3.1625 \(\int \frac{1}{(a+\frac{b}{x})^2 x} \, dx\)

Optimal. Leaf size=23 \[ \frac{b}{a^2 (a x+b)}+\frac{\log (a x+b)}{a^2} \]

[Out]

b/(a^2*(b + a*x)) + Log[b + a*x]/a^2

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Rubi [A]  time = 0.0147991, antiderivative size = 23, 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} \[ \frac{b}{a^2 (a x+b)}+\frac{\log (a x+b)}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^2*x),x]

[Out]

b/(a^2*(b + a*x)) + Log[b + a*x]/a^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{1}{\left (a+\frac{b}{x}\right )^2 x} \, dx &=\int \frac{x}{(b+a x)^2} \, dx\\ &=\int \left (-\frac{b}{a (b+a x)^2}+\frac{1}{a (b+a x)}\right ) \, dx\\ &=\frac{b}{a^2 (b+a x)}+\frac{\log (b+a x)}{a^2}\\ \end{align*}

Mathematica [A]  time = 0.0060767, size = 20, normalized size = 0.87 \[ \frac{\frac{b}{a x+b}+\log (a x+b)}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^2*x),x]

[Out]

(b/(b + a*x) + Log[b + a*x])/a^2

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Maple [A]  time = 0.005, size = 24, normalized size = 1. \begin{align*}{\frac{b}{{a}^{2} \left ( ax+b \right ) }}+{\frac{\ln \left ( ax+b \right ) }{{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01971, size = 35, normalized size = 1.52 \begin{align*} \frac{b}{a^{3} x + a^{2} b} + \frac{\log \left (a x + b\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^2/x,x, algorithm="maxima")

[Out]

b/(a^3*x + a^2*b) + log(a*x + b)/a^2

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Fricas [A]  time = 1.41431, size = 62, normalized size = 2.7 \begin{align*} \frac{{\left (a x + b\right )} \log \left (a x + b\right ) + b}{a^{3} x + a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^2/x,x, algorithm="fricas")

[Out]

((a*x + b)*log(a*x + b) + b)/(a^3*x + a^2*b)

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Sympy [A]  time = 0.277199, size = 20, normalized size = 0.87 \begin{align*} \frac{b}{a^{3} x + a^{2} b} + \frac{\log{\left (a x + b \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**2/x,x)

[Out]

b/(a**3*x + a**2*b) + log(a*x + b)/a**2

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Giac [A]  time = 1.09878, size = 32, normalized size = 1.39 \begin{align*} \frac{\log \left ({\left | a x + b \right |}\right )}{a^{2}} + \frac{b}{{\left (a x + b\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^2/x,x, algorithm="giac")

[Out]

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