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

Optimal. Leaf size=46 \[ \frac{\left (a+b x^2\right )^{3/2} (A b-a B)}{3 b^2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0362037, antiderivative size = 46, 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} \[ \frac{\left (a+b x^2\right )^{3/2} (A b-a B)}{3 b^2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*x^2]*(A + B*x^2),x]

[Out]

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

Mathematica [A]  time = 0.0225728, size = 34, normalized size = 0.74 \[ \frac{\left (a+b x^2\right )^{3/2} \left (-2 a B+5 A b+3 b B x^2\right )}{15 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 31, normalized size = 0.7 \begin{align*}{\frac{3\,bB{x}^{2}+5\,Ab-2\,Ba}{15\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}} \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]

1/15*(b*x^2+a)^(3/2)*(3*B*b*x^2+5*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 = 1.81207, size = 113, normalized size = 2.46 \begin{align*} \frac{{\left (3 \, B b^{2} x^{4} - 2 \, B a^{2} + 5 \, A a b +{\left (B a b + 5 \, A b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{15 \, 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]

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

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Sympy [A]  time = 0.329264, size = 110, normalized size = 2.39 \begin{align*} \begin{cases} \frac{A a \sqrt{a + b x^{2}}}{3 b} + \frac{A x^{2} \sqrt{a + b x^{2}}}{3} - \frac{2 B a^{2} \sqrt{a + b x^{2}}}{15 b^{2}} + \frac{B a x^{2} \sqrt{a + b x^{2}}}{15 b} + \frac{B x^{4} \sqrt{a + b x^{2}}}{5} & \text{for}\: b \neq 0 \\\sqrt{a} \left (\frac{A x^{2}}{2} + \frac{B x^{4}}{4}\right ) & \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*a*sqrt(a + b*x**2)/(3*b) + A*x**2*sqrt(a + b*x**2)/3 - 2*B*a**2*sqrt(a + b*x**2)/(15*b**2) + B*a*
x**2*sqrt(a + b*x**2)/(15*b) + B*x**4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*(A*x**2/2 + B*x**4/4), True))

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Giac [A]  time = 1.11125, size = 63, normalized size = 1.37 \begin{align*} \frac{5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A + \frac{{\left (3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a\right )} B}{b}}{15 \, b} \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]

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