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]

3/2*x/a^2-1/2*x^3/a/(a*x^2+b)-3/2*arctan(x*a^(1/2)/b^(1/2))*b^(1/2)/a^(5/2)

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.93 \[ -\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{5/2}}+\frac {b x}{2 a^2 \left (a x^2+b\right )}+\frac {x}{a^2} \]

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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fricas [A]  time = 0.90, size = 136, normalized size = 2.47 \[ \left [\frac {4 \, a x^{3} + 3 \, {\left (a x^{2} + b\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - b}{a x^{2} + b}\right ) + 6 \, b x}{4 \, {\left (a^{3} x^{2} + a^{2} b\right )}}, \frac {2 \, a x^{3} - 3 \, {\left (a x^{2} + b\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a x \sqrt {\frac {b}{a}}}{b}\right ) + 3 \, b x}{2 \, {\left (a^{3} x^{2} + a^{2} b\right )}}\right ] \]

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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giac [A]  time = 0.18, size = 42, normalized size = 0.76 \[ -\frac {3 \, b \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} + \frac {x}{a^{2}} + \frac {b x}{2 \, {\left (a x^{2} + b\right )} a^{2}} \]

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)

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maple [A]  time = 0.01, size = 43, normalized size = 0.78 \[ \frac {b x}{2 \left (a \,x^{2}+b \right ) a^{2}}-\frac {3 b \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a^{2}}+\frac {x}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.98, size = 45, normalized size = 0.82 \[ \frac {b x}{2 \, {\left (a^{3} x^{2} + a^{2} b\right )}} - \frac {3 \, b \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} + \frac {x}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.10, size = 43, normalized size = 0.78 \[ \frac {x}{a^2}+\frac {b\,x}{2\,\left (a^3\,x^2+b\,a^2\right )}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{2\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/x^2)^2,x)

[Out]

x/a^2 + (b*x)/(2*(a^2*b + a^3*x^2)) - (3*b^(1/2)*atan((a^(1/2)*x)/b^(1/2)))/(2*a^(5/2))

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sympy [A]  time = 0.28, size = 83, normalized size = 1.51 \[ \frac {b x}{2 a^{3} x^{2} + 2 a^{2} b} + \frac {3 \sqrt {- \frac {b}{a^{5}}} \log {\left (- a^{2} \sqrt {- \frac {b}{a^{5}}} + x \right )}}{4} - \frac {3 \sqrt {- \frac {b}{a^{5}}} \log {\left (a^{2} \sqrt {- \frac {b}{a^{5}}} + x \right )}}{4} + \frac {x}{a^{2}} \]

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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