3.17.25 \(\int \frac {1}{(a+\frac {b}{x})^2 x} \, dx\) [1625]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 23, 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, 45} \begin {gather*} \frac {b}{a^2 (a x+b)}+\frac {\log (a x+b)}{a^2} \end {gather*}

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 45

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])

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]

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*}

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Mathematica [A]
time = 0.01, size = 20, normalized size = 0.87 \begin {gather*} \frac {\frac {b}{b+a x}+\log (b+a x)}{a^2} \end {gather*}

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 = 0.02, size = 24, normalized size = 1.04

method	result	size
default	\(\frac {b}{a^{2} \left (a x +b \right )}+\frac {\ln \left (a x +b \right )}{a^{2}}\)	\(24\)
norman	\(\frac {b}{a^{2} \left (a x +b \right )}+\frac {\ln \left (a x +b \right )}{a^{2}}\)	\(24\)
risch	\(\frac {b}{a^{2} \left (a x +b \right )}+\frac {\ln \left (a x +b \right )}{a^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+1/x*b)^2/x,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 26, normalized size = 1.13 \begin {gather*} \frac {b}{a^{3} x + a^{2} b} + \frac {\log \left (a x + b\right )}{a^{2}} \end {gather*}

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 = 0.38, size = 28, normalized size = 1.22 \begin {gather*} \frac {{\left (a x + b\right )} \log \left (a x + b\right ) + b}{a^{3} x + a^{2} b} \end {gather*}

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 = 0.05, size = 20, normalized size = 0.87 \begin {gather*} \frac {b}{a^{3} x + a^{2} b} + \frac {\log {\left (a x + b \right )}}{a^{2}} \end {gather*}

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 = 0.48, size = 24, normalized size = 1.04 \begin {gather*} \frac {\log \left ({\left | a x + b \right |}\right )}{a^{2}} + \frac {b}{{\left (a x + b\right )} a^{2}} \end {gather*}

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)

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Mupad [B]
time = 1.06, size = 23, normalized size = 1.00 \begin {gather*} \frac {\ln \left (b+a\,x\right )}{a^2}+\frac {b}{a^2\,\left (b+a\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b/x)^2),x)

[Out]

log(b + a*x)/a^2 + b/(a^2*(b + a*x))

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