3.1567 \(\int \frac{(a+\frac{b}{x})^2}{x^3} \, dx\)

Optimal. Leaf size=30 \[ -\frac{a^2}{2 x^2}-\frac{2 a b}{3 x^3}-\frac{b^2}{4 x^4} \]

[Out]

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

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Rubi [A]  time = 0.01058, 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, 43} \[ -\frac{a^2}{2 x^2}-\frac{2 a b}{3 x^3}-\frac{b^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^2/x^3,x]

[Out]

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0030134, size = 30, normalized size = 1. \[ -\frac{a^2}{2 x^2}-\frac{2 a b}{3 x^3}-\frac{b^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^2/x^3,x]

[Out]

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

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Maple [A]  time = 0.004, size = 25, normalized size = 0.8 \begin{align*} -{\frac{{b}^{2}}{4\,{x}^{4}}}-{\frac{2\,ab}{3\,{x}^{3}}}-{\frac{{a}^{2}}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^2/x^3,x)

[Out]

-1/4*b^2/x^4-2/3*a*b/x^3-1/2/x^2*a^2

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Maxima [A]  time = 0.96483, size = 32, normalized size = 1.07 \begin{align*} -\frac{6 \, a^{2} x^{2} + 8 \, a b x + 3 \, b^{2}}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^2/x^3,x, algorithm="maxima")

[Out]

-1/12*(6*a^2*x^2 + 8*a*b*x + 3*b^2)/x^4

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Fricas [A]  time = 1.41532, size = 55, normalized size = 1.83 \begin{align*} -\frac{6 \, a^{2} x^{2} + 8 \, a b x + 3 \, b^{2}}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^2/x^3,x, algorithm="fricas")

[Out]

-1/12*(6*a^2*x^2 + 8*a*b*x + 3*b^2)/x^4

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Sympy [A]  time = 0.302161, size = 26, normalized size = 0.87 \begin{align*} - \frac{6 a^{2} x^{2} + 8 a b x + 3 b^{2}}{12 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**2/x**3,x)

[Out]

-(6*a**2*x**2 + 8*a*b*x + 3*b**2)/(12*x**4)

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Giac [A]  time = 1.18694, size = 32, normalized size = 1.07 \begin{align*} -\frac{6 \, a^{2} x^{2} + 8 \, a b x + 3 \, b^{2}}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^2/x^3,x, algorithm="giac")

[Out]

-1/12*(6*a^2*x^2 + 8*a*b*x + 3*b^2)/x^4