3.5.80 \(\int \frac {x^{11}}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=253 \[ \frac {a^3}{b^3 \sqrt [3]{a+b x^3} (b c-a d)}-\frac {\left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^3 d^2}-\frac {a \left (a+b x^3\right )^{2/3}}{2 b^3 d}+\frac {\left (a+b x^3\right )^{5/3}}{5 b^3 d}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{8/3} (b c-a d)^{4/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} (b c-a d)^{4/3}}-\frac {c^3 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{8/3} (b c-a d)^{4/3}} \]

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Rubi [A]  time = 0.34, antiderivative size = 253, 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, 87, 43, 56, 617, 204, 31} \begin {gather*} \frac {a^3}{b^3 \sqrt [3]{a+b x^3} (b c-a d)}-\frac {\left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^3 d^2}-\frac {a \left (a+b x^3\right )^{2/3}}{2 b^3 d}+\frac {\left (a+b x^3\right )^{5/3}}{5 b^3 d}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{8/3} (b c-a d)^{4/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} (b c-a d)^{4/3}}-\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^{8/3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^11/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

a^3/(b^3*(b*c - a*d)*(a + b*x^3)^(1/3)) - (a*(a + b*x^3)^(2/3))/(2*b^3*d) - ((b*c + a*d)*(a + b*x^3)^(2/3))/(2
*b^3*d^2) + (a + b*x^3)^(5/3)/(5*b^3*d) - (c^3*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sq
rt[3]])/(Sqrt[3]*d^(8/3)*(b*c - a*d)^(4/3)) + (c^3*Log[c + d*x^3])/(6*d^(8/3)*(b*c - a*d)^(4/3)) - (c^3*Log[(b
*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(8/3)*(b*c - a*d)^(4/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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 )^{4/3} \left (c+d x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^3}{(a+b x)^{4/3} (c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {a^3}{b^2 (b c-a d) (a+b x)^{4/3}}+\frac {-b c-a d}{b^2 d^2 \sqrt [3]{a+b x}}+\frac {x}{b d \sqrt [3]{a+b x}}-\frac {c^3}{d^2 (-b c+a d) \sqrt [3]{a+b x} (c+d x)}\right ) \, dx,x,x^3\right )\\ &=\frac {a^3}{b^3 (b c-a d) \sqrt [3]{a+b x^3}}-\frac {(b c+a d) \left (a+b x^3\right )^{2/3}}{2 b^3 d^2}+\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt [3]{a+b x}} \, dx,x,x^3\right )}{3 b d}+\frac {c^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )}{3 d^2 (b c-a d)}\\ &=\frac {a^3}{b^3 (b c-a d) \sqrt [3]{a+b x^3}}-\frac {(b c+a d) \left (a+b x^3\right )^{2/3}}{2 b^3 d^2}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{8/3} (b c-a d)^{4/3}}+\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b \sqrt [3]{a+b x}}+\frac {(a+b x)^{2/3}}{b}\right ) \, dx,x,x^3\right )}{3 b d}-\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^{8/3} (b c-a d)^{4/3}}+\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^3 (b c-a d)}\\ &=\frac {a^3}{b^3 (b c-a d) \sqrt [3]{a+b x^3}}-\frac {a \left (a+b x^3\right )^{2/3}}{2 b^3 d}-\frac {(b c+a d) \left (a+b x^3\right )^{2/3}}{2 b^3 d^2}+\frac {\left (a+b x^3\right )^{5/3}}{5 b^3 d}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{8/3} (b c-a d)^{4/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} (b c-a d)^{4/3}}+\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^{8/3} (b c-a d)^{4/3}}\\ &=\frac {a^3}{b^3 (b c-a d) \sqrt [3]{a+b x^3}}-\frac {a \left (a+b x^3\right )^{2/3}}{2 b^3 d}-\frac {(b c+a d) \left (a+b x^3\right )^{2/3}}{2 b^3 d^2}+\frac {\left (a+b x^3\right )^{5/3}}{5 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^{8/3} (b c-a d)^{4/3}}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{8/3} (b c-a d)^{4/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} (b c-a d)^{4/3}}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 147, normalized size = 0.58 \begin {gather*} \frac {18 a^3 d^3+3 a^2 b d^2 \left (2 d x^3-c\right )+10 b^3 c^3 \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )-a b^2 d \left (5 c^2+c d x^3+2 d^2 x^6\right )+b^3 c \left (-10 c^2-5 c d x^3+2 d^2 x^6\right )}{10 b^3 d^3 \sqrt [3]{a+b x^3} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^11/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

(18*a^3*d^3 + 3*a^2*b*d^2*(-c + 2*d*x^3) + b^3*c*(-10*c^2 - 5*c*d*x^3 + 2*d^2*x^6) - a*b^2*d*(5*c^2 + c*d*x^3
+ 2*d^2*x^6) + 10*b^3*c^3*Hypergeometric2F1[-1/3, 1, 2/3, (d*(a + b*x^3))/(-(b*c) + a*d)])/(10*b^3*d^3*(b*c -
a*d)*(a + b*x^3)^(1/3))

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IntegrateAlgebraic [A]  time = 0.50, size = 322, normalized size = 1.27 \begin {gather*} -\frac {-18 a^3 d^2+3 a^2 b c d-6 a^2 b d^2 x^3+5 a b^2 c^2+a b^2 c d x^3+2 a b^2 d^2 x^6+5 b^3 c^2 x^3-2 b^3 c d x^6}{10 b^3 d^2 \sqrt [3]{a+b x^3} (b c-a d)}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{8/3} (b c-a d)^{4/3}}+\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^{8/3} (b c-a d)^{4/3}}-\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^{8/3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^11/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

-1/10*(5*a*b^2*c^2 + 3*a^2*b*c*d - 18*a^3*d^2 + 5*b^3*c^2*x^3 + a*b^2*c*d*x^3 - 6*a^2*b*d^2*x^3 - 2*b^3*c*d*x^
6 + 2*a*b^2*d^2*x^6)/(b^3*d^2*(b*c - a*d)*(a + b*x^3)^(1/3)) - (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^(8/3)*(b*c - a*d)^(4/3)) - (c^3*Log[(b*c - a*d)^(1/3) + d^(1/3
)*(a + b*x^3)^(1/3)])/(3*d^(8/3)*(b*c - a*d)^(4/3)) + (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^(8/3)*(b*c - a*d)^(4/3))

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fricas [B]  time = 0.73, size = 1141, normalized size = 4.51

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[-1/30*(15*sqrt(1/3)*(a*b^4*c^4*d - a^2*b^3*c^3*d^2 + (b^5*c^4*d - a*b^4*c^3*d^2)*x^3)*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)) - 5*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2
- a*d^3)^(2/3)*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^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2 - a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3))
+ 3*(5*a*b^3*c^3*d^2 - 2*a^2*b^2*c^2*d^3 - 21*a^3*b*c*d^4 + 18*a^4*d^5 - 2*(b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2
*b^2*d^5)*x^6 + (5*b^4*c^3*d^2 - 4*a*b^3*c^2*d^3 - 7*a^2*b^2*c*d^4 + 6*a^3*b*d^5)*x^3)*(b*x^3 + a)^(2/3))/(a*b
^5*c^2*d^4 - 2*a^2*b^4*c*d^5 + a^3*b^3*d^6 + (b^6*c^2*d^4 - 2*a*b^5*c*d^5 + a^2*b^4*d^6)*x^3), 1/30*(30*sqrt(1
/3)*(a*b^4*c^4*d - a^2*b^3*c^3*d^2 + (b^5*c^4*d - a*b^4*c^3*d^2)*x^3)*sqrt((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) + 5*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2 - a*d^3)^(2/3)*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^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2 - a*d^3)^(2/3)*log((b*x
^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3)) - 3*(5*a*b^3*c^3*d^2 - 2*a^2*b^2*c^2*d^3 - 21*a^3*b*c*d^4 + 18*a^4*
d^5 - 2*(b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)*x^6 + (5*b^4*c^3*d^2 - 4*a*b^3*c^2*d^3 - 7*a^2*b^2*c*d^4 +
6*a^3*b*d^5)*x^3)*(b*x^3 + a)^(2/3))/(a*b^5*c^2*d^4 - 2*a^2*b^4*c*d^5 + a^3*b^3*d^6 + (b^6*c^2*d^4 - 2*a*b^5*
c*d^5 + a^2*b^4*d^6)*x^3)]

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giac [A]  time = 0.26, size = 372, normalized size = 1.47 \begin {gather*} -\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{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^{2} c^{2} d^{4} - 2 \, \sqrt {3} a b c d^{5} + \sqrt {3} a^{2} d^{6}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{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^{2} c^{2} d^{4} - 2 \, a b c d^{5} + a^{2} d^{6}\right )}} - \frac {c^{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^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}} + \frac {a^{3}}{{\left (b^{4} c - a b^{3} d\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{13} c d^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{12} d^{4} + 10 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b^{12} d^{4}}{10 \, b^{15} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-(-b*c*d^2 + a*d^3)^(2/3)*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^2*c^2*d^4 - 2*sqrt(3)*a*b*c*d^5 + sqrt(3)*a^2*d^6) + 1/6*(-b*c*d^2 + a*d^3)^(2/3)*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^2*c^2*d^4 - 2*a*b*c
*d^5 + a^2*d^6) - 1/3*c^3*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^2*c^2
*d^2 - 2*a*b*c*d^3 + a^2*d^4) + a^3/((b^4*c - a*b^3*d)*(b*x^3 + a)^(1/3)) - 1/10*(5*(b*x^3 + a)^(2/3)*b^13*c*d
^3 - 2*(b*x^3 + a)^(5/3)*b^12*d^4 + 10*(b*x^3 + a)^(2/3)*a*b^12*d^4)/(b^15*d^5)

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11}}{\left (b \,x^{3}+a \right )^{\frac {4}{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)^(4/3)/(d*x^3+c),x)

[Out]

int(x^11/(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^11/(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 = 5.16, size = 493, normalized size = 1.95 \begin {gather*} \frac {{\left (b\,x^3+a\right )}^{5/3}}{5\,b^3\,d}-\left (\frac {3\,a}{2\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{2\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{2/3}-\frac {a^3}{b^3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}-\frac {c^3\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^6\,d^4-b\,c^7\,d^3\right )-\frac {c^6\,\left (9\,a^4\,d^{12}-36\,a^3\,b\,c\,d^{11}+54\,a^2\,b^2\,c^2\,d^{10}-36\,a\,b^3\,c^3\,d^9+9\,b^4\,c^4\,d^8\right )}{9\,d^{16/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )}{3\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^6\,d^4-b\,c^7\,d^3\right )-\frac {{\left (c^3+\sqrt {3}\,c^3\,1{}\mathrm {i}\right )}^2\,\left (9\,a^4\,d^{12}-36\,a^3\,b\,c\,d^{11}+54\,a^2\,b^2\,c^2\,d^{10}-36\,a\,b^3\,c^3\,d^9+9\,b^4\,c^4\,d^8\right )}{36\,d^{16/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (c^3+\sqrt {3}\,c^3\,1{}\mathrm {i}\right )}{6\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{4/3}}-\frac {c^3\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^6\,d^4-b\,c^7\,d^3\right )-\frac {c^6\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (9\,a^4\,d^{12}-36\,a^3\,b\,c\,d^{11}+54\,a^2\,b^2\,c^2\,d^{10}-36\,a\,b^3\,c^3\,d^9+9\,b^4\,c^4\,d^8\right )}{9\,d^{16/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{4/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/((a + b*x^3)^(4/3)*(c + d*x^3)),x)

[Out]

(a + b*x^3)^(5/3)/(5*b^3*d) - ((3*a)/(2*b^3*d) + (b^4*c - a*b^3*d)/(2*b^6*d^2))*(a + b*x^3)^(2/3) - a^3/(b^3*(
a + b*x^3)^(1/3)*(a*d - b*c)) - (c^3*log((a + b*x^3)^(1/3)*(a*c^6*d^4 - b*c^7*d^3) - (c^6*(9*a^4*d^12 + 9*b^4*
c^4*d^8 - 36*a*b^3*c^3*d^9 + 54*a^2*b^2*c^2*d^10 - 36*a^3*b*c*d^11))/(9*d^(16/3)*(a*d - b*c)^(8/3))))/(3*d^(8/
3)*(a*d - b*c)^(4/3)) + (log((a + b*x^3)^(1/3)*(a*c^6*d^4 - b*c^7*d^3) - ((3^(1/2)*c^3*1i + c^3)^2*(9*a^4*d^12
+ 9*b^4*c^4*d^8 - 36*a*b^3*c^3*d^9 + 54*a^2*b^2*c^2*d^10 - 36*a^3*b*c*d^11))/(36*d^(16/3)*(a*d - b*c)^(8/3)))
*(3^(1/2)*c^3*1i + c^3))/(6*d^(8/3)*(a*d - b*c)^(4/3)) - (c^3*log((a + b*x^3)^(1/3)*(a*c^6*d^4 - b*c^7*d^3) -
(c^6*((3^(1/2)*1i)/2 - 1/2)^2*(9*a^4*d^12 + 9*b^4*c^4*d^8 - 36*a*b^3*c^3*d^9 + 54*a^2*b^2*c^2*d^10 - 36*a^3*b*
c*d^11))/(9*d^(16/3)*(a*d - b*c)^(8/3)))*((3^(1/2)*1i)/2 - 1/2))/(3*d^(8/3)*(a*d - b*c)^(4/3))

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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 {4}{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)**(4/3)/(d*x**3+c),x)

[Out]

Integral(x**11/((a + b*x**3)**(4/3)*(c + d*x**3)), x)

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