3.228 \(\int \frac{x^8 (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=103 \[ -\frac{2 a^2 (A b-a B)}{3 b^4 \sqrt{a+b x^3}}+\frac{2 \left (a+b x^3\right )^{3/2} (A b-3 a B)}{9 b^4}-\frac{2 a \sqrt{a+b x^3} (2 A b-3 a B)}{3 b^4}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b^4} \]

[Out]

(-2*a^2*(A*b - a*B))/(3*b^4*Sqrt[a + b*x^3]) - (2*a*(2*A*b - 3*a*B)*Sqrt[a + b*x^3])/(3*b^4) + (2*(A*b - 3*a*B
)*(a + b*x^3)^(3/2))/(9*b^4) + (2*B*(a + b*x^3)^(5/2))/(15*b^4)

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Rubi [A]  time = 0.0767542, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {446, 77} \[ -\frac{2 a^2 (A b-a B)}{3 b^4 \sqrt{a+b x^3}}+\frac{2 \left (a+b x^3\right )^{3/2} (A b-3 a B)}{9 b^4}-\frac{2 a \sqrt{a+b x^3} (2 A b-3 a B)}{3 b^4}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^3))/(a + b*x^3)^(3/2),x]

[Out]

(-2*a^2*(A*b - a*B))/(3*b^4*Sqrt[a + b*x^3]) - (2*a*(2*A*b - 3*a*B)*Sqrt[a + b*x^3])/(3*b^4) + (2*(A*b - 3*a*B
)*(a + b*x^3)^(3/2))/(9*b^4) + (2*B*(a + b*x^3)^(5/2))/(15*b^4)

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2 (A+B x)}{(a+b x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{a^2 (-A b+a B)}{b^3 (a+b x)^{3/2}}+\frac{a (-2 A b+3 a B)}{b^3 \sqrt{a+b x}}+\frac{(A b-3 a B) \sqrt{a+b x}}{b^3}+\frac{B (a+b x)^{3/2}}{b^3}\right ) \, dx,x,x^3\right )\\ &=-\frac{2 a^2 (A b-a B)}{3 b^4 \sqrt{a+b x^3}}-\frac{2 a (2 A b-3 a B) \sqrt{a+b x^3}}{3 b^4}+\frac{2 (A b-3 a B) \left (a+b x^3\right )^{3/2}}{9 b^4}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0515247, size = 77, normalized size = 0.75 \[ \frac{2 \left (-8 a^2 b \left (5 A-3 B x^3\right )+48 a^3 B-2 a b^2 x^3 \left (10 A+3 B x^3\right )+b^3 x^6 \left (5 A+3 B x^3\right )\right )}{45 b^4 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^3))/(a + b*x^3)^(3/2),x]

[Out]

(2*(48*a^3*B - 8*a^2*b*(5*A - 3*B*x^3) + b^3*x^6*(5*A + 3*B*x^3) - 2*a*b^2*x^3*(10*A + 3*B*x^3)))/(45*b^4*Sqrt
[a + b*x^3])

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Maple [A]  time = 0.007, size = 77, normalized size = 0.8 \begin{align*} -{\frac{-6\,B{x}^{9}{b}^{3}-10\,A{b}^{3}{x}^{6}+12\,Ba{b}^{2}{x}^{6}+40\,Aa{b}^{2}{x}^{3}-48\,B{a}^{2}b{x}^{3}+80\,A{a}^{2}b-96\,B{a}^{3}}{45\,{b}^{4}}{\frac{1}{\sqrt{b{x}^{3}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x)

[Out]

-2/45/(b*x^3+a)^(1/2)*(-3*B*b^3*x^9-5*A*b^3*x^6+6*B*a*b^2*x^6+20*A*a*b^2*x^3-24*B*a^2*b*x^3+40*A*a^2*b-48*B*a^
3)/b^4

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Maxima [A]  time = 0.943429, size = 157, normalized size = 1.52 \begin{align*} \frac{2}{15} \, B{\left (\frac{{\left (b x^{3} + a\right )}^{\frac{5}{2}}}{b^{4}} - \frac{5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a}{b^{4}} + \frac{15 \, \sqrt{b x^{3} + a} a^{2}}{b^{4}} + \frac{5 \, a^{3}}{\sqrt{b x^{3} + a} b^{4}}\right )} + \frac{2}{9} \, A{\left (\frac{{\left (b x^{3} + a\right )}^{\frac{3}{2}}}{b^{3}} - \frac{6 \, \sqrt{b x^{3} + a} a}{b^{3}} - \frac{3 \, a^{2}}{\sqrt{b x^{3} + a} b^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="maxima")

[Out]

2/15*B*((b*x^3 + a)^(5/2)/b^4 - 5*(b*x^3 + a)^(3/2)*a/b^4 + 15*sqrt(b*x^3 + a)*a^2/b^4 + 5*a^3/(sqrt(b*x^3 + a
)*b^4)) + 2/9*A*((b*x^3 + a)^(3/2)/b^3 - 6*sqrt(b*x^3 + a)*a/b^3 - 3*a^2/(sqrt(b*x^3 + a)*b^3))

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Fricas [A]  time = 1.75572, size = 186, normalized size = 1.81 \begin{align*} \frac{2 \,{\left (3 \, B b^{3} x^{9} -{\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 48 \, B a^{3} - 40 \, A a^{2} b + 4 \,{\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{3}\right )} \sqrt{b x^{3} + a}}{45 \,{\left (b^{5} x^{3} + a b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

2/45*(3*B*b^3*x^9 - (6*B*a*b^2 - 5*A*b^3)*x^6 + 48*B*a^3 - 40*A*a^2*b + 4*(6*B*a^2*b - 5*A*a*b^2)*x^3)*sqrt(b*
x^3 + a)/(b^5*x^3 + a*b^4)

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Sympy [A]  time = 3.70818, size = 175, normalized size = 1.7 \begin{align*} \begin{cases} - \frac{16 A a^{2}}{9 b^{3} \sqrt{a + b x^{3}}} - \frac{8 A a x^{3}}{9 b^{2} \sqrt{a + b x^{3}}} + \frac{2 A x^{6}}{9 b \sqrt{a + b x^{3}}} + \frac{32 B a^{3}}{15 b^{4} \sqrt{a + b x^{3}}} + \frac{16 B a^{2} x^{3}}{15 b^{3} \sqrt{a + b x^{3}}} - \frac{4 B a x^{6}}{15 b^{2} \sqrt{a + b x^{3}}} + \frac{2 B x^{9}}{15 b \sqrt{a + b x^{3}}} & \text{for}\: b \neq 0 \\\frac{\frac{A x^{9}}{9} + \frac{B x^{12}}{12}}{a^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(B*x**3+A)/(b*x**3+a)**(3/2),x)

[Out]

Piecewise((-16*A*a**2/(9*b**3*sqrt(a + b*x**3)) - 8*A*a*x**3/(9*b**2*sqrt(a + b*x**3)) + 2*A*x**6/(9*b*sqrt(a
+ b*x**3)) + 32*B*a**3/(15*b**4*sqrt(a + b*x**3)) + 16*B*a**2*x**3/(15*b**3*sqrt(a + b*x**3)) - 4*B*a*x**6/(15
*b**2*sqrt(a + b*x**3)) + 2*B*x**9/(15*b*sqrt(a + b*x**3)), Ne(b, 0)), ((A*x**9/9 + B*x**12/12)/a**(3/2), True
))

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Giac [A]  time = 1.14203, size = 131, normalized size = 1.27 \begin{align*} \frac{2 \,{\left (3 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} B - 15 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} B a + 45 \, \sqrt{b x^{3} + a} B a^{2} + 5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} A b - 30 \, \sqrt{b x^{3} + a} A a b + \frac{15 \,{\left (B a^{3} - A a^{2} b\right )}}{\sqrt{b x^{3} + a}}\right )}}{45 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

2/45*(3*(b*x^3 + a)^(5/2)*B - 15*(b*x^3 + a)^(3/2)*B*a + 45*sqrt(b*x^3 + a)*B*a^2 + 5*(b*x^3 + a)^(3/2)*A*b -
30*sqrt(b*x^3 + a)*A*a*b + 15*(B*a^3 - A*a^2*b)/sqrt(b*x^3 + a))/b^4