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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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 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])

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]

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 = 0.02, 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)

________________________________________________________________________________________

fricas [A]  time = 1.16, size = 29, normalized size = 0.67 \[ \frac {{\left (B b x^{2} - 2 \, B a + 3 \, A b\right )} \sqrt {b x^{2} + a}}{3 \, b^{2}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.31, size = 38, normalized size = 0.88 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{3 \, b^{2}} - \frac {\sqrt {b x^{2} + a} {\left (B a - A b\right )}}{b^{2}} \]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 30, normalized size = 0.70 \[ \frac {\sqrt {b \,x^{2}+a}\, \left (B b \,x^{2}+3 A b -2 B a \right )}{3 b^{2}} \]

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

________________________________________________________________________________________

maxima [A]  time = 1.00, size = 49, normalized size = 1.14 \[ \frac {\sqrt {b x^{2} + a} B x^{2}}{3 \, b} - \frac {2 \, \sqrt {b x^{2} + a} B a}{3 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A}{b} \]

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]

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

________________________________________________________________________________________

mupad [B]  time = 0.57, size = 34, normalized size = 0.79 \[ \left (\frac {3\,A\,b-2\,B\,a}{3\,b^2}+\frac {B\,x^2}{3\,b}\right )\,\sqrt {b\,x^2+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.65, size = 70, normalized size = 1.63 \[ \begin {cases} \frac {A \sqrt {a + b x^{2}}}{b} - \frac {2 B a \sqrt {a + b x^{2}}}{3 b^{2}} + \frac {B x^{2} \sqrt {a + b x^{2}}}{3 b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{2}}{2} + \frac {B x^{4}}{4}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

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))

________________________________________________________________________________________