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

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

[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))

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Rubi [A]  time = 0.322075, antiderivative size = 266, normalized size of antiderivative = 1., 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} \[ \frac{\left (a+b x^3\right )^{5/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{5 b^3 d^3}-\frac{\left (a+b x^3\right )^{8/3} (2 a d+b c)}{8 b^3 d^2}+\frac{\left (a+b x^3\right )^{11/3}}{11 b^3 d}-\frac{c^3 \left (a+b x^3\right )^{2/3}}{2 d^4}+\frac{c^3 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{14/3}}-\frac{c^3 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3}}-\frac{c^3 (b c-a d)^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{14/3}} \]

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 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 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 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 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]

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 31

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

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*}

Mathematica [C]  time = 0.105893, size = 148, normalized size = 0.56 \[ \frac{\left (a+b x^3\right )^{2/3} \left (3 a^2 b d^2 \left (11 c-4 d x^3\right )+18 a^3 d^3+2 a b^2 d \left (44 c^2-11 c d x^3+5 d^2 x^6\right )+220 b^3 c^3 \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{d \left (b x^3+a\right )}{a d-b c}\right )+b^3 \left (88 c^2 d x^3-220 c^3-55 c d^2 x^6+40 d^3 x^9\right )\right )}{440 b^3 d^4} \]

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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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{11}}{d{x}^{3}+c} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 1.88651, size = 1033, normalized size = 3.88 \begin{align*} -\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{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.21092, size = 552, normalized size = 2.08 \begin{align*} -\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{align*}

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)