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

Optimal. Leaf size=99 \[ -\frac{a^2 (A b-a B)}{b^4 \sqrt{a+b x^2}}+\frac{\left (a+b x^2\right )^{3/2} (A b-3 a B)}{3 b^4}-\frac{a \sqrt{a+b x^2} (2 A b-3 a B)}{b^4}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^4} \]

[Out]

-((a^2*(A*b - a*B))/(b^4*Sqrt[a + b*x^2])) - (a*(2*A*b - 3*a*B)*Sqrt[a + b*x^2])/b^4 + ((A*b - 3*a*B)*(a + b*x
^2)^(3/2))/(3*b^4) + (B*(a + b*x^2)^(5/2))/(5*b^4)

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Rubi [A]  time = 0.0761042, antiderivative size = 99, 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} \[ -\frac{a^2 (A b-a B)}{b^4 \sqrt{a+b x^2}}+\frac{\left (a+b x^2\right )^{3/2} (A b-3 a B)}{3 b^4}-\frac{a \sqrt{a+b x^2} (2 A b-3 a B)}{b^4}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*(A*b - a*B))/(b^4*Sqrt[a + b*x^2])) - (a*(2*A*b - 3*a*B)*Sqrt[a + b*x^2])/b^4 + ((A*b - 3*a*B)*(a + b*x
^2)^(3/2))/(3*b^4) + (B*(a + b*x^2)^(5/2))/(5*b^4)

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^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 (A+B x)}{(a+b x)^{3/2}} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{a^2 (-A b+a B)}{b^3 (a+b x)^{3/2}}+\frac{a (-2 A b+3 a B)}{b^3 \sqrt{a+b x}}+\frac{(A b-3 a B) \sqrt{a+b x}}{b^3}+\frac{B (a+b x)^{3/2}}{b^3}\right ) \, dx,x,x^2\right )\\ &=-\frac{a^2 (A b-a B)}{b^4 \sqrt{a+b x^2}}-\frac{a (2 A b-3 a B) \sqrt{a+b x^2}}{b^4}+\frac{(A b-3 a B) \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0505392, size = 77, normalized size = 0.78 \[ \frac{-8 a^2 b \left (5 A-3 B x^2\right )+48 a^3 B-2 a b^2 x^2 \left (10 A+3 B x^2\right )+b^3 x^4 \left (5 A+3 B x^2\right )}{15 b^4 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*a^3*B - 8*a^2*b*(5*A - 3*B*x^2) + b^3*x^4*(5*A + 3*B*x^2) - 2*a*b^2*x^2*(10*A + 3*B*x^2))/(15*b^4*Sqrt[a +
b*x^2])

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Maple [A]  time = 0.006, size = 77, normalized size = 0.8 \begin{align*} -{\frac{-3\,{x}^{6}B{b}^{3}-5\,A{b}^{3}{x}^{4}+6\,Ba{b}^{2}{x}^{4}+20\,Aa{b}^{2}{x}^{2}-24\,B{a}^{2}b{x}^{2}+40\,A{a}^{2}b-48\,B{a}^{3}}{15\,{b}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(-3*B*b^3*x^6-5*A*b^3*x^4+6*B*a*b^2*x^4+20*A*a*b^2*x^2-24*B*a^2*b*x^2+40*A*a^2*b-48*B*a^3)/(b*x^2+a)^(1/
2)/b^4

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61189, size = 186, normalized size = 1.88 \begin{align*} \frac{{\left (3 \, B b^{3} x^{6} -{\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} x^{4} + 48 \, B a^{3} - 40 \, A a^{2} b + 4 \,{\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{15 \,{\left (b^{5} x^{2} + a b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*B*b^3*x^6 - (6*B*a*b^2 - 5*A*b^3)*x^4 + 48*B*a^3 - 40*A*a^2*b + 4*(6*B*a^2*b - 5*A*a*b^2)*x^2)*sqrt(b*
x^2 + a)/(b^5*x^2 + a*b^4)

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Sympy [A]  time = 1.52397, size = 172, normalized size = 1.74 \begin{align*} \begin{cases} - \frac{8 A a^{2}}{3 b^{3} \sqrt{a + b x^{2}}} - \frac{4 A a x^{2}}{3 b^{2} \sqrt{a + b x^{2}}} + \frac{A x^{4}}{3 b \sqrt{a + b x^{2}}} + \frac{16 B a^{3}}{5 b^{4} \sqrt{a + b x^{2}}} + \frac{8 B a^{2} x^{2}}{5 b^{3} \sqrt{a + b x^{2}}} - \frac{2 B a x^{4}}{5 b^{2} \sqrt{a + b x^{2}}} + \frac{B x^{6}}{5 b \sqrt{a + b x^{2}}} & \text{for}\: b \neq 0 \\\frac{\frac{A x^{6}}{6} + \frac{B x^{8}}{8}}{a^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-8*A*a**2/(3*b**3*sqrt(a + b*x**2)) - 4*A*a*x**2/(3*b**2*sqrt(a + b*x**2)) + A*x**4/(3*b*sqrt(a + b
*x**2)) + 16*B*a**3/(5*b**4*sqrt(a + b*x**2)) + 8*B*a**2*x**2/(5*b**3*sqrt(a + b*x**2)) - 2*B*a*x**4/(5*b**2*s
qrt(a + b*x**2)) + B*x**6/(5*b*sqrt(a + b*x**2)), Ne(b, 0)), ((A*x**6/6 + B*x**8/8)/a**(3/2), True))

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Giac [A]  time = 1.1427, size = 131, normalized size = 1.32 \begin{align*} \frac{3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} B - 15 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a + 45 \, \sqrt{b x^{2} + a} B a^{2} + 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A b - 30 \, \sqrt{b x^{2} + a} A a b + \frac{15 \,{\left (B a^{3} - A a^{2} b\right )}}{\sqrt{b x^{2} + a}}}{15 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*(b*x^2 + a)^(5/2)*B - 15*(b*x^2 + a)^(3/2)*B*a + 45*sqrt(b*x^2 + a)*B*a^2 + 5*(b*x^2 + a)^(3/2)*A*b -
30*sqrt(b*x^2 + a)*A*a*b + 15*(B*a^3 - A*a^2*b)/sqrt(b*x^2 + a))/b^4