\(\int \frac {1}{x (-1+b x)} \, dx\) [280]

Optimal result

Integrand size = 11, antiderivative size = 12 \[ \int \frac {1}{x (-1+b x)} \, dx=-\log (x)+\log (1-b x) \]

[Out]

-ln(x)+ln(-b*x+1)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {36, 29, 31} \[ \int \frac {1}{x (-1+b x)} \, dx=\log (1-b x)-\log (x) \]

[In]

Int[1/(x*(-1 + b*x)),x]

[Out]

-Log[x] + Log[1 - b*x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rubi steps \begin{align*} \text {integral}& = b \int \frac {1}{-1+b x} \, dx-\int \frac {1}{x} \, dx \\ & = -\log (x)+\log (1-b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (-1+b x)} \, dx=-\log (x)+\log (1-b x) \]

[In]

Integrate[1/(x*(-1 + b*x)),x]

[Out]

-Log[x] + Log[1 - b*x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00

method	result	size
default	\(-\ln \left (x \right )+\ln \left (b x -1\right )\)	\(12\)
norman	\(-\ln \left (x \right )+\ln \left (b x -1\right )\)	\(12\)
parallelrisch	\(-\ln \left (x \right )+\ln \left (b x -1\right )\)	\(12\)
risch	\(-\ln \left (x \right )+\ln \left (-b x +1\right )\)	\(13\)
meijerg	\(-\ln \left (x \right )-\ln \left (-b \right )+\ln \left (-b x +1\right )\)	\(19\)

[In]

int(1/x/(b*x-1),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(b*x-1)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x (-1+b x)} \, dx=\log \left (b x - 1\right ) - \log \left (x\right ) \]

[In]

integrate(1/x/(b*x-1),x, algorithm="fricas")

[Out]

log(b*x - 1) - log(x)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x (-1+b x)} \, dx=- \log {\left (x \right )} + \log {\left (x - \frac {1}{b} \right )} \]

[In]

integrate(1/x/(b*x-1),x)

[Out]

-log(x) + log(x - 1/b)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x (-1+b x)} \, dx=\log \left (b x - 1\right ) - \log \left (x\right ) \]

[In]

integrate(1/x/(b*x-1),x, algorithm="maxima")

[Out]

log(b*x - 1) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x (-1+b x)} \, dx=\log \left ({\left | b x - 1 \right |}\right ) - \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/x/(b*x-1),x, algorithm="giac")

[Out]

log(abs(b*x - 1)) - log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x (-1+b x)} \, dx=-2\,\mathrm {atanh}\left (2\,b\,x-1\right ) \]

[In]

int(1/(x*(b*x - 1)),x)

[Out]

-2*atanh(2*b*x - 1)