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

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

[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))

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Rubi [A]  time = 0.243404, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 88, 56, 617, 204, 31} \[ \frac{\left (a+b x^3\right )^{2/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{2 b^3 d^3}-\frac{\left (a+b x^3\right )^{5/3} (2 a d+b c)}{5 b^3 d^2}+\frac{\left (a+b x^3\right )^{8/3}}{8 b^3 d}-\frac{c^3 \log \left (c+d x^3\right )}{6 d^{11/3} \sqrt [3]{b c-a d}}+\frac{c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3} \sqrt [3]{b c-a d}}+\frac{c^3 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{11/3} \sqrt [3]{b c-a d}} \]

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 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 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}}{\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*}

Mathematica [C]  time = 0.0933834, size = 145, normalized size = 0.59 \[ -\frac{\left (a+b x^3\right )^{2/3} \left (3 a^2 b d^2 \left (c-2 d x^3\right )+9 a^3 d^3+a b^2 d \left (8 c^2-2 c d x^3+5 d^2 x^6\right )+20 b^3 c^3 \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{d \left (b x^3+a\right )}{a d-b c}\right )+b^3 c \left (-20 c^2+8 c d x^3-5 d^2 x^6\right )\right )}{40 b^3 d^3 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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)]))
/(40*b^3*d^3*(b*c - a*d))

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{11}}{d{x}^{3}+c}{\frac{1}{\sqrt [3]{b{x}^{3}+a}}}}\, dx \end{align*}

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., 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)^(1/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03154, size = 1904, normalized size = 7.8 \begin{align*} \text{result too large to display} \end{align*}

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

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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Giac [A]  time = 1.25669, size = 501, normalized size = 2.05 \begin{align*} \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{align*}

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)