3.245 \(\int \frac{x^5 (A+B x^3)}{(a+b x^3)^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0555171, 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} \[ -\frac{2 (A b-2 a B)}{3 b^3 \sqrt{a+b x^3}}+\frac{2 a (A b-a B)}{9 b^3 \left (a+b x^3\right )^{3/2}}+\frac{2 B \sqrt{a+b x^3}}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^3))/(a + b*x^3)^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.0374402, size = 54, normalized size = 0.74 \[ \frac{16 a^2 B-4 a b \left (A-6 B x^3\right )+6 b^2 x^3 \left (B x^3-A\right )}{9 b^3 \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(A + B*x^3))/(a + b*x^3)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.006, size = 53, normalized size = 0.7 \begin{align*} -{\frac{-6\,{b}^{2}B{x}^{6}+6\,A{x}^{3}{b}^{2}-24\,B{x}^{3}ab+4\,abA-16\,{a}^{2}B}{9\,{b}^{3}} \left ( b{x}^{3}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9/(b*x^3+a)^(3/2)*(-3*B*b^2*x^6+3*A*b^2*x^3-12*B*a*b*x^3+2*A*a*b-8*B*a^2)/b^3

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Maxima [A]  time = 0.925425, size = 113, normalized size = 1.55 \begin{align*} \frac{2}{9} \, B{\left (\frac{3 \, \sqrt{b x^{3} + a}}{b^{3}} + \frac{6 \, a}{\sqrt{b x^{3} + a} b^{3}} - \frac{a^{2}}{{\left (b x^{3} + a\right )}^{\frac{3}{2}} b^{3}}\right )} - \frac{2}{9} \, A{\left (\frac{3}{\sqrt{b x^{3} + a} b^{2}} - \frac{a}{{\left (b x^{3} + a\right )}^{\frac{3}{2}} 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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.7427, size = 155, normalized size = 2.12 \begin{align*} \frac{2 \,{\left (3 \, B b^{2} x^{6} + 3 \,{\left (4 \, B a b - A b^{2}\right )} x^{3} + 8 \, B a^{2} - 2 \, A a b\right )} \sqrt{b x^{3} + a}}{9 \,{\left (b^{5} x^{6} + 2 \, a b^{4} x^{3} + a^{2} b^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

2/9*(3*B*b^2*x^6 + 3*(4*B*a*b - A*b^2)*x^3 + 8*B*a^2 - 2*A*a*b)*sqrt(b*x^3 + a)/(b^5*x^6 + 2*a*b^4*x^3 + a^2*b
^3)

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Sympy [A]  time = 2.41853, size = 240, normalized size = 3.29 \begin{align*} \begin{cases} - \frac{4 A a b}{9 a b^{3} \sqrt{a + b x^{3}} + 9 b^{4} x^{3} \sqrt{a + b x^{3}}} - \frac{6 A b^{2} x^{3}}{9 a b^{3} \sqrt{a + b x^{3}} + 9 b^{4} x^{3} \sqrt{a + b x^{3}}} + \frac{16 B a^{2}}{9 a b^{3} \sqrt{a + b x^{3}} + 9 b^{4} x^{3} \sqrt{a + b x^{3}}} + \frac{24 B a b x^{3}}{9 a b^{3} \sqrt{a + b x^{3}} + 9 b^{4} x^{3} \sqrt{a + b x^{3}}} + \frac{6 B b^{2} x^{6}}{9 a b^{3} \sqrt{a + b x^{3}} + 9 b^{4} x^{3} \sqrt{a + b x^{3}}} & \text{for}\: b \neq 0 \\\frac{\frac{A x^{6}}{6} + \frac{B x^{9}}{9}}{a^{\frac{5}{2}}} & \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)**(5/2),x)

[Out]

Piecewise((-4*A*a*b/(9*a*b**3*sqrt(a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x**3)) - 6*A*b**2*x**3/(9*a*b**3*sqrt(
a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x**3)) + 16*B*a**2/(9*a*b**3*sqrt(a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x*
*3)) + 24*B*a*b*x**3/(9*a*b**3*sqrt(a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x**3)) + 6*B*b**2*x**6/(9*a*b**3*sqrt
(a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x**3)), Ne(b, 0)), ((A*x**6/6 + B*x**9/9)/a**(5/2), True))

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Giac [A]  time = 1.13762, size = 82, normalized size = 1.12 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{b x^{3} + a} B + \frac{6 \,{\left (b x^{3} + a\right )} B a - B a^{2} - 3 \,{\left (b x^{3} + a\right )} A b + A a b}{{\left (b x^{3} + a\right )}^{\frac{3}{2}}}\right )}}{9 \, 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)^(5/2),x, algorithm="giac")

[Out]

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