3.1565 \(\int \frac{(a+\frac{b}{x})^2}{x} \, dx\)

Optimal. Leaf size=24 \[ a^2 \log (x)-\frac{2 a b}{x}-\frac{b^2}{2 x^2} \]

[Out]

-b^2/(2*x^2) - (2*a*b)/x + a^2*Log[x]

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Rubi [A]  time = 0.0105294, antiderivative size = 24, 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} \[ a^2 \log (x)-\frac{2 a b}{x}-\frac{b^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^2/x,x]

[Out]

-b^2/(2*x^2) - (2*a*b)/x + a^2*Log[x]

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

Mathematica [A]  time = 0.0034823, size = 24, normalized size = 1. \[ a^2 \log (x)-\frac{2 a b}{x}-\frac{b^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^2/x,x]

[Out]

-b^2/(2*x^2) - (2*a*b)/x + a^2*Log[x]

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Maple [A]  time = 0.005, size = 23, normalized size = 1. \begin{align*} -{\frac{{b}^{2}}{2\,{x}^{2}}}-2\,{\frac{ab}{x}}+{a}^{2}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b^2/x^2-2*a*b/x+a^2*ln(x)

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Maxima [A]  time = 0.952182, size = 28, normalized size = 1.17 \begin{align*} a^{2} \log \left (x\right ) - \frac{4 \, a b x + b^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(x) - 1/2*(4*a*b*x + b^2)/x^2

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Fricas [A]  time = 1.44283, size = 59, normalized size = 2.46 \begin{align*} \frac{2 \, a^{2} x^{2} \log \left (x\right ) - 4 \, a b x - b^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^2*x^2*log(x) - 4*a*b*x - b^2)/x^2

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Sympy [A]  time = 0.291155, size = 20, normalized size = 0.83 \begin{align*} a^{2} \log{\left (x \right )} - \frac{4 a b x + b^{2}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*log(x) - (4*a*b*x + b**2)/(2*x**2)

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Giac [A]  time = 1.15572, size = 30, normalized size = 1.25 \begin{align*} a^{2} \log \left ({\left | x \right |}\right ) - \frac{4 \, a b x + b^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(abs(x)) - 1/2*(4*a*b*x + b^2)/x^2