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

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

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Rubi [A]  time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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]))))

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]]

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*}

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Mathematica [A]  time = 0.05, size = 54, normalized size = 0.74 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.05, size = 56, normalized size = 0.77 \begin {gather*} \frac {2 \left (8 a^2 B-2 a A b+12 a b B x^3-3 A b^2 x^3+3 b^2 B x^6\right )}{9 b^3 \left (a+b x^3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 75, normalized size = 1.03 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.20, size = 63, normalized size = 0.86 \begin {gather*} \frac {2 \, \sqrt {b x^{3} + a} B}{3 \, b^{3}} + \frac {2 \, {\left (6 \, {\left (b x^{3} + a\right )} B a - B a^{2} - 3 \, {\left (b x^{3} + a\right )} A b + A a b\right )}}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3}} \end {gather*}

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]

2/3*sqrt(b*x^3 + a)*B/b^3 + 2/9*(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
)

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maple [A]  time = 0.05, size = 53, normalized size = 0.73 \begin {gather*} -\frac {2 \left (-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}\right )}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{3}} \end {gather*}

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 = 0.49, size = 84, normalized size = 1.15 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 2.76, size = 60, normalized size = 0.82 \begin {gather*} \frac {6\,B\,{\left (b\,x^3+a\right )}^2-2\,B\,a^2-6\,A\,b\,\left (b\,x^3+a\right )+12\,B\,a\,\left (b\,x^3+a\right )+2\,A\,a\,b}{9\,b^3\,{\left (b\,x^3+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.42, size = 240, normalized size = 3.29 \begin {gather*} \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 {gather*}

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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