^3),x)

[Out]

((3*a^2)/(4*b^3*d) + (((3*a)/(b^3*d) + (b^4*c - a*b^3*d)/(b^6*d^2))*(b^4*c - a*b^3*d))/(4*b^3*d))*(a + b*x^3)^
(4/3) - ((3*a)/(7*b^3*d) + (b^4*c - a*b^3*d)/(7*b^6*d^2))*(a + b*x^3)^(7/3) - (a + b*x^3)^(1/3)*(a^3/(b^3*d) +
(((3*a^2)/(b^3*d) + (((3*a)/(b^3*d) + (b^4*c - a*b^3*d)/(b^6*d^2))*(b^4*c - a*b^3*d))/(b^3*d))*(b^4*c - a*b^3
*d))/(b^3*d)) + (a + b*x^3)^(10/3)/(10*b^3*d) - (c^3*log((a*d - b*c)^(1/3) - d^(1/3)*(a + b*x^3)^(1/3))*(a*d -
b*c)^(1/3))/(3*d^(13/3)) - (c^3*log((3*(a + b*x^3)^(1/3)*(b*c^4 - a*c^3*d))/d^2 + (3*c^3*((3^(1/2)*1i)/2 - 1/
2)*(a*d - b*c)^(4/3))/d^(7/3))*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(1/3))/(3*d^(13/3)) + (c^3*log((3*(a + b*x^3
)^(1/3)*(b*c^4 - a*c^3*d))/d^2 - (9*c^3*((3^(1/2)*1i)/6 + 1/6)*(a*d - b*c)^(4/3))/d^(7/3))*((3^(1/2)*1i)/6 + 1
/6)*(a*d - b*c)^(1/3))/d^(13/3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11} \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**11*(b*x**3+a)**(1/3)/(d*x**3+c),x)

[Out]

Integral(x**11*(a + b*x**3)**(1/3)/(c + d*x**3), x)

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