3.5.68 \(\int \frac {x^{11}}{(a+b x^3)^{2/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=241 \[ \frac {\sqrt [3]{a+b x^3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{b^3 d^3}-\frac {\left (a+b x^3\right )^{4/3} (2 a d+b c)}{4 b^3 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^3 d}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{10/3} (b c-a d)^{2/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3} (b c-a d)^{2/3}}+\frac {c^3 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{10/3} (b c-a d)^{2/3}} \]

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Rubi [A]  time = 0.26, antiderivative size = 241, 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, 58, 617, 204, 31} \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{b^3 d^3}-\frac {\left (a+b x^3\right )^{4/3} (2 a d+b c)}{4 b^3 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^3 d}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{10/3} (b c-a d)^{2/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3} (b c-a d)^{2/3}}+\frac {c^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^{10/3} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^11/((a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^3}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {b^2 c^2+a b c d+a^2 d^2}{b^2 d^3 (a+b x)^{2/3}}+\frac {(-b c-2 a d) \sqrt [3]{a+b x}}{b^2 d^2}+\frac {(a+b x)^{4/3}}{b^2 d}-\frac {c^3}{d^3 (a+b x)^{2/3} (c+d x)}\right ) \, dx,x,x^3\right )\\ &=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \sqrt [3]{a+b x^3}}{b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{4/3}}{4 b^3 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^3 d}-\frac {c^3 \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 d^3}\\ &=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \sqrt [3]{a+b x^3}}{b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{4/3}}{4 b^3 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^3 d}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{10/3} (b c-a d)^{2/3}}-\frac {c^3 \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} (b c-a d)^{2/3}}-\frac {c^3 \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} \sqrt [3]{b c-a d}}\\ &=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \sqrt [3]{a+b x^3}}{b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{4/3}}{4 b^3 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^3 d}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{10/3} (b c-a d)^{2/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3} (b c-a d)^{2/3}}-\frac {c^3 \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} (b c-a d)^{2/3}}\\ &=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \sqrt [3]{a+b x^3}}{b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{4/3}}{4 b^3 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^3 d}+\frac {c^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^{10/3} (b c-a d)^{2/3}}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{10/3} (b c-a d)^{2/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3} (b c-a d)^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 251, normalized size = 1.04 \begin {gather*} \frac {\frac {84 \sqrt [3]{a+b x^3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{b^3}-\frac {21 d \left (a+b x^3\right )^{4/3} (2 a d+b c)}{b^3}+\frac {12 d^2 \left (a+b x^3\right )^{7/3}}{b^3}+\frac {14 c^3 \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} (b c-a d)^{2/3}}}{84 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^11/((a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.50, size = 284, normalized size = 1.18 \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (18 a^2 d^2+21 a b c d-6 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^3 d^3}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{10/3} (b c-a d)^{2/3}}+\frac {c^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^{10/3} (b c-a d)^{2/3}}+\frac {c^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^{10/3} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^11/((a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

((a + b*x^3)^(1/3)*(28*b^2*c^2 + 21*a*b*c*d + 18*a^2*d^2 - 7*b^2*c*d*x^3 - 6*a*b*d^2*x^3 + 4*b^2*d^2*x^6))/(28
*b^3*d^3) + (c^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)*(b*c - a*d)^(2/3)) - (c^3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*d^(10/3)*(b*c - a*d)^(2/3
)) + (c^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)*(b*c - a*d)^(2/3))

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fricas [B]  time = 0.83, size = 1322, normalized size = 5.49

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[1/84*(14*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b^3*c^3*log(-(b*x^3 + a)^(2/3)*(b*c*d - a*d^2) + (-b^2*c^
2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)*(b*c - a*d) + (-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*(b*x^3 + a)^(1/3))
- 28*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b^3*c^3*log(-(b*x^3 + a)^(1/3)*(b*c*d - a*d^2) - (-b^2*c^2*d
+ 2*a*b*c*d^2 - a^2*d^3)^(2/3)) - 42*sqrt(1/3)*(b^4*c^4*d - a*b^3*c^3*d^2)*sqrt((-b^2*c^2*d + 2*a*b*c*d^2 - a^
2*d^3)^(1/3)/d)*log((b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2 - 2*(b^2*c*d - a*b*d^2)*x^3 - 3*sqrt(1/3)*(2*(b*x^3 + a)^
(2/3)*(b*c*d - a*d^2) + (-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)*(b*c - a*d) + (-b^2*c^2*d + 2*a*b*c*d^2 - a
^2*d^3)^(2/3)*(b*x^3 + a)^(1/3))*sqrt((-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)/d) - 3*(-b^2*c^2*d + 2*a*b*c*
d^2 - a^2*d^3)^(1/3)*(b*x^3 + a)^(1/3)*(b*c - a*d))/(d*x^3 + c)) + 3*(28*b^4*c^4*d - 35*a*b^3*c^3*d^2 + 4*a^2*
b^2*c^2*d^3 - 15*a^3*b*c*d^4 + 18*a^4*d^5 + 4*(b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)*x^6 - (7*b^4*c^3*d^2
- 8*a*b^3*c^2*d^3 - 5*a^2*b^2*c*d^4 + 6*a^3*b*d^5)*x^3)*(b*x^3 + a)^(1/3))/(b^5*c^2*d^4 - 2*a*b^4*c*d^5 + a^2
*b^3*d^6), 1/84*(14*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b^3*c^3*log(-(b*x^3 + a)^(2/3)*(b*c*d - a*d^2)
+ (-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)*(b*c - a*d) + (-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*(b*x^3 +
a)^(1/3)) - 28*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b^3*c^3*log(-(b*x^3 + a)^(1/3)*(b*c*d - a*d^2) - (-
b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)) - 84*sqrt(1/3)*(b^4*c^4*d - a*b^3*c^3*d^2)*sqrt(-(-b^2*c^2*d + 2*a*b
*c*d^2 - a^2*d^3)^(1/3)/d)*arctan(sqrt(1/3)*((-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)*(b*c - a*d) + 2*(-b^2*
c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*(b*x^3 + a)^(1/3))*sqrt(-(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3)/d)/(b
^2*c^2 - 2*a*b*c*d + a^2*d^2)) + 3*(28*b^4*c^4*d - 35*a*b^3*c^3*d^2 + 4*a^2*b^2*c^2*d^3 - 15*a^3*b*c*d^4 + 18*
a^4*d^5 + 4*(b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)*x^6 - (7*b^4*c^3*d^2 - 8*a*b^3*c^2*d^3 - 5*a^2*b^2*c*d
^4 + 6*a^3*b*d^5)*x^3)*(b*x^3 + a)^(1/3))/(b^5*c^2*d^4 - 2*a*b^4*c*d^5 + a^2*b^3*d^6)]

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giac [A]  time = 0.26, size = 372, normalized size = 1.54 \begin {gather*} \frac {b^{24} c^{3} d^{4} \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^{25} c d^{7} - a b^{24} d^{8}\right )}} - \frac {{\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 )}{\sqrt {3} b c d^{4} - \sqrt {3} a 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 \, {\left (b c d^{4} - a d^{5}\right )}} + \frac {28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{20} c^{2} d^{4} - 7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{19} c d^{5} + 28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a b^{19} c d^{5} + 4 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{18} d^{6} - 14 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a b^{18} d^{6} + 28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2} b^{18} d^{6}}{28 \, b^{21} d^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")

[Out]

1/3*b^24*c^3*d^4*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^25*c*d^7 - a*b
^24*d^8) - (-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))/(sqrt(3)*b*c*d^4 - sqrt(3)*a*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))/(b*c*d^4 - a*d^5) + 1/28*(28*(b*x^3 + a)
^(1/3)*b^20*c^2*d^4 - 7*(b*x^3 + a)^(4/3)*b^19*c*d^5 + 28*(b*x^3 + a)^(1/3)*a*b^19*c*d^5 + 4*(b*x^3 + a)^(7/3)
*b^18*d^6 - 14*(b*x^3 + a)^(4/3)*a*b^18*d^6 + 28*(b*x^3 + a)^(1/3)*a^2*b^18*d^6)/(b^21*d^7)

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (d \,x^{3}+c \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(b*x^3+a)^(2/3)/(d*x^3+c),x)

[Out]

int(x^11/(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 \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)^(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 = 5.01, size = 331, normalized size = 1.37 \begin {gather*} \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\,x^3+a\right )}^{1/3}-\left (\frac {3\,a}{4\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{4\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{4/3}+\frac {{\left (b\,x^3+a\right )}^{7/3}}{7\,b^3\,d}+\frac {\ln \left (\frac {3\,c^3\,{\left (b\,x^3+a\right )}^{1/3}}{d}+\frac {3\,c^3\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{2\,d^{4/3}}\right )\,\left (c^3+\sqrt {3}\,c^3\,1{}\mathrm {i}\right )}{6\,d^{10/3}\,{\left (a\,d-b\,c\right )}^{2/3}}-\frac {c^3\,\ln \left (\frac {3\,c^3\,{\left (b\,x^3+a\right )}^{1/3}}{d}-\frac {3\,c^3\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{4/3}}\right )}{3\,d^{10/3}\,{\left (a\,d-b\,c\right )}^{2/3}}-\frac {c^3\,\ln \left (\frac {3\,c^3\,{\left (b\,x^3+a\right )}^{1/3}}{d}-\frac {3\,c^3\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{4/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,d^{10/3}\,{\left (a\,d-b\,c\right )}^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/((a + b*x^3)^(2/3)*(c + d*x^3)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11}}{\left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(b*x**3+a)**(2/3)/(d*x**3+c),x)

[Out]

Integral(x**11/((a + b*x**3)**(2/3)*(c + d*x**3)), x)

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