3.214 \(\int \frac{x^5 (A+B x^3)}{\sqrt{a+b x^3}} \, dx\)

Optimal. Leaf size=73 \[ \frac{2 \left (a+b x^3\right )^{3/2} (A b-2 a B)}{9 b^3}-\frac{2 a \sqrt{a+b x^3} (A b-a B)}{3 b^3}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0546994, antiderivative size = 73, 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} \[ \frac{2 \left (a+b x^3\right )^{3/2} (A b-2 a B)}{9 b^3}-\frac{2 a \sqrt{a+b x^3} (A b-a B)}{3 b^3}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^3))/Sqrt[a + b*x^3],x]

[Out]

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

Mathematica [A]  time = 0.039873, size = 56, normalized size = 0.77 \[ \frac{2 \sqrt{a+b x^3} \left (8 a^2 B-2 a b \left (5 A+2 B x^3\right )+b^2 x^3 \left (5 A+3 B x^3\right )\right )}{45 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(A + B*x^3))/Sqrt[a + b*x^3],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 53, normalized size = 0.7 \begin{align*} -{\frac{-6\,{b}^{2}B{x}^{6}-10\,A{x}^{3}{b}^{2}+8\,B{x}^{3}ab+20\,abA-16\,{a}^{2}B}{45\,{b}^{3}}\sqrt{b{x}^{3}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^3+A)/(b*x^3+a)^(1/2),x)

[Out]

-2/45*(b*x^3+a)^(1/2)*(-3*B*b^2*x^6-5*A*b^2*x^3+4*B*a*b*x^3+10*A*a*b-8*B*a^2)/b^3

________________________________________________________________________________________

Maxima [A]  time = 0.929006, size = 112, normalized size = 1.53 \begin{align*} \frac{2}{45} \, B{\left (\frac{3 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}}}{b^{3}} - \frac{10 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a}{b^{3}} + \frac{15 \, \sqrt{b x^{3} + a} a^{2}}{b^{3}}\right )} + \frac{2}{9} \, A{\left (\frac{{\left (b x^{3} + a\right )}^{\frac{3}{2}}}{b^{2}} - \frac{3 \, \sqrt{b x^{3} + a} a}{b^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

2/45*B*(3*(b*x^3 + a)^(5/2)/b^3 - 10*(b*x^3 + a)^(3/2)*a/b^3 + 15*sqrt(b*x^3 + a)*a^2/b^3) + 2/9*A*((b*x^3 + a
)^(3/2)/b^2 - 3*sqrt(b*x^3 + a)*a/b^2)

________________________________________________________________________________________

Fricas [A]  time = 1.6849, size = 117, normalized size = 1.6 \begin{align*} \frac{2 \,{\left (3 \, B b^{2} x^{6} -{\left (4 \, B a b - 5 \, A b^{2}\right )} x^{3} + 8 \, B a^{2} - 10 \, A a b\right )} \sqrt{b x^{3} + a}}{45 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.59172, size = 124, normalized size = 1.7 \begin{align*} \begin{cases} - \frac{4 A a \sqrt{a + b x^{3}}}{9 b^{2}} + \frac{2 A x^{3} \sqrt{a + b x^{3}}}{9 b} + \frac{16 B a^{2} \sqrt{a + b x^{3}}}{45 b^{3}} - \frac{8 B a x^{3} \sqrt{a + b x^{3}}}{45 b^{2}} + \frac{2 B x^{6} \sqrt{a + b x^{3}}}{15 b} & \text{for}\: b \neq 0 \\\frac{\frac{A x^{6}}{6} + \frac{B x^{9}}{9}}{\sqrt{a}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**3+A)/(b*x**3+a)**(1/2),x)

[Out]

Piecewise((-4*A*a*sqrt(a + b*x**3)/(9*b**2) + 2*A*x**3*sqrt(a + b*x**3)/(9*b) + 16*B*a**2*sqrt(a + b*x**3)/(45
*b**3) - 8*B*a*x**3*sqrt(a + b*x**3)/(45*b**2) + 2*B*x**6*sqrt(a + b*x**3)/(15*b), Ne(b, 0)), ((A*x**6/6 + B*x
**9/9)/sqrt(a), True))

________________________________________________________________________________________

Giac [A]  time = 1.11859, size = 99, normalized size = 1.36 \begin{align*} \frac{2 \,{\left (3 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} B - 10 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} B a + 15 \, \sqrt{b x^{3} + a} B a^{2} + 5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} A b - 15 \, \sqrt{b x^{3} + a} A a b\right )}}{45 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

2/45*(3*(b*x^3 + a)^(5/2)*B - 10*(b*x^3 + a)^(3/2)*B*a + 15*sqrt(b*x^3 + a)*B*a^2 + 5*(b*x^3 + a)^(3/2)*A*b -
15*sqrt(b*x^3 + a)*A*a*b)/b^3