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

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

[Out]

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

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Rubi [A]  time = 0.0188691, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {453, 191} \[ -\frac{x (2 A b-a B)}{a^2 \sqrt{a+b x^2}}-\frac{A}{a x \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[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{A+B x^2}{x^2 \left (a+b x^2\right )^{3/2}} \, dx &=-\frac{A}{a x \sqrt{a+b x^2}}-\frac{(2 A b-a B) \int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx}{a}\\ &=-\frac{A}{a x \sqrt{a+b x^2}}-\frac{(2 A b-a B) x}{a^2 \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0116595, size = 36, normalized size = 0.77 \[ \frac{-a A+a B x^2-2 A b x^2}{a^2 x \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 36, normalized size = 0.8 \begin{align*} -{\frac{2\,Ab{x}^{2}-Ba{x}^{2}+Aa}{x{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49501, size = 85, normalized size = 1.81 \begin{align*} \frac{{\left ({\left (B a - 2 \, A b\right )} x^{2} - A a\right )} \sqrt{b x^{2} + a}}{a^{2} b x^{3} + a^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.80439, size = 68, normalized size = 1.45 \begin{align*} A \left (- \frac{1}{a \sqrt{b} x^{2} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{2 \sqrt{b}}{a^{2} \sqrt{\frac{a}{b x^{2}} + 1}}\right ) + \frac{B x}{a^{\frac{3}{2}} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12052, size = 77, normalized size = 1.64 \begin{align*} \frac{2 \, A \sqrt{b}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )} a} + \frac{{\left (B a - A b\right )} x}{\sqrt{b x^{2} + a} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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