3.1827 \(\int \frac{(a+\frac{b}{x^2})^2}{x^4} \, dx\)

Optimal. Leaf size=30 \[ -\frac{a^2}{3 x^3}-\frac{2 a b}{5 x^5}-\frac{b^2}{7 x^7} \]

[Out]

-b^2/(7*x^7) - (2*a*b)/(5*x^5) - a^2/(3*x^3)

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Rubi [A]  time = 0.011996, antiderivative size = 30, normalized size of antiderivative = 1., 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 = {263, 270} \[ -\frac{a^2}{3 x^3}-\frac{2 a b}{5 x^5}-\frac{b^2}{7 x^7} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^2/x^4,x]

[Out]

-b^2/(7*x^7) - (2*a*b)/(5*x^5) - a^2/(3*x^3)

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 270

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^4} \, dx &=\int \frac{\left (b+a x^2\right )^2}{x^8} \, dx\\ &=\int \left (\frac{b^2}{x^8}+\frac{2 a b}{x^6}+\frac{a^2}{x^4}\right ) \, dx\\ &=-\frac{b^2}{7 x^7}-\frac{2 a b}{5 x^5}-\frac{a^2}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0007825, size = 30, normalized size = 1. \[ -\frac{a^2}{3 x^3}-\frac{2 a b}{5 x^5}-\frac{b^2}{7 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^2/x^4,x]

[Out]

-b^2/(7*x^7) - (2*a*b)/(5*x^5) - a^2/(3*x^3)

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Maple [A]  time = 0.003, size = 25, normalized size = 0.8 \begin{align*} -{\frac{{b}^{2}}{7\,{x}^{7}}}-{\frac{2\,ab}{5\,{x}^{5}}}-{\frac{{a}^{2}}{3\,{x}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*b^2/x^7-2/5*a*b/x^5-1/3*a^2/x^3

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Maxima [A]  time = 1.00278, size = 35, normalized size = 1.17 \begin{align*} -\frac{35 \, a^{2} x^{4} + 42 \, a b x^{2} + 15 \, b^{2}}{105 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(35*a^2*x^4 + 42*a*b*x^2 + 15*b^2)/x^7

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Fricas [A]  time = 1.41135, size = 63, normalized size = 2.1 \begin{align*} -\frac{35 \, a^{2} x^{4} + 42 \, a b x^{2} + 15 \, b^{2}}{105 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(35*a^2*x^4 + 42*a*b*x^2 + 15*b^2)/x^7

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Sympy [A]  time = 0.336951, size = 27, normalized size = 0.9 \begin{align*} - \frac{35 a^{2} x^{4} + 42 a b x^{2} + 15 b^{2}}{105 x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(35*a**2*x**4 + 42*a*b*x**2 + 15*b**2)/(105*x**7)

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Giac [A]  time = 1.15992, size = 35, normalized size = 1.17 \begin{align*} -\frac{35 \, a^{2} x^{4} + 42 \, a b x^{2} + 15 \, b^{2}}{105 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(35*a^2*x^4 + 42*a*b*x^2 + 15*b^2)/x^7