\(\int \frac {1}{(a+\frac {b}{x^2})^2 x^4} \, dx\) [1868]

Optimal result

Integrand size = 13, antiderivative size = 45 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^4} \, dx=\frac {x}{2 b \left (b+a x^2\right )}+\frac {\arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 \sqrt {a} b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {269, 205, 211} \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^4} \, dx=\frac {\arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 \sqrt {a} b^{3/2}}+\frac {x}{2 b \left (a x^2+b\right )} \]

[In]

Int[1/((a + b/x^2)^2*x^4),x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

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

Rule 269

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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^4} \, dx=\frac {x}{2 b \left (b+a x^2\right )}+\frac {\arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 \sqrt {a} b^{3/2}} \]

[In]

Integrate[1/((a + b/x^2)^2*x^4),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {x}{2 b \left (a \,x^{2}+b \right )}+\frac {\arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 b \sqrt {a b}}\)	\(36\)
risch	\(\frac {x}{2 b \left (a \,x^{2}+b \right )}-\frac {\ln \left (a x +\sqrt {-a b}\right )}{4 \sqrt {-a b}\, b}+\frac {\ln \left (-a x +\sqrt {-a b}\right )}{4 \sqrt {-a b}\, b}\)	\(62\)

[In]

int(1/(a+b/x^2)^2/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.67 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^4} \, dx=\left [\frac {2 \, a b x - {\left (a x^{2} + b\right )} \sqrt {-a b} \log \left (\frac {a x^{2} - 2 \, \sqrt {-a b} x - b}{a x^{2} + b}\right )}{4 \, {\left (a^{2} b^{2} x^{2} + a b^{3}\right )}}, \frac {a b x + {\left (a x^{2} + b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{b}\right )}{2 \, {\left (a^{2} b^{2} x^{2} + a b^{3}\right )}}\right ] \]

[In]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (36) = 72\).

Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.73 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^4} \, dx=\frac {x}{2 a b x^{2} + 2 b^{2}} - \frac {\sqrt {- \frac {1}{a b^{3}}} \log {\left (- b^{2} \sqrt {- \frac {1}{a b^{3}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a b^{3}}} \log {\left (b^{2} \sqrt {- \frac {1}{a b^{3}}} + x \right )}}{4} \]

[In]

integrate(1/(a+b/x**2)**2/x**4,x)

[Out]

x/(2*a*b*x**2 + 2*b**2) - sqrt(-1/(a*b**3))*log(-b**2*sqrt(-1/(a*b**3)) + x)/4 + sqrt(-1/(a*b**3))*log(b**2*sq
rt(-1/(a*b**3)) + x)/4

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^4} \, dx=\frac {x}{2 \, {\left (a b x^{2} + b^{2}\right )}} + \frac {\arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b} \]

[In]

integrate(1/(a+b/x^2)^2/x^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^4} \, dx=\frac {\arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b} + \frac {x}{2 \, {\left (a x^{2} + b\right )} b} \]

[In]

integrate(1/(a+b/x^2)^2/x^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^4} \, dx=\frac {x}{2\,b\,\left (a\,x^2+b\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{2\,\sqrt {a}\,b^{3/2}} \]

[In]

int(1/(x^4*(a + b/x^2)^2),x)

[Out]

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