\(\int \frac {1}{2-b x} \, dx\) [1544]

Optimal result

Integrand size = 8, antiderivative size = 12 \[ \int \frac {1}{2-b x} \, dx=-\frac {\log (2-b x)}{b} \]

[Out]

-ln(-b*x+2)/b

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {31} \[ \int \frac {1}{2-b x} \, dx=-\frac {\log (2-b x)}{b} \]

[In]

Int[(2 - b*x)^(-1),x]

[Out]

-(Log[2 - b*x]/b)

Rule 31

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\log (2-b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2-b x} \, dx=-\frac {\log (2-b x)}{b} \]

[In]

Integrate[(2 - b*x)^(-1),x]

[Out]

-(Log[2 - b*x]/b)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00

method	result	size
norman	\(-\frac {\ln \left (b x -2\right )}{b}\)	\(12\)
risch	\(-\frac {\ln \left (b x -2\right )}{b}\)	\(12\)
parallelrisch	\(-\frac {\ln \left (b x -2\right )}{b}\)	\(12\)
default	\(-\frac {\ln \left (-b x +2\right )}{b}\)	\(13\)
meijerg	\(-\frac {\ln \left (-\frac {b x}{2}+1\right )}{b}\)	\(13\)

[In]

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

[Out]

-1/b*ln(b*x-2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {1}{2-b x} \, dx=-\frac {\log \left (b x - 2\right )}{b} \]

[In]

integrate(1/(-b*x+2),x, algorithm="fricas")

[Out]

-log(b*x - 2)/b

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{2-b x} \, dx=- \frac {\log {\left (b x - 2 \right )}}{b} \]

[In]

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

[Out]

-log(b*x - 2)/b

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {1}{2-b x} \, dx=-\frac {\log \left (b x - 2\right )}{b} \]

[In]

integrate(1/(-b*x+2),x, algorithm="maxima")

[Out]

-log(b*x - 2)/b

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2-b x} \, dx=-\frac {\log \left ({\left | b x - 2 \right |}\right )}{b} \]

[In]

integrate(1/(-b*x+2),x, algorithm="giac")

[Out]

-log(abs(b*x - 2))/b

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {1}{2-b x} \, dx=-\frac {\ln \left (b\,x-2\right )}{b} \]

[In]

int(-1/(b*x - 2),x)

[Out]

-log(b*x - 2)/b