\(\int \frac {-1-2 x+2 \log (\log (2))}{-x+\log (\log (2))} \, dx\) [7617]

Optimal result

Integrand size = 20, antiderivative size = 18 \[ \int \frac {-1-2 x+2 \log (\log (2))}{-x+\log (\log (2))} \, dx=-\frac {52}{25}+e^5+2 x+\log (x-\log (\log (2))) \]

[Out]

2*x+ln(x-ln(ln(2)))-52/25+exp(5)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {45} \[ \int \frac {-1-2 x+2 \log (\log (2))}{-x+\log (\log (2))} \, dx=2 x+\log (x-\log (\log (2))) \]

[In]

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

[Out]

2*x + Log[x - Log[Log[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])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-1-2 x+2 \log (\log (2))}{-x+\log (\log (2))} \, dx=2 (x-\log (\log (2)))+\log (x-\log (\log (2))) \]

[In]

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

[Out]

2*(x - Log[Log[2]]) + Log[x - Log[Log[2]]]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72

method	result	size
default	\(2 x +\ln \left (x -\ln \left (\ln \left (2\right )\right )\right )\)	\(13\)
norman	\(2 x +\ln \left (\ln \left (\ln \left (2\right )\right )-x \right )\)	\(13\)
risch	\(2 x +\ln \left (x -\ln \left (\ln \left (2\right )\right )\right )\)	\(13\)
parallelrisch	\(2 x +\ln \left (x -\ln \left (\ln \left (2\right )\right )\right )\)	\(13\)
meijerg	\(-2 \ln \left (\ln \left (2\right )\right ) \ln \left (1-\frac {x}{\ln \left (\ln \left (2\right )\right )}\right )-2 \ln \left (\ln \left (2\right )\right ) \left (-\frac {x}{\ln \left (\ln \left (2\right )\right )}-\ln \left (1-\frac {x}{\ln \left (\ln \left (2\right )\right )}\right )\right )+\ln \left (1-\frac {x}{\ln \left (\ln \left (2\right )\right )}\right )\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {-1-2 x+2 \log (\log (2))}{-x+\log (\log (2))} \, dx=2 \, x + \log \left (x - \log \left (\log \left (2\right )\right )\right ) \]

[In]

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

[Out]

2*x + log(x - log(log(2)))

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {-1-2 x+2 \log (\log (2))}{-x+\log (\log (2))} \, dx=2 x + \log {\left (x - \log {\left (\log {\left (2 \right )} \right )} \right )} \]

[In]

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

[Out]

2*x + log(x - log(log(2)))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {-1-2 x+2 \log (\log (2))}{-x+\log (\log (2))} \, dx=2 \, x + \log \left (x - \log \left (\log \left (2\right )\right )\right ) \]

[In]

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

[Out]

2*x + log(x - log(log(2)))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {-1-2 x+2 \log (\log (2))}{-x+\log (\log (2))} \, dx=2 \, x + \log \left ({\left | x - \log \left (\log \left (2\right )\right ) \right |}\right ) \]

[In]

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

[Out]

2*x + log(abs(x - log(log(2))))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {-1-2 x+2 \log (\log (2))}{-x+\log (\log (2))} \, dx=2\,x+\ln \left (x-\ln \left (\ln \left (2\right )\right )\right ) \]

[In]

int((2*x - 2*log(log(2)) + 1)/(x - log(log(2))),x)

[Out]

2*x + log(x - log(log(2)))