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

Optimal. Leaf size=21 \[ \frac {1}{2} \log (1-2 x)-\frac {1}{2} \log (1+2 x) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 0, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \frac {1}{2} \log (1-2 x)-\frac {1}{2} \log (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[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*} \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)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 23, normalized size = 1.10 \begin {gather*} 2 \left (\frac {1}{4} \log (1-2 x)-\frac {1}{4} \log (1+2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*(Log[1 - 2*x]/4 - Log[1 + 2*x]/4)

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Maple [A]
time = 0.20, size = 18, normalized size = 0.86

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 17, normalized size = 0.81 \begin {gather*} -\frac {1}{2} \, \log \left (2 \, x + 1\right ) + \frac {1}{2} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 0.39, size = 17, normalized size = 0.81 \begin {gather*} -\frac {1}{2} \, \log \left (2 \, x + 1\right ) + \frac {1}{2} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]
time = 0.03, size = 15, normalized size = 0.71 \begin {gather*} \frac {\log {\left (x - \frac {1}{2} \right )}}{2} - \frac {\log {\left (x + \frac {1}{2} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 1/2)/2 - log(x + 1/2)/2

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Giac [A]
time = 3.39, size = 19, normalized size = 0.90 \begin {gather*} -\frac {1}{2} \, \log \left ({\left | 2 \, x + 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Mupad [B]
time = 0.15, size = 6, normalized size = 0.29 \begin {gather*} -\mathrm {atanh}\left (2\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x - 1) - 1/(2*x + 1),x)

[Out]

-atanh(2*x)

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