3.19.25 \(\int \frac {(a+\frac {b}{x^2})^2}{x^2} \, dx\) [1825]

Optimal. Leaf size=28 \[ -\frac {b^2}{5 x^5}-\frac {2 a b}{3 x^3}-\frac {a^2}{x} \]

[Out]

-1/5*b^2/x^5-2/3*a*b/x^3-a^2/x

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {269, 276} \begin {gather*} -\frac {a^2}{x}-\frac {2 a b}{3 x^3}-\frac {b^2}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^2/x^2,x]

[Out]

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

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]

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*} \int \frac {\left (a+\frac {b}{x^2}\right )^2}{x^2} \, dx &=\int \frac {\left (b+a x^2\right )^2}{x^6} \, dx\\ &=\int \left (\frac {b^2}{x^6}+\frac {2 a b}{x^4}+\frac {a^2}{x^2}\right ) \, dx\\ &=-\frac {b^2}{5 x^5}-\frac {2 a b}{3 x^3}-\frac {a^2}{x}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 28, normalized size = 1.00 \begin {gather*} -\frac {b^2}{5 x^5}-\frac {2 a b}{3 x^3}-\frac {a^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^2/x^2,x]

[Out]

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

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Maple [A]
time = 0.02, size = 25, normalized size = 0.89

method	result	size
default	\(-\frac {b^{2}}{5 x^{5}}-\frac {2 a b}{3 x^{3}}-\frac {a^{2}}{x}\)	\(25\)
norman	\(\frac {-a^{2} x^{4}-\frac {2}{3} a b \,x^{2}-\frac {1}{5} b^{2}}{x^{5}}\)	\(26\)
risch	\(\frac {-a^{2} x^{4}-\frac {2}{3} a b \,x^{2}-\frac {1}{5} b^{2}}{x^{5}}\)	\(26\)
gosper	\(-\frac {15 a^{2} x^{4}+10 a b \,x^{2}+3 b^{2}}{15 x^{5}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*b^2/x^5-2/3*a*b/x^3-a^2/x

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Maxima [A]
time = 0.29, size = 26, normalized size = 0.93 \begin {gather*} -\frac {15 \, a^{2} x^{4} + 10 \, a b x^{2} + 3 \, b^{2}}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*a^2*x^4 + 10*a*b*x^2 + 3*b^2)/x^5

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Fricas [A]
time = 0.36, size = 26, normalized size = 0.93 \begin {gather*} -\frac {15 \, a^{2} x^{4} + 10 \, a b x^{2} + 3 \, b^{2}}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*a^2*x^4 + 10*a*b*x^2 + 3*b^2)/x^5

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Sympy [A]
time = 0.07, size = 27, normalized size = 0.96 \begin {gather*} \frac {- 15 a^{2} x^{4} - 10 a b x^{2} - 3 b^{2}}{15 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-15*a**2*x**4 - 10*a*b*x**2 - 3*b**2)/(15*x**5)

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Giac [A]
time = 0.63, size = 26, normalized size = 0.93 \begin {gather*} -\frac {15 \, a^{2} x^{4} + 10 \, a b x^{2} + 3 \, b^{2}}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*a^2*x^4 + 10*a*b*x^2 + 3*b^2)/x^5

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Mupad [B]
time = 0.04, size = 25, normalized size = 0.89 \begin {gather*} -\frac {a^2\,x^4+\frac {2\,a\,b\,x^2}{3}+\frac {b^2}{5}}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^2/x^2,x)

[Out]

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

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