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

Optimal. Leaf size=203 \[ -\frac{a^2}{b^2 \sqrt [3]{a+b x^3} (b c-a d)}+\frac{\left (a+b x^3\right )^{2/3}}{2 b^2 d}-\frac{c^2 \log \left (c+d x^3\right )}{6 d^{5/3} (b c-a d)^{4/3}}+\frac{c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3} (b c-a d)^{4/3}}+\frac{c^2 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{5/3} (b c-a d)^{4/3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.241595, antiderivative size = 203, 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, 87, 56, 617, 204, 31} \[ -\frac{a^2}{b^2 \sqrt [3]{a+b x^3} (b c-a d)}+\frac{\left (a+b x^3\right )^{2/3}}{2 b^2 d}-\frac{c^2 \log \left (c+d x^3\right )}{6 d^{5/3} (b c-a d)^{4/3}}+\frac{c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3} (b c-a d)^{4/3}}+\frac{c^2 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} d^{5/3} (b c-a d)^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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

Mathematica [C]  time = 0.0599008, size = 101, normalized size = 0.5 \[ \frac{-3 a^2 d^2-2 b^2 c^2 \, _2F_1\left (-\frac{1}{3},1;\frac{2}{3};\frac{d \left (b x^3+a\right )}{a d-b c}\right )+a b d \left (c-d x^3\right )+b^2 c \left (2 c+d x^3\right )}{2 b^2 d^2 \sqrt [3]{a+b x^3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*d^2 + a*b*d*(c - d*x^3) + b^2*c*(2*c + d*x^3) - 2*b^2*c^2*Hypergeometric2F1[-1/3, 1, 2/3, (d*(a + b*x^
3))/(-(b*c) + a*d)])/(2*b^2*d^2*(b*c - a*d)*(a + b*x^3)^(1/3))

________________________________________________________________________________________

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 [B]  time = 1.54935, size = 2159, normalized size = 10.64 \begin{align*} \text{result too large to display} \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]

[-1/6*(3*sqrt(1/3)*(a*b^3*c^3*d - a^2*b^2*c^2*d^2 + (b^4*c^3*d - a*b^3*c^2*d^2)*x^3)*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)) + (b^3*c^2*x^3 + a*b^2*c^2)*(-b*c*d^2
+ a*d^3)^(2/3)*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)) - 2*(b^3*c^2*x^3 + a*b^2*c^2)*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/
3)) - 3*(a*b^2*c^2*d^2 - 4*a^2*b*c*d^3 + 3*a^3*d^4 + (b^3*c^2*d^2 - 2*a*b^2*c*d^3 + a^2*b*d^4)*x^3)*(b*x^3 + a
)^(2/3))/(a*b^4*c^2*d^3 - 2*a^2*b^3*c*d^4 + a^3*b^2*d^5 + (b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*x^3), -1
/6*(6*sqrt(1/3)*(a*b^3*c^3*d - a^2*b^2*c^2*d^2 + (b^4*c^3*d - a*b^3*c^2*d^2)*x^3)*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) + (b^3*c^2*x^3 + a*b^2*c^2)*(-b*c*d^2 + a*d^3)^(2/3)*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)) - 2*(b^3*c^2*x^3 + a*b^2*c^2)*(-b*c*d^2 + a*d
^3)^(2/3)*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/3)) - 3*(a*b^2*c^2*d^2 - 4*a^2*b*c*d^3 + 3*a^3*d^4 +
(b^3*c^2*d^2 - 2*a*b^2*c*d^3 + a^2*b*d^4)*x^3)*(b*x^3 + a)^(2/3))/(a*b^4*c^2*d^3 - 2*a^2*b^3*c*d^4 + a^3*b^2*
d^5 + (b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*x^3)]

________________________________________________________________________________________

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

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

________________________________________________________________________________________

Giac [A]  time = 1.19971, size = 439, normalized size = 2.16 \begin{align*} \frac{{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} c^{2} \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^{2} c^{2} d^{3} - 2 \, \sqrt{3} a b c d^{4} + \sqrt{3} a^{2} d^{5}} - \frac{{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} c^{2} \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^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )}} + \frac{c^{2} \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^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )}} - \frac{a^{2}}{{\left (b^{3} c - a b^{2} d\right )}{\left (b x^{3} + a\right )}^{\frac{1}{3}}} + \frac{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{2 \, b^{2} d} \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]

(-b*c*d^2 + a*d^3)^(2/3)*c^2*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^2*c^2*d^3 - 2*sqrt(3)*a*b*c*d^4 + sqrt(3)*a^2*d^5) - 1/6*(-b*c*d^2 + a*d^3)^(2/3)*c^2*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^2*c^2*d^3 - 2*a*b*c*
d^4 + a^2*d^5) + 1/3*c^2*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^2*c^2*
d - 2*a*b*c*d^2 + a^2*d^3) - a^2/((b^3*c - a*b^2*d)*(b*x^3 + a)^(1/3)) + 1/2*(b*x^3 + a)^(2/3)/(b^2*d)