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

Optimal. Leaf size=43 \[ \frac{\sqrt{a+b x^2} (A b-a B)}{b^2}+\frac{B \left (a+b x^2\right )^{3/2}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0335874, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {444, 43} \[ \frac{\sqrt{a+b x^2} (A b-a B)}{b^2}+\frac{B \left (a+b x^2\right )^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.0217856, size = 33, normalized size = 0.77 \[ \frac{\sqrt{a+b x^2} \left (-2 a B+3 A b+b B x^2\right )}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 30, normalized size = 0.7 \begin{align*}{\frac{bB{x}^{2}+3\,Ab-2\,Ba}{3\,{b}^{2}}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56539, size = 69, normalized size = 1.6 \begin{align*} \frac{{\left (B b x^{2} - 2 \, B a + 3 \, A b\right )} \sqrt{b x^{2} + a}}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.506212, size = 70, normalized size = 1.63 \begin{align*} \begin{cases} \frac{A \sqrt{a + b x^{2}}}{b} - \frac{2 B a \sqrt{a + b x^{2}}}{3 b^{2}} + \frac{B x^{2} \sqrt{a + b x^{2}}}{3 b} & \text{for}\: b \neq 0 \\\frac{\frac{A x^{2}}{2} + \frac{B x^{4}}{4}}{\sqrt{a}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11937, size = 58, normalized size = 1.35 \begin{align*} \frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}} B - 3 \, \sqrt{b x^{2} + a} B a + 3 \, \sqrt{b x^{2} + a} A b}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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