3.1866 \(\int \frac {1}{(a+\frac {b}{x^2})^2 x^2} \, dx\)

Optimal. Leaf size=45 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{3/2} \sqrt {b}}-\frac {x}{2 a \left (a x^2+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {263, 288, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{3/2} \sqrt {b}}-\frac {x}{2 a \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^2*x^2),x]

[Out]

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

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]

Rule 263

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 45, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{3/2} \sqrt {b}}-\frac {x}{2 a \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^2*x^2),x]

[Out]

-1/2*x/(a*(b + a*x^2)) + ArcTan[(Sqrt[a]*x)/Sqrt[b]]/(2*a^(3/2)*Sqrt[b])

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fricas [A]  time = 0.91, size = 120, normalized size = 2.67 \[ \left [-\frac {2 \, a b x + {\left (a x^{2} + b\right )} \sqrt {-a b} \log \left (\frac {a x^{2} - 2 \, \sqrt {-a b} x - b}{a x^{2} + b}\right )}{4 \, {\left (a^{3} b x^{2} + a^{2} b^{2}\right )}}, -\frac {a b x - {\left (a x^{2} + b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{b}\right )}{2 \, {\left (a^{3} b x^{2} + a^{2} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 35, normalized size = 0.78 \[ \frac {\arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a} - \frac {x}{2 \, {\left (a x^{2} + b\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 36, normalized size = 0.80 \[ -\frac {x}{2 \left (a \,x^{2}+b \right ) a}+\frac {\arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.99, size = 36, normalized size = 0.80 \[ -\frac {x}{2 \, {\left (a^{2} x^{2} + a b\right )}} + \frac {\arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 33, normalized size = 0.73 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{2\,a^{3/2}\,\sqrt {b}}-\frac {x}{2\,a\,\left (a\,x^2+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b/x^2)^2),x)

[Out]

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

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sympy [B]  time = 0.22, size = 78, normalized size = 1.73 \[ - \frac {x}{2 a^{2} x^{2} + 2 a b} - \frac {\sqrt {- \frac {1}{a^{3} b}} \log {\left (- a b \sqrt {- \frac {1}{a^{3} b}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a^{3} b}} \log {\left (a b \sqrt {- \frac {1}{a^{3} b}} + x \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(2*a**2*x**2 + 2*a*b) - sqrt(-1/(a**3*b))*log(-a*b*sqrt(-1/(a**3*b)) + x)/4 + sqrt(-1/(a**3*b))*log(a*b*sqr
t(-1/(a**3*b)) + x)/4

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