3.557 \(\int \frac{x^3 (A+B x^2)}{\sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=71 \[ \frac{\left (a+b x^2\right )^{3/2} (A b-2 a B)}{3 b^3}-\frac{a \sqrt{a+b x^2} (A b-a B)}{b^3}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0549245, antiderivative size = 71, 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} \[ \frac{\left (a+b x^2\right )^{3/2} (A b-2 a B)}{3 b^3}-\frac{a \sqrt{a+b x^2} (A b-a B)}{b^3}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0355832, size = 56, normalized size = 0.79 \[ \frac{\sqrt{a+b x^2} \left (8 a^2 B-2 a b \left (5 A+2 B x^2\right )+b^2 x^2 \left (5 A+3 B x^2\right )\right )}{15 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 53, normalized size = 0.8 \begin{align*} -{\frac{-3\,{b}^{2}B{x}^{4}-5\,A{b}^{2}{x}^{2}+4\,Bab{x}^{2}+10\,abA-8\,{a}^{2}B}{15\,{b}^{3}}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(b*x^2+a)^(1/2)*(-3*B*b^2*x^4-5*A*b^2*x^2+4*B*a*b*x^2+10*A*a*b-8*B*a^2)/b^3

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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^3*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63704, size = 117, normalized size = 1.65 \begin{align*} \frac{{\left (3 \, B b^{2} x^{4} + 8 \, B a^{2} - 10 \, A a b -{\left (4 \, B a b - 5 \, A b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{15 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*B*b^2*x^4 + 8*B*a^2 - 10*A*a*b - (4*B*a*b - 5*A*b^2)*x^2)*sqrt(b*x^2 + a)/b^3

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Sympy [A]  time = 0.843588, size = 121, normalized size = 1.7 \begin{align*} \begin{cases} - \frac{2 A a \sqrt{a + b x^{2}}}{3 b^{2}} + \frac{A x^{2} \sqrt{a + b x^{2}}}{3 b} + \frac{8 B a^{2} \sqrt{a + b x^{2}}}{15 b^{3}} - \frac{4 B a x^{2} \sqrt{a + b x^{2}}}{15 b^{2}} + \frac{B x^{4} \sqrt{a + b x^{2}}}{5 b} & \text{for}\: b \neq 0 \\\frac{\frac{A x^{4}}{4} + \frac{B x^{6}}{6}}{\sqrt{a}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x**2+A)/(b*x**2+a)**(1/2),x)

[Out]

Piecewise((-2*A*a*sqrt(a + b*x**2)/(3*b**2) + A*x**2*sqrt(a + b*x**2)/(3*b) + 8*B*a**2*sqrt(a + b*x**2)/(15*b*
*3) - 4*B*a*x**2*sqrt(a + b*x**2)/(15*b**2) + B*x**4*sqrt(a + b*x**2)/(5*b), Ne(b, 0)), ((A*x**4/4 + B*x**6/6)
/sqrt(a), True))

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Giac [A]  time = 1.1498, size = 99, normalized size = 1.39 \begin{align*} \frac{3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} B - 10 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a + 15 \, \sqrt{b x^{2} + a} B a^{2} + 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A b - 15 \, \sqrt{b x^{2} + a} A a b}{15 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/15*(3*(b*x^2 + a)^(5/2)*B - 10*(b*x^2 + a)^(3/2)*B*a + 15*sqrt(b*x^2 + a)*B*a^2 + 5*(b*x^2 + a)^(3/2)*A*b -
15*sqrt(b*x^2 + a)*A*a*b)/b^3