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

Optimal. Leaf size=266 \[ \frac {\left (a+b x^3\right )^{5/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{5 b^3 d^3}-\frac {\left (a+b x^3\right )^{8/3} (2 a d+b c)}{8 b^3 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^3 d}+\frac {c^3 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{14/3}}-\frac {c^3 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3}}-\frac {c^3 (b c-a d)^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{14/3}}-\frac {c^3 \left (a+b x^3\right )^{2/3}}{2 d^4} \]

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Rubi [A]  time = 0.32, antiderivative size = 266, 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, 56, 617, 204, 31} \begin {gather*} \frac {\left (a+b x^3\right )^{5/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{5 b^3 d^3}-\frac {\left (a+b x^3\right )^{8/3} (2 a d+b c)}{8 b^3 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^3 d}-\frac {c^3 \left (a+b x^3\right )^{2/3}}{2 d^4}+\frac {c^3 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{14/3}}-\frac {c^3 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3}}-\frac {c^3 (b c-a d)^{2/3} \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^{14/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^3*(a + b*x^3)^(2/3))/(2*d^4) + ((b^2*c^2 + a*b*c*d + a^2*d^2)*(a + b*x^3)^(5/3))/(5*b^3*d^3) - ((b*c + 2*a
*d)*(a + b*x^3)^(8/3))/(8*b^3*d^2) + (a + b*x^3)^(11/3)/(11*b^3*d) - (c^3*(b*c - a*d)^(2/3)*ArcTan[(1 - (2*d^(
1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(14/3)) + (c^3*(b*c - a*d)^(2/3)*Log[c + d*x^3
])/(6*d^(14/3)) - (c^3*(b*c - a*d)^(2/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(14/3))

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 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} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, 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 {\left (b^2 c^2+a b c d+a^2 d^2\right ) (a+b x)^{2/3}}{b^2 d^3}+\frac {(-b c-2 a d) (a+b x)^{5/3}}{b^2 d^2}+\frac {(a+b x)^{8/3}}{b^2 d}-\frac {c^3 (a+b x)^{2/3}}{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 )^{5/3}}{5 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{8/3}}{8 b^3 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^3 d}-\frac {c^3 \operatorname {Subst}\left (\int \frac {(a+b x)^{2/3}}{c+d x} \, dx,x,x^3\right )}{3 d^3}\\ &=-\frac {c^3 \left (a+b x^3\right )^{2/3}}{2 d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{8/3}}{8 b^3 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^3 d}+\frac {\left (c^3 (b c-a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )}{3 d^4}\\ &=-\frac {c^3 \left (a+b x^3\right )^{2/3}}{2 d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{8/3}}{8 b^3 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^3 d}+\frac {c^3 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{14/3}}-\frac {\left (c^3 (b c-a d)^{2/3}\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^{14/3}}+\frac {\left (c^3 (b c-a d)\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^5}\\ &=-\frac {c^3 \left (a+b x^3\right )^{2/3}}{2 d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{8/3}}{8 b^3 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^3 d}+\frac {c^3 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{14/3}}-\frac {c^3 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3}}+\frac {\left (c^3 (b c-a d)^{2/3}\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^{14/3}}\\ &=-\frac {c^3 \left (a+b x^3\right )^{2/3}}{2 d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{8/3}}{8 b^3 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^3 d}-\frac {c^3 (b c-a d)^{2/3} \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^{14/3}}+\frac {c^3 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{14/3}}-\frac {c^3 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3}}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 148, normalized size = 0.56 \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (18 a^3 d^3+3 a^2 b d^2 \left (11 c-4 d x^3\right )+220 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 )+2 a b^2 d \left (44 c^2-11 c d x^3+5 d^2 x^6\right )+b^3 \left (-220 c^3+88 c^2 d x^3-55 c d^2 x^6+40 d^3 x^9\right )\right )}{440 b^3 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^(2/3)*(18*a^3*d^3 + 3*a^2*b*d^2*(11*c - 4*d*x^3) + 2*a*b^2*d*(44*c^2 - 11*c*d*x^3 + 5*d^2*x^6) +
b^3*(-220*c^3 + 88*c^2*d*x^3 - 55*c*d^2*x^6 + 40*d^3*x^9) + 220*b^3*c^3*Hypergeometric2F1[2/3, 1, 5/3, (d*(a +
b*x^3))/(-(b*c) + a*d)]))/(440*b^3*d^4)

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IntegrateAlgebraic [A]  time = 0.65, size = 340, normalized size = 1.28 \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (18 a^3 d^3+33 a^2 b c d^2-12 a^2 b d^3 x^3+88 a b^2 c^2 d-22 a b^2 c d^2 x^3+10 a b^2 d^3 x^6-220 b^3 c^3+88 b^3 c^2 d x^3-55 b^3 c d^2 x^6+40 b^3 d^3 x^9\right )}{440 b^3 d^4}-\frac {c^3 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{14/3}}+\frac {c^3 (b c-a d)^{2/3} \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^{14/3}}-\frac {c^3 (b c-a d)^{2/3} \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^{14/3}} \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)^(2/3)*(-220*b^3*c^3 + 88*a*b^2*c^2*d + 33*a^2*b*c*d^2 + 18*a^3*d^3 + 88*b^3*c^2*d*x^3 - 22*a*b^2*
c*d^2*x^3 - 12*a^2*b*d^3*x^3 - 55*b^3*c*d^2*x^6 + 10*a*b^2*d^3*x^6 + 40*b^3*d^3*x^9))/(440*b^3*d^4) - (c^3*(b*
c - a*d)^(2/3)*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^(14/3
)) - (c^3*(b*c - a*d)^(2/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*d^(14/3)) + (c^3*(b*c - a*d
)^(2/3)*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
^(14/3))

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fricas [B]  time = 0.69, size = 455, normalized size = 1.71 \begin {gather*} -\frac {440 \, \sqrt {3} b^{3} c^{3} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} + \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 220 \, b^{3} c^{3} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )} + {\left (b c - a d\right )} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}}\right ) - 440 \, b^{3} c^{3} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left (-d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right ) - 3 \, {\left (40 \, b^{3} d^{3} x^{9} - 5 \, {\left (11 \, b^{3} c d^{2} - 2 \, a b^{2} d^{3}\right )} x^{6} - 220 \, b^{3} c^{3} + 88 \, a b^{2} c^{2} d + 33 \, a^{2} b c d^{2} + 18 \, a^{3} d^{3} + 2 \, {\left (44 \, b^{3} c^{2} d - 11 \, a b^{2} c d^{2} - 6 \, a^{2} b d^{3}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{1320 \, b^{3} d^{4}} \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="fricas")

[Out]

-1/1320*(440*sqrt(3)*b^3*c^3*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b*x^3 + a)^(
1/3)*d*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3) + sqrt(3)*(b*c - a*d))/(b*c - a*d)) + 220*b^3*c^3*(-(b^2*c
^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*log((b*x^3 + a)^(1/3)*d*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b
*x^3 + a)^(2/3)*(b*c - a*d) + (b*c - a*d)*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)) - 440*b^3*c^3*(-(b^2*c
^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*log(-d*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b*x^3 + a)^(1/3)*(
b*c - a*d)) - 3*(40*b^3*d^3*x^9 - 5*(11*b^3*c*d^2 - 2*a*b^2*d^3)*x^6 - 220*b^3*c^3 + 88*a*b^2*c^2*d + 33*a^2*b
*c*d^2 + 18*a^3*d^3 + 2*(44*b^3*c^2*d - 11*a*b^2*c*d^2 - 6*a^2*b*d^3)*x^3)*(b*x^3 + a)^(2/3))/(b^3*d^4)

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giac [A]  time = 0.29, size = 409, normalized size = 1.54 \begin {gather*} -\frac {{\left (b^{37} c^{4} d^{7} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} - a b^{36} c^{3} d^{8} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\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^{37} c d^{11} - a b^{36} d^{12}\right )}} - \frac {\sqrt {3} {\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 )}{3 \, 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 \, d^{6}} - \frac {220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{33} c^{3} d^{7} - 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{32} c^{2} d^{8} + 55 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{31} c d^{9} - 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a b^{31} c d^{9} - 40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} b^{30} d^{10} + 110 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a b^{30} d^{10} - 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2} b^{30} d^{10}}{440 \, b^{33} d^{11}} \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^37*c^4*d^7*(-(b*c - a*d)/d)^(1/3) - a*b^36*c^3*d^8*(-(b*c - a*d)/d)^(1/3))*(-(b*c - a*d)/d)^(1/3)*log(
abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^37*c*d^11 - a*b^36*d^12) - 1/3*sqrt(3)*(-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))/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))/d^6 - 1/440*(220*(b*x^3 + a)^(2/3)*b^33*c^3*d^7 - 88*(b*x^3 + a)^(5/3)*b^32*c^2*d^8 + 55*(b*x^3 + a)
^(8/3)*b^31*c*d^9 - 88*(b*x^3 + a)^(5/3)*a*b^31*c*d^9 - 40*(b*x^3 + a)^(11/3)*b^30*d^10 + 110*(b*x^3 + a)^(8/3
)*a*b^30*d^10 - 88*(b*x^3 + a)^(5/3)*a^2*b^30*d^10)/(b^33*d^11)

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{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)^(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 = 5.13, size = 490, normalized size = 1.84 \begin {gather*} \left (\frac {3\,a^2}{5\,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 )}{5\,b^3\,d}\right )\,{\left (b\,x^3+a\right )}^{5/3}-\left (\frac {3\,a}{8\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{8\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{8/3}-{\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {a^3}{2\,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 )}{2\,b^3\,d}\right )+\frac {{\left (b\,x^3+a\right )}^{11/3}}{11\,b^3\,d}-\frac {c^3\,\ln \left (\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c^6\,d^2-2\,a\,b\,c^7\,d+b^2\,c^8\right )}{d^7}-\frac {c^6\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{9\,d^{28/3}}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{14/3}}-\frac {c^3\,\ln \left (\frac {c^6\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{22/3}}+\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^7}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{14/3}}+\frac {c^3\,\ln \left (\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^7}-\frac {c^6\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{7/3}}{4\,d^{22/3}}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{d^{14/3}} \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)/(5*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))/(5*b^3*d))*(a + b*x^3)^
(5/3) - ((3*a)/(8*b^3*d) + (b^4*c - a*b^3*d)/(8*b^6*d^2))*(a + b*x^3)^(8/3) - (a + b*x^3)^(2/3)*(a^3/(2*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))/(2*b^3*d)) + (a + b*x^3)^(11/3)/(11*b^3*d) - (c^3*log(((a + b*x^3)^(1/3)*(b^2*c^8 + a^2*c^6*d^2 - 2*a*b
*c^7*d))/d^7 - (c^6*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(9*d^(28/3)))*(a*d - b*c)^(2/3))/(3*d^(14/3)) - (
c^3*log((c^6*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(7/3))/d^(22/3) + (c^6*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^7)*(
(3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(2/3))/(3*d^(14/3)) + (c^3*log((c^6*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^7 - (
c^6*(3^(1/2)*1i + 1)^2*(a*d - b*c)^(7/3))/(4*d^(22/3)))*((3^(1/2)*1i)/6 + 1/6)*(a*d - b*c)^(2/3))/d^(14/3)

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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}}}{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)**(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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