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

Optimal. Leaf size=47 \[ \frac{B x}{3 a b \sqrt{a+b x^2}}-\frac{A+B x}{3 b \left (a+b x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0144857, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {778, 191} \[ \frac{B x}{3 a b \sqrt{a+b x^2}}-\frac{A+B x}{3 b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.0171072, size = 32, normalized size = 0.68 \[ \frac{b B x^3-a A}{3 a b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 29, normalized size = 0.6 \begin{align*} -{\frac{-bB{x}^{3}+Aa}{3\,ab} \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+A)/(b*x^2+a)^(5/2),x)

[Out]

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

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Maxima [A]  time = 0.991677, size = 69, normalized size = 1.47 \begin{align*} -\frac{B x}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} b} + \frac{B x}{3 \, \sqrt{b x^{2} + a} a b} - \frac{A}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.54832, size = 99, normalized size = 2.11 \begin{align*} \frac{{\left (B b x^{3} - A a\right )} \sqrt{b x^{2} + a}}{3 \,{\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 9.03744, size = 95, normalized size = 2.02 \begin{align*} A \left (\begin{cases} - \frac{1}{3 a b \sqrt{a + b x^{2}} + 3 b^{2} x^{2} \sqrt{a + b x^{2}}} & \text{for}\: b \neq 0 \\\frac{x^{2}}{2 a^{\frac{5}{2}}} & \text{otherwise} \end{cases}\right ) + \frac{B x^{3}}{3 a^{\frac{5}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{3}{2}} b x^{2} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*Piecewise((-1/(3*a*b*sqrt(a + b*x**2) + 3*b**2*x**2*sqrt(a + b*x**2)), Ne(b, 0)), (x**2/(2*a**(5/2)), True))
+ B*x**3/(3*a**(5/2)*sqrt(1 + b*x**2/a) + 3*a**(3/2)*b*x**2*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.16797, size = 35, normalized size = 0.74 \begin{align*} \frac{\frac{B x^{3}}{a} - \frac{A}{b}}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(B*x^3/a - A/b)/(b*x^2 + a)^(3/2)