\(\int \frac {3-12 \log (2)}{\log (2)} \, dx\) [7680]

Optimal result

Integrand size = 11, antiderivative size = 20 \[ \int \frac {3-12 \log (2)}{\log (2)} \, dx=5+2 (7 (5-x)+x)+\frac {3 x}{\log (2)} \]

[Out]

75-12*x+3*x/ln(2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {8} \[ \int \frac {3-12 \log (2)}{\log (2)} \, dx=\frac {3 x (1-4 \log (2))}{\log (2)} \]

[In]

Int[(3 - 12*Log[2])/Log[2],x]

[Out]

(3*x*(1 - 4*Log[2]))/Log[2]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps \begin{align*} \text {integral}& = \frac {3 x (1-4 \log (2))}{\log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {3-12 \log (2)}{\log (2)} \, dx=-12 x+\frac {3 x}{\log (2)} \]

[In]

Integrate[(3 - 12*Log[2])/Log[2],x]

[Out]

-12*x + (3*x)/Log[2]

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60

method	result	size
risch	\(-12 x +\frac {3 x}{\ln \left (2\right )}\)	\(12\)
parallelrisch	\(\frac {\left (-12 \ln \left (2\right )+3\right ) x}{\ln \left (2\right )}\)	\(13\)
default	\(\frac {3 \left (-4 \ln \left (2\right )+1\right ) x}{\ln \left (2\right )}\)	\(14\)
norman	\(-\frac {3 \left (4 \ln \left (2\right )-1\right ) x}{\ln \left (2\right )}\)	\(14\)

[In]

int((-12*ln(2)+3)/ln(2),x,method=_RETURNVERBOSE)

[Out]

-12*x+3*x/ln(2)

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {3-12 \log (2)}{\log (2)} \, dx=-\frac {3 \, {\left (4 \, x \log \left (2\right ) - x\right )}}{\log \left (2\right )} \]

[In]

integrate((-12*log(2)+3)/log(2),x, algorithm="fricas")

[Out]

-3*(4*x*log(2) - x)/log(2)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.50 \[ \int \frac {3-12 \log (2)}{\log (2)} \, dx=\frac {x \left (3 - 12 \log {\left (2 \right )}\right )}{\log {\left (2 \right )}} \]

[In]

integrate((-12*ln(2)+3)/ln(2),x)

[Out]

x*(3 - 12*log(2))/log(2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {3-12 \log (2)}{\log (2)} \, dx=-\frac {3 \, x {\left (4 \, \log \left (2\right ) - 1\right )}}{\log \left (2\right )} \]

[In]

integrate((-12*log(2)+3)/log(2),x, algorithm="maxima")

[Out]

-3*x*(4*log(2) - 1)/log(2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {3-12 \log (2)}{\log (2)} \, dx=-\frac {3 \, x {\left (4 \, \log \left (2\right ) - 1\right )}}{\log \left (2\right )} \]

[In]

integrate((-12*log(2)+3)/log(2),x, algorithm="giac")

[Out]

-3*x*(4*log(2) - 1)/log(2)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {3-12 \log (2)}{\log (2)} \, dx=-\frac {x\,\left (12\,\ln \left (2\right )-3\right )}{\ln \left (2\right )} \]

[In]

int(-(12*log(2) - 3)/log(2),x)

[Out]

-(x*(12*log(2) - 3))/log(2)