\(\int (\frac {1}{-1+2 x}-\frac {1}{1+2 x}) \, dx\) [422]

Optimal result

Integrand size = 17, antiderivative size = 21 \[ \int \left (\frac {1}{-1+2 x}-\frac {1}{1+2 x}\right ) \, dx=\frac {1}{2} \log (1-2 x)-\frac {1}{2} \log (1+2 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \left (\frac {1}{-1+2 x}-\frac {1}{1+2 x}\right ) \, dx=\frac {1}{2} \log (1-2 x)-\frac {1}{2} \log (2 x+1) \]

[In]

Int[(-1 + 2*x)^(-1) - (1 + 2*x)^(-1),x]

[Out]

Log[1 - 2*x]/2 - Log[1 + 2*x]/2

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \log (1-2 x)-\frac {1}{2} \log (1+2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \left (\frac {1}{-1+2 x}-\frac {1}{1+2 x}\right ) \, dx=2 \left (\frac {1}{4} \log (1-2 x)-\frac {1}{4} \log (1+2 x)\right ) \]

[In]

Integrate[(-1 + 2*x)^(-1) - (1 + 2*x)^(-1),x]

[Out]

2*(Log[1 - 2*x]/4 - Log[1 + 2*x]/4)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {\ln \left (x -\frac {1}{2}\right )}{2}-\frac {\ln \left (x +\frac {1}{2}\right )}{2}\)	\(14\)
default	\(-\frac {\ln \left (1+2 x \right )}{2}+\frac {\ln \left (2 x -1\right )}{2}\)	\(18\)
norman	\(-\frac {\ln \left (1+2 x \right )}{2}+\frac {\ln \left (2 x -1\right )}{2}\)	\(18\)
meijerg	\(\frac {\ln \left (1-2 x \right )}{2}-\frac {\ln \left (1+2 x \right )}{2}\)	\(18\)
risch	\(-\frac {\ln \left (1+2 x \right )}{2}+\frac {\ln \left (2 x -1\right )}{2}\)	\(18\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (\frac {1}{-1+2 x}-\frac {1}{1+2 x}\right ) \, dx=-\frac {1}{2} \, \log \left (2 \, x + 1\right ) + \frac {1}{2} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/2*log(2*x + 1) + 1/2*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \left (\frac {1}{-1+2 x}-\frac {1}{1+2 x}\right ) \, dx=\frac {\log {\left (x - \frac {1}{2} \right )}}{2} - \frac {\log {\left (x + \frac {1}{2} \right )}}{2} \]

[In]

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

[Out]

log(x - 1/2)/2 - log(x + 1/2)/2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (\frac {1}{-1+2 x}-\frac {1}{1+2 x}\right ) \, dx=-\frac {1}{2} \, \log \left (2 \, x + 1\right ) + \frac {1}{2} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/2*log(2*x + 1) + 1/2*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \left (\frac {1}{-1+2 x}-\frac {1}{1+2 x}\right ) \, dx=-\frac {1}{2} \, \log \left ({\left | 2 \, x + 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-1/2*log(abs(2*x + 1)) + 1/2*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.29 \[ \int \left (\frac {1}{-1+2 x}-\frac {1}{1+2 x}\right ) \, dx=-\mathrm {atanh}\left (2\,x\right ) \]

[In]

int(1/(2*x - 1) - 1/(2*x + 1),x)

[Out]

-atanh(2*x)