\(\int (a+\frac {b}{x^2})^2 \, dx\) [1823]

Optimal result

Integrand size = 9, antiderivative size = 23 \[ \int \left (a+\frac {b}{x^2}\right )^2 \, dx=-\frac {b^2}{3 x^3}-\frac {2 a b}{x}+a^2 x \]

[Out]

-1/3*b^2/x^3-2*a*b/x+a^2*x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {199, 276} \[ \int \left (a+\frac {b}{x^2}\right )^2 \, dx=a^2 x-\frac {2 a b}{x}-\frac {b^2}{3 x^3} \]

[In]

Int[(a + b/x^2)^2,x]

[Out]

-1/3*b^2/x^3 - (2*a*b)/x + a^2*x

Rule 199

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 276

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (b+a x^2\right )^2}{x^4} \, dx \\ & = \int \left (a^2+\frac {b^2}{x^4}+\frac {2 a b}{x^2}\right ) \, dx \\ & = -\frac {b^2}{3 x^3}-\frac {2 a b}{x}+a^2 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{x^2}\right )^2 \, dx=-\frac {b^2}{3 x^3}-\frac {2 a b}{x}+a^2 x \]

[In]

Integrate[(a + b/x^2)^2,x]

[Out]

-1/3*b^2/x^3 - (2*a*b)/x + a^2*x

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {b^{2}}{3 x^{3}}-\frac {2 a b}{x}+a^{2} x\)	\(22\)
risch	\(a^{2} x +\frac {-2 a b \,x^{2}-\frac {1}{3} b^{2}}{x^{3}}\)	\(24\)
norman	\(\frac {a^{2} x^{4}-2 a b \,x^{2}-\frac {1}{3} b^{2}}{x^{3}}\)	\(25\)
gosper	\(\frac {3 a^{2} x^{4}-6 a b \,x^{2}-b^{2}}{3 x^{3}}\)	\(27\)
parallelrisch	\(\frac {3 a^{2} x^{4}-6 a b \,x^{2}-b^{2}}{3 x^{3}}\)	\(27\)

[In]

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

[Out]

-1/3*b^2/x^3-2*a*b/x+a^2*x

Fricas [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \left (a+\frac {b}{x^2}\right )^2 \, dx=\frac {3 \, a^{2} x^{4} - 6 \, a b x^{2} - b^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

1/3*(3*a^2*x^4 - 6*a*b*x^2 - b^2)/x^3

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \left (a+\frac {b}{x^2}\right )^2 \, dx=a^{2} x + \frac {- 6 a b x^{2} - b^{2}}{3 x^{3}} \]

[In]

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

[Out]

a**2*x + (-6*a*b*x**2 - b**2)/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \left (a+\frac {b}{x^2}\right )^2 \, dx=a^{2} x - \frac {2 \, a b}{x} - \frac {b^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

a^2*x - 2*a*b/x - 1/3*b^2/x^3

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \left (a+\frac {b}{x^2}\right )^2 \, dx=a^{2} x - \frac {6 \, a b x^{2} + b^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

a^2*x - 1/3*(6*a*b*x^2 + b^2)/x^3

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \left (a+\frac {b}{x^2}\right )^2 \, dx=a^2\,x-\frac {\frac {b^2}{3}+2\,a\,b\,x^2}{x^3} \]

[In]

int((a + b/x^2)^2,x)

[Out]

a^2*x - (b^2/3 + 2*a*b*x^2)/x^3