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

Optimal. Leaf size=244 \[ \frac {\left (a+b x^3\right )^{2/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{2 b^3 d^3}-\frac {\left (a+b x^3\right )^{5/3} (2 a d+b c)}{5 b^3 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^3 d}-\frac {c^3 \log \left (c+d x^3\right )}{6 d^{11/3} \sqrt [3]{b c-a d}}+\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3} \sqrt [3]{b c-a d}}+\frac {c^3 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{11/3} \sqrt [3]{b c-a d}} \]

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Rubi [A]  time = 0.24, antiderivative size = 244, 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, 56, 617, 204, 31} \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{2 b^3 d^3}-\frac {\left (a+b x^3\right )^{5/3} (2 a d+b c)}{5 b^3 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^3 d}-\frac {c^3 \log \left (c+d x^3\right )}{6 d^{11/3} \sqrt [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 )}{2 d^{11/3} \sqrt [3]{b c-a 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^{11/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*c^2 + a*b*c*d + a^2*d^2)*(a + b*x^3)^(2/3))/(2*b^3*d^3) - ((b*c + 2*a*d)*(a + b*x^3)^(5/3))/(5*b^3*d^2)
+ (a + b*x^3)^(8/3)/(8*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]])/(S
qrt[3]*d^(11/3)*(b*c - a*d)^(1/3)) - (c^3*Log[c + d*x^3])/(6*d^(11/3)*(b*c - a*d)^(1/3)) + (c^3*Log[(b*c - a*d
)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(11/3)*(b*c - a*d)^(1/3))

Rule 31

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

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 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} \left (c+d x^3\right )} \, 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 {b^2 c^2+a b c d+a^2 d^2}{b^2 d^3 \sqrt [3]{a+b x}}+\frac {(-b c-2 a d) (a+b x)^{2/3}}{b^2 d^2}+\frac {(a+b x)^{5/3}}{b^2 d}-\frac {c^3}{d^3 \sqrt [3]{a+b x} (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 )^{2/3}}{2 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{5/3}}{5 b^3 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^3 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^3}\\ &=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{5/3}}{5 b^3 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^3 d}-\frac {c^3 \log \left (c+d x^3\right )}{6 d^{11/3} \sqrt [3]{b c-a d}}-\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^4}+\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^{11/3} \sqrt [3]{b c-a d}}\\ &=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{5/3}}{5 b^3 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^3 d}-\frac {c^3 \log \left (c+d x^3\right )}{6 d^{11/3} \sqrt [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 )}{2 d^{11/3} \sqrt [3]{b c-a d}}-\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^{11/3} \sqrt [3]{b c-a d}}\\ &=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{5/3}}{5 b^3 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 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^{11/3} \sqrt [3]{b c-a d}}-\frac {c^3 \log \left (c+d x^3\right )}{6 d^{11/3} \sqrt [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 )}{2 d^{11/3} \sqrt [3]{b c-a d}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 145, normalized size = 0.59 \begin {gather*} -\frac {\left (a+b x^3\right )^{2/3} \left (9 a^3 d^3+3 a^2 b d^2 \left (c-2 d x^3\right )+20 b^3 c^3 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )+a b^2 d \left (8 c^2-2 c d x^3+5 d^2 x^6\right )+b^3 c \left (-20 c^2+8 c d x^3-5 d^2 x^6\right )\right )}{40 b^3 d^3 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/40*((a + b*x^3)^(2/3)*(9*a^3*d^3 + 3*a^2*b*d^2*(c - 2*d*x^3) + b^3*c*(-20*c^2 + 8*c*d*x^3 - 5*d^2*x^6) + a*
b^2*d*(8*c^2 - 2*c*d*x^3 + 5*d^2*x^6) + 20*b^3*c^3*Hypergeometric2F1[2/3, 1, 5/3, (d*(a + b*x^3))/(-(b*c) + a*
d)]))/(b^3*d^3*(b*c - a*d))

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IntegrateAlgebraic [A]  time = 0.42, size = 284, normalized size = 1.16 \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (9 a^2 d^2+12 a b c d-6 a b d^2 x^3+20 b^2 c^2-8 b^2 c d x^3+5 b^2 d^2 x^6\right )}{40 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^{11/3} \sqrt [3]{b c-a d}}-\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^{11/3} \sqrt [3]{b c-a d}}+\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^{11/3} \sqrt [3]{b c-a d}} \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)^(2/3)*(20*b^2*c^2 + 12*a*b*c*d + 9*a^2*d^2 - 8*b^2*c*d*x^3 - 6*a*b*d^2*x^3 + 5*b^2*d^2*x^6))/(40*
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^(11/
3)*(b*c - a*d)^(1/3)) + (c^3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*d^(11/3)*(b*c - a*d)^(1/3)
) - (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^(11/3)*(b*c - a*d)^(1/3))

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fricas [A]  time = 0.74, size = 873, normalized size = 3.58 \begin {gather*} \left [-\frac {20 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} b^{3} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}}\right ) - 40 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} b^{3} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right ) - 60 \, \sqrt {\frac {1}{3}} {\left (b^{4} c^{4} d - a b^{3} c^{3} d^{2}\right )} \sqrt {-\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \log \left (\frac {2 \, b d^{2} x^{3} - b c d + 3 \, a d^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c d - a d^{2}\right )} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right )} \sqrt {-\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} - 3 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{d x^{3} + c}\right ) - 3 \, {\left (20 \, b^{3} c^{3} d^{2} - 8 \, a b^{2} c^{2} d^{3} - 3 \, a^{2} b c d^{4} - 9 \, a^{3} d^{5} + 5 \, {\left (b^{3} c d^{4} - a b^{2} d^{5}\right )} x^{6} - 2 \, {\left (4 \, b^{3} c^{2} d^{3} - a b^{2} c d^{4} - 3 \, a^{2} b d^{5}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{120 \, {\left (b^{4} c d^{5} - a b^{3} d^{6}\right )}}, -\frac {20 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} b^{3} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}}\right ) - 40 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} b^{3} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right ) + 120 \, \sqrt {\frac {1}{3}} {\left (b^{4} c^{4} d - a b^{3} c^{3} d^{2}\right )} \sqrt {\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} d - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right )} \sqrt {\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}}}{d}\right ) - 3 \, {\left (20 \, b^{3} c^{3} d^{2} - 8 \, a b^{2} c^{2} d^{3} - 3 \, a^{2} b c d^{4} - 9 \, a^{3} d^{5} + 5 \, {\left (b^{3} c d^{4} - a b^{2} d^{5}\right )} x^{6} - 2 \, {\left (4 \, b^{3} c^{2} d^{3} - a b^{2} c d^{4} - 3 \, a^{2} b d^{5}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{120 \, {\left (b^{4} c d^{5} - a b^{3} d^{6}\right )}}\right ] \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]

[-1/120*(20*(b*c*d^2 - a*d^3)^(2/3)*b^3*c^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)) - 40*(b*c*d^2 - a*d^3)^(2/3)*b^3*c^3*log((b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d
^3)^(1/3)) - 60*sqrt(1/3)*(b^4*c^4*d - a*b^3*c^3*d^2)*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)) - 3*(20*b^3*c^3*d^2 - 8*a*b^2*c^2*d^3 - 3*a^2*b*c*d^4 - 9*a^3*d^5 + 5*(
b^3*c*d^4 - a*b^2*d^5)*x^6 - 2*(4*b^3*c^2*d^3 - a*b^2*c*d^4 - 3*a^2*b*d^5)*x^3)*(b*x^3 + a)^(2/3))/(b^4*c*d^5
- a*b^3*d^6), -1/120*(20*(b*c*d^2 - a*d^3)^(2/3)*b^3*c^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)) - 40*(b*c*d^2 - a*d^3)^(2/3)*b^3*c^3*log((b*x^3 + a)^(1/3)*d + (
b*c*d^2 - a*d^3)^(1/3)) + 120*sqrt(1/3)*(b^4*c^4*d - a*b^3*c^3*d^2)*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
) - 3*(20*b^3*c^3*d^2 - 8*a*b^2*c^2*d^3 - 3*a^2*b*c*d^4 - 9*a^3*d^5 + 5*(b^3*c*d^4 - a*b^2*d^5)*x^6 - 2*(4*b^3
*c^2*d^3 - a*b^2*c*d^4 - 3*a^2*b*d^5)*x^3)*(b*x^3 + a)^(2/3))/(b^4*c*d^5 - a*b^3*d^6)]

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giac [A]  time = 0.28, size = 371, normalized size = 1.52 \begin {gather*} \frac {b^{27} c^{3} d^{5} \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^{28} c d^{8} - a b^{27} d^{9}\right )}} + \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 c d^{5} - \sqrt {3} a 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 c d^{5} - a d^{6}\right )}} + \frac {20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{23} c^{2} d^{5} - 8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{22} c d^{6} + 20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b^{22} c d^{6} + 5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{21} d^{7} - 16 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a b^{21} d^{7} + 20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b^{21} d^{7}}{40 \, b^{24} d^{8}} \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^27*c^3*d^5*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^28*c*d^8 - a*b
^27*d^9) + (-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*c*d^5 - sqrt(3)*a*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*c*d^5 - a*d^6) + 1/40*(20*(b*x^3 + a)
^(2/3)*b^23*c^2*d^5 - 8*(b*x^3 + a)^(5/3)*b^22*c*d^6 + 20*(b*x^3 + a)^(2/3)*a*b^22*c*d^6 + 5*(b*x^3 + a)^(8/3)
*b^21*d^7 - 16*(b*x^3 + a)^(5/3)*a*b^21*d^7 + 20*(b*x^3 + a)^(2/3)*a^2*b^21*d^7)/(b^24*d^8)

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11}}{\left (b \,x^{3}+a \right )^{\frac {1}{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)^(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 = 5.09, size = 339, normalized size = 1.39 \begin {gather*} \left (\frac {3\,a^2}{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 )}{2\,b^3\,d}\right )\,{\left (b\,x^3+a\right )}^{2/3}-\left (\frac {3\,a}{5\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{5\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{5/3}+\frac {{\left (b\,x^3+a\right )}^{8/3}}{8\,b^3\,d}-\frac {c^3\,\ln \left (\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}}{d^5}+\frac {b\,c^7-a\,c^6\,d}{d^{16/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )}{3\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{1/3}}+\frac {\ln \left (\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}}{d^5}-\frac {c^6\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{16/3}}\right )\,\left (c^3+\sqrt {3}\,c^3\,1{}\mathrm {i}\right )}{6\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{1/3}}-\frac {c^3\,\ln \left (\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}}{d^5}+\frac {c^6\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{16/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{1/3}} \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)/(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))/(2*b^3*d))*(a + b*x^3)^
(2/3) - ((3*a)/(5*b^3*d) + (b^4*c - a*b^3*d)/(5*b^6*d^2))*(a + b*x^3)^(5/3) + (a + b*x^3)^(8/3)/(8*b^3*d) - (c
^3*log((c^6*(a + b*x^3)^(1/3))/d^5 + (b*c^7 - a*c^6*d)/(d^(16/3)*(a*d - b*c)^(2/3))))/(3*d^(11/3)*(a*d - b*c)^
(1/3)) + (log((c^6*(a + b*x^3)^(1/3))/d^5 - (c^6*(3^(1/2)*1i + 1)^2*(a*d - b*c)^(1/3))/(4*d^(16/3)))*(3^(1/2)*
c^3*1i + c^3))/(6*d^(11/3)*(a*d - b*c)^(1/3)) - (c^3*log((c^6*(a + b*x^3)^(1/3))/d^5 + (c^6*((3^(1/2)*1i)/2 +
1/2)*(a*d - b*c)^(1/3))/d^(16/3))*((3^(1/2)*1i)/2 - 1/2))/(3*d^(11/3)*(a*d - b*c)^(1/3))

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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}} \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)**(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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