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

Optimal. Leaf size=46 \[ -\frac {2 (A b-a B)}{9 b^2 \left (a+b x^3\right )^{3/2}}-\frac {2 B}{3 b^2 \sqrt {a+b x^3}} \]

[Out]

-2/9*(A*b-B*a)/b^2/(b*x^3+a)^(3/2)-2/3*B/b^2/(b*x^3+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 46, 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 = {444, 43} \[ -\frac {2 (A b-a B)}{9 b^2 \left (a+b x^3\right )^{3/2}}-\frac {2 B}{3 b^2 \sqrt {a+b x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 33, normalized size = 0.72 \[ -\frac {2 \left (2 a B+A b+3 b B x^3\right )}{9 b^2 \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 52, normalized size = 1.13 \[ -\frac {2 \, {\left (3 \, B b x^{3} + 2 \, B a + A b\right )} \sqrt {b x^{3} + a}}{9 \, {\left (b^{4} x^{6} + 2 \, a b^{3} x^{3} + a^{2} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 32, normalized size = 0.70 \[ -\frac {2 \, {\left (3 \, {\left (b x^{3} + a\right )} B - B a + A b\right )}}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 30, normalized size = 0.65 \[ -\frac {2 \left (3 B b \,x^{3}+A b +2 B a \right )}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 49, normalized size = 1.07 \[ -\frac {2}{9} \, B {\left (\frac {3}{\sqrt {b x^{3} + a} b^{2}} - \frac {a}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2}}\right )} - \frac {2 \, A}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.68, size = 33, normalized size = 0.72 \[ -\frac {2\,A\,b-2\,B\,a+6\,B\,\left (b\,x^3+a\right )}{9\,b^2\,{\left (b\,x^3+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.34, size = 144, normalized size = 3.13 \[ \begin {cases} - \frac {2 A b}{9 a b^{2} \sqrt {a + b x^{3}} + 9 b^{3} x^{3} \sqrt {a + b x^{3}}} - \frac {4 B a}{9 a b^{2} \sqrt {a + b x^{3}} + 9 b^{3} x^{3} \sqrt {a + b x^{3}}} - \frac {6 B b x^{3}}{9 a b^{2} \sqrt {a + b x^{3}} + 9 b^{3} x^{3} \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{3}}{3} + \frac {B x^{6}}{6}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________