3.1864 \(\int \frac{1}{(a+\frac{b}{x^2})^2} \, dx\)

Optimal. Leaf size=55 \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 a^{5/2}}+\frac{3 x}{2 a^2}-\frac{x^3}{2 a \left (a x^2+b\right )} \]

[Out]

(3*x)/(2*a^2) - x^3/(2*a*(b + a*x^2)) - (3*Sqrt[b]*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*a^(5/2))

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Rubi [A]  time = 0.0181359, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {193, 288, 321, 205} \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 a^{5/2}}+\frac{3 x}{2 a^2}-\frac{x^3}{2 a \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(-2),x]

[Out]

(3*x)/(2*a^2) - x^3/(2*a*(b + a*x^2)) - (3*Sqrt[b]*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*a^(5/2))

Rule 193

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x^2}\right )^2} \, dx &=\int \frac{x^4}{\left (b+a x^2\right )^2} \, dx\\ &=-\frac{x^3}{2 a \left (b+a x^2\right )}+\frac{3 \int \frac{x^2}{b+a x^2} \, dx}{2 a}\\ &=\frac{3 x}{2 a^2}-\frac{x^3}{2 a \left (b+a x^2\right )}-\frac{(3 b) \int \frac{1}{b+a x^2} \, dx}{2 a^2}\\ &=\frac{3 x}{2 a^2}-\frac{x^3}{2 a \left (b+a x^2\right )}-\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0310112, size = 51, normalized size = 0.93 \[ \frac{b x}{2 a^2 \left (a x^2+b\right )}-\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 a^{5/2}}+\frac{x}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(-2),x]

[Out]

x/a^2 + (b*x)/(2*a^2*(b + a*x^2)) - (3*Sqrt[b]*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*a^(5/2))

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Maple [A]  time = 0.008, size = 43, normalized size = 0.8 \begin{align*}{\frac{x}{{a}^{2}}}+{\frac{bx}{2\,{a}^{2} \left ( a{x}^{2}+b \right ) }}-{\frac{3\,b}{2\,{a}^{2}}\arctan \left ({ax{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(1/(a+b/x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47755, size = 285, normalized size = 5.18 \begin{align*} \left [\frac{4 \, a x^{3} + 3 \,{\left (a x^{2} + b\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{a x^{2} - 2 \, a x \sqrt{-\frac{b}{a}} - b}{a x^{2} + b}\right ) + 6 \, b x}{4 \,{\left (a^{3} x^{2} + a^{2} b\right )}}, \frac{2 \, a x^{3} - 3 \,{\left (a x^{2} + b\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a x \sqrt{\frac{b}{a}}}{b}\right ) + 3 \, b x}{2 \,{\left (a^{3} x^{2} + a^{2} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*a*x^3 + 3*(a*x^2 + b)*sqrt(-b/a)*log((a*x^2 - 2*a*x*sqrt(-b/a) - b)/(a*x^2 + b)) + 6*b*x)/(a^3*x^2 + a
^2*b), 1/2*(2*a*x^3 - 3*(a*x^2 + b)*sqrt(b/a)*arctan(a*x*sqrt(b/a)/b) + 3*b*x)/(a^3*x^2 + a^2*b)]

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Sympy [A]  time = 0.524379, size = 83, normalized size = 1.51 \begin{align*} \frac{b x}{2 a^{3} x^{2} + 2 a^{2} b} + \frac{3 \sqrt{- \frac{b}{a^{5}}} \log{\left (- a^{2} \sqrt{- \frac{b}{a^{5}}} + x \right )}}{4} - \frac{3 \sqrt{- \frac{b}{a^{5}}} \log{\left (a^{2} \sqrt{- \frac{b}{a^{5}}} + x \right )}}{4} + \frac{x}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/(2*a**3*x**2 + 2*a**2*b) + 3*sqrt(-b/a**5)*log(-a**2*sqrt(-b/a**5) + x)/4 - 3*sqrt(-b/a**5)*log(a**2*sqrt(
-b/a**5) + x)/4 + x/a**2

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Giac [A]  time = 1.15568, size = 57, normalized size = 1.04 \begin{align*} -\frac{3 \, b \arctan \left (\frac{a x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{2}} + \frac{x}{a^{2}} + \frac{b x}{2 \,{\left (a x^{2} + b\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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