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

Optimal. Leaf size=68 \[ -\frac {A b-2 a B}{b^3 \sqrt {a+b x^2}}+\frac {a (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}+\frac {B \sqrt {a+b x^2}}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 68, 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} \[ -\frac {A b-2 a B}{b^3 \sqrt {a+b x^2}}+\frac {a (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}+\frac {B \sqrt {a+b x^2}}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 54, normalized size = 0.79 \[ \frac {8 a^2 B-2 a b \left (A-6 B x^2\right )+3 b^2 x^2 \left (B x^2-A\right )}{3 b^3 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.88, size = 75, normalized size = 1.10 \[ \frac {{\left (3 \, B b^{2} x^{4} + 8 \, B a^{2} - 2 \, A a b + 3 \, {\left (4 \, B a b - A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.34, size = 62, normalized size = 0.91 \[ \frac {\sqrt {b x^{2} + a} B}{b^{3}} + \frac {6 \, {\left (b x^{2} + a\right )} B a - B a^{2} - 3 \, {\left (b x^{2} + a\right )} A b + A a b}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b*x^2 + a)*B/b^3 + 1/3*(6*(b*x^2 + a)*B*a - B*a^2 - 3*(b*x^2 + a)*A*b + A*a*b)/((b*x^2 + a)^(3/2)*b^3)

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maple [A]  time = 0.01, size = 53, normalized size = 0.78 \[ -\frac {-3 B \,b^{2} x^{4}+3 A \,b^{2} x^{2}-12 B a b \,x^{2}+2 a b A -8 a^{2} B}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-3*B*b^2*x^4+3*A*b^2*x^2-12*B*a*b*x^2+2*A*a*b-8*B*a^2)/(b*x^2+a)^(3/2)/b^3

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maxima [A]  time = 1.07, size = 89, normalized size = 1.31 \[ \frac {B x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {4 \, B a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} - \frac {A x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {8 \, B a^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} - \frac {2 \, A a}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.67, size = 59, normalized size = 0.87 \[ \frac {3\,B\,{\left (b\,x^2+a\right )}^2-B\,a^2-3\,A\,b\,\left (b\,x^2+a\right )+6\,B\,a\,\left (b\,x^2+a\right )+A\,a\,b}{3\,b^3\,{\left (b\,x^2+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.97, size = 240, normalized size = 3.53 \[ \begin {cases} - \frac {2 A a b}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} - \frac {3 A b^{2} x^{2}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} + \frac {8 B a^{2}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} + \frac {12 B a b x^{2}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} + \frac {3 B b^{2} x^{4}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{4}}{4} + \frac {B x^{6}}{6}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*A*a*b/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2)) - 3*A*b**2*x**2/(3*a*b**3*sqrt(
a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2)) + 8*B*a**2/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**
2)) + 12*B*a*b*x**2/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2)) + 3*B*b**2*x**4/(3*a*b**3*sqrt(
a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2)), Ne(b, 0)), ((A*x**4/4 + B*x**6/6)/a**(5/2), True))

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