3.696 \(\int \frac{x^8 (a+b x^3)^{4/3}}{c+d x^3} \, dx\)

Optimal. Leaf size=251 \[ -\frac{\left (a+b x^3\right )^{7/3} (a d+b c)}{7 b^2 d^2}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^2 d}+\frac{c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac{c^2 \sqrt [3]{a+b x^3} (b c-a d)}{d^4}-\frac{c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac{c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}-\frac{c^2 (b c-a d)^{4/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{13/3}} \]

[Out]

-((c^2*(b*c - a*d)*(a + b*x^3)^(1/3))/d^4) + (c^2*(a + b*x^3)^(4/3))/(4*d^3) - ((b*c + a*d)*(a + b*x^3)^(7/3))
/(7*b^2*d^2) + (a + b*x^3)^(10/3)/(10*b^2*d) - (c^2*(b*c - a*d)^(4/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3)
)/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(13/3)) - (c^2*(b*c - a*d)^(4/3)*Log[c + d*x^3])/(6*d^(13/3)) + (c^2
*(b*c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(13/3))

________________________________________________________________________________________

Rubi [A]  time = 0.362667, antiderivative size = 251, normalized size of antiderivative = 1., 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, 88, 50, 58, 617, 204, 31} \[ -\frac{\left (a+b x^3\right )^{7/3} (a d+b c)}{7 b^2 d^2}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^2 d}+\frac{c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac{c^2 \sqrt [3]{a+b x^3} (b c-a d)}{d^4}-\frac{c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac{c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}-\frac{c^2 (b c-a d)^{4/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{13/3}} \]

Antiderivative was successfully verified.

[In]

Int[(x^8*(a + b*x^3)^(4/3))/(c + d*x^3),x]

[Out]

-((c^2*(b*c - a*d)*(a + b*x^3)^(1/3))/d^4) + (c^2*(a + b*x^3)^(4/3))/(4*d^3) - ((b*c + a*d)*(a + b*x^3)^(7/3))
/(7*b^2*d^2) + (a + b*x^3)^(10/3)/(10*b^2*d) - (c^2*(b*c - a*d)^(4/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3)
)/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(13/3)) - (c^2*(b*c - a*d)^(4/3)*Log[c + d*x^3])/(6*d^(13/3)) + (c^2
*(b*c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(13/3))

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 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d
*x)^(1/3)], x] + Dist[3/(2*b*q^2), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x
] && NegQ[(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^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2 (a+b x)^{4/3}}{c+d x} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{(-b c-a d) (a+b x)^{4/3}}{b d^2}+\frac{(a+b x)^{7/3}}{b d}+\frac{c^2 (a+b x)^{4/3}}{d^2 (c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac{(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^2 d}+\frac{c^2 \operatorname{Subst}\left (\int \frac{(a+b x)^{4/3}}{c+d x} \, dx,x,x^3\right )}{3 d^2}\\ &=\frac{c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac{(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac{\left (c^2 (b c-a d)\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x}}{c+d x} \, dx,x,x^3\right )}{3 d^3}\\ &=-\frac{c^2 (b c-a d) \sqrt [3]{a+b x^3}}{d^4}+\frac{c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac{(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^2 d}+\frac{\left (c^2 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 d^4}\\ &=-\frac{c^2 (b c-a d) \sqrt [3]{a+b x^3}}{d^4}+\frac{c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac{(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac{c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac{\left (c^2 (b c-a d)^{4/3}\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^{13/3}}+\frac{\left (c^2 (b c-a d)^{5/3}\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^{14/3}}\\ &=-\frac{c^2 (b c-a d) \sqrt [3]{a+b x^3}}{d^4}+\frac{c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac{(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac{c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac{c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}+\frac{\left (c^2 (b c-a d)^{4/3}\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^{13/3}}\\ &=-\frac{c^2 (b c-a d) \sqrt [3]{a+b x^3}}{d^4}+\frac{c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac{(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac{\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac{c^2 (b c-a d)^{4/3} \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^{13/3}}-\frac{c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac{c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}\\ \end{align*}

Mathematica [A]  time = 0.432544, size = 258, normalized size = 1.03 \[ \frac{-\frac{60 d \left (a+b x^3\right )^{7/3} (a d+b c)}{b^2}+\frac{42 d^2 \left (a+b x^3\right )^{10/3}}{b^2}-\frac{70 c^2 (b c-a d) \left (\sqrt [3]{b c-a d} \left (\log \left (-\sqrt [3]{d} \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )-2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )\right )+6 \sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{d^{4/3}}+105 c^2 \left (a+b x^3\right )^{4/3}}{420 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(a + b*x^3)^(4/3))/(c + d*x^3),x]

[Out]

(105*c^2*(a + b*x^3)^(4/3) - (60*d*(b*c + a*d)*(a + b*x^3)^(7/3))/b^2 + (42*d^2*(a + b*x^3)^(10/3))/b^2 - (70*
c^2*(b*c - a*d)*(6*d^(1/3)*(a + b*x^3)^(1/3) + (b*c - a*d)^(1/3)*(2*Sqrt[3]*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)
^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - 2*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + 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)])))/d^(4/3))/(420*d^3)

________________________________________________________________________________________

Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{8}}{d{x}^{3}+c} \left ( b{x}^{3}+a \right ) ^{{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(x^8*(b*x^3+a)^(4/3)/(d*x^3+c),x)

________________________________________________________________________________________

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^8*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.48239, size = 829, normalized size = 3.3 \begin{align*} \frac{140 \, \sqrt{3}{\left (b^{3} c^{3} - a b^{2} c^{2} d\right )} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} \arctan \left (-\frac{2 \, \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}} d \left (-\frac{b c - a d}{d}\right )^{\frac{2}{3}} - \sqrt{3}{\left (b c - a d\right )}}{3 \,{\left (b c - a d\right )}}\right ) + 70 \,{\left (b^{3} c^{3} - a b^{2} c^{2} d\right )} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{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 ) - 140 \,{\left (b^{3} c^{3} - a b^{2} c^{2} d\right )} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{1}{3}} - \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}\right ) + 3 \,{\left (14 \, b^{3} d^{3} x^{9} - 2 \,{\left (10 \, b^{3} c d^{2} - 11 \, a b^{2} d^{3}\right )} x^{6} - 140 \, b^{3} c^{3} + 175 \, a b^{2} c^{2} d - 20 \, a^{2} b c d^{2} - 6 \, a^{3} d^{3} +{\left (35 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 2 \, a^{2} b d^{3}\right )} x^{3}\right )}{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{420 \, b^{2} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

1/420*(140*sqrt(3)*(b^3*c^3 - a*b^2*c^2*d)*(-(b*c - a*d)/d)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*d*(
-(b*c - a*d)/d)^(2/3) - sqrt(3)*(b*c - a*d))/(b*c - a*d)) + 70*(b^3*c^3 - a*b^2*c^2*d)*(-(b*c - a*d)/d)^(1/3)*
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)) - 140*(b^3*c^3 - a*
b^2*c^2*d)*(-(b*c - a*d)/d)^(1/3)*log((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)) + 3*(14*b^3*d^3*x^9 - 2*(10*
b^3*c*d^2 - 11*a*b^2*d^3)*x^6 - 140*b^3*c^3 + 175*a*b^2*c^2*d - 20*a^2*b*c*d^2 - 6*a^3*d^3 + (35*b^3*c^2*d - 4
0*a*b^2*c*d^2 + 2*a^2*b*d^3)*x^3)*(b*x^3 + a)^(1/3))/(b^2*d^4)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.19153, size = 532, normalized size = 2.12 \begin{align*} -\frac{{\left (b^{24} c^{4} d^{6} - 2 \, a b^{23} c^{3} d^{7} + a^{2} b^{22} c^{2} d^{8}\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^{23} c d^{10} - a b^{22} d^{11}\right )}} + \frac{\sqrt{3}{\left (b c^{3} - a c^{2} d\right )}{\left (-b c d^{2} + a d^{3}\right )}^{\frac{1}{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^{5}} + \frac{{\left (b c^{3} - a c^{2} d\right )}{\left (-b c d^{2} + a d^{3}\right )}^{\frac{1}{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^{5}} - \frac{140 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} b^{21} c^{3} d^{6} - 35 \,{\left (b x^{3} + a\right )}^{\frac{4}{3}} b^{20} c^{2} d^{7} - 140 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} a b^{20} c^{2} d^{7} + 20 \,{\left (b x^{3} + a\right )}^{\frac{7}{3}} b^{19} c d^{8} - 14 \,{\left (b x^{3} + a\right )}^{\frac{10}{3}} b^{18} d^{9} + 20 \,{\left (b x^{3} + a\right )}^{\frac{7}{3}} a b^{18} d^{9}}{140 \, b^{20} d^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-1/3*(b^24*c^4*d^6 - 2*a*b^23*c^3*d^7 + a^2*b^22*c^2*d^8)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (
-(b*c - a*d)/d)^(1/3)))/(b^23*c*d^10 - a*b^22*d^11) + 1/3*sqrt(3)*(b*c^3 - a*c^2*d)*(-b*c*d^2 + a*d^3)^(1/3)*a
rctan(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^5 + 1/6*(b*c^3 - a*
c^2*d)*(-b*c*d^2 + a*d^3)^(1/3)*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^5 - 1/140*(140*(b*x^3 + a)^(1/3)*b^21*c^3*d^6 - 35*(b*x^3 + a)^(4/3)*b^20*c^2*d^7 - 140*(b*x^3
+ a)^(1/3)*a*b^20*c^2*d^7 + 20*(b*x^3 + a)^(7/3)*b^19*c*d^8 - 14*(b*x^3 + a)^(10/3)*b^18*d^9 + 20*(b*x^3 + a)^
(7/3)*a*b^18*d^9)/(b^20*d^10)