\(\int \frac {(-12+2 x) \log (3)+\log (4)}{\log (4)} \, dx\) [3270]

Optimal result

Integrand size = 16, antiderivative size = 19 \[ \int \frac {(-12+2 x) \log (3)+\log (4)}{\log (4)} \, dx=2+x+\frac {\left (-1+(6-x)^2\right ) \log (3)}{\log (4)} \]

[Out]

x+2+1/2*((6-x)^2-1)/ln(2)*ln(3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12} \[ \int \frac {(-12+2 x) \log (3)+\log (4)}{\log (4)} \, dx=x+\frac {(6-x)^2 \log (3)}{\log (4)} \]

[In]

Int[((-12 + 2*x)*Log[3] + Log[4])/Log[4],x]

[Out]

x + ((6 - x)^2*Log[3])/Log[4]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps \begin{align*} \text {integral}& = \frac {\int ((-12+2 x) \log (3)+\log (4)) \, dx}{\log (4)} \\ & = x+\frac {(6-x)^2 \log (3)}{\log (4)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {(-12+2 x) \log (3)+\log (4)}{\log (4)} \, dx=x-\frac {12 x \log (3)}{\log (4)}+\frac {x^2 \log (3)}{\log (4)} \]

[In]

Integrate[((-12 + 2*x)*Log[3] + Log[4])/Log[4],x]

[Out]

x - (12*x*Log[3])/Log[4] + (x^2*Log[3])/Log[4]

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11

method	result	size
gosper	\(\frac {x \left (x \ln \left (3\right )-12 \ln \left (3\right )+2 \ln \left (2\right )\right )}{2 \ln \left (2\right )}\)	\(21\)
risch	\(\frac {\ln \left (3\right ) x^{2}}{2 \ln \left (2\right )}-\frac {6 x \ln \left (3\right )}{\ln \left (2\right )}+x\)	\(23\)
parallelrisch	\(\frac {\ln \left (3\right ) \left (x^{2}-12 x \right )+2 x \ln \left (2\right )}{2 \ln \left (2\right )}\)	\(23\)
default	\(\frac {x^{2} \ln \left (3\right )-12 x \ln \left (3\right )+2 x \ln \left (2\right )}{2 \ln \left (2\right )}\)	\(24\)
norman	\(-\frac {\left (6 \ln \left (3\right )-\ln \left (2\right )\right ) x}{\ln \left (2\right )}+\frac {\ln \left (3\right ) x^{2}}{2 \ln \left (2\right )}\)	\(29\)

[In]

int(1/2*(2*ln(2)+(2*x-12)*ln(3))/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/2*x*(x*ln(3)-12*ln(3)+2*ln(2))/ln(2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {(-12+2 x) \log (3)+\log (4)}{\log (4)} \, dx=\frac {{\left (x^{2} - 12 \, x\right )} \log \left (3\right ) + 2 \, x \log \left (2\right )}{2 \, \log \left (2\right )} \]

[In]

integrate(1/2*(2*log(2)+(2*x-12)*log(3))/log(2),x, algorithm="fricas")

[Out]

1/2*((x^2 - 12*x)*log(3) + 2*x*log(2))/log(2)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {(-12+2 x) \log (3)+\log (4)}{\log (4)} \, dx=\frac {x^{2} \log {\left (3 \right )}}{2 \log {\left (2 \right )}} + \frac {x \left (- 6 \log {\left (3 \right )} + \log {\left (2 \right )}\right )}{\log {\left (2 \right )}} \]

[In]

integrate(1/2*(2*ln(2)+(2*x-12)*ln(3))/ln(2),x)

[Out]

x**2*log(3)/(2*log(2)) + x*(-6*log(3) + log(2))/log(2)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {(-12+2 x) \log (3)+\log (4)}{\log (4)} \, dx=\frac {{\left (x^{2} - 12 \, x\right )} \log \left (3\right ) + 2 \, x \log \left (2\right )}{2 \, \log \left (2\right )} \]

[In]

integrate(1/2*(2*log(2)+(2*x-12)*log(3))/log(2),x, algorithm="maxima")

[Out]

1/2*((x^2 - 12*x)*log(3) + 2*x*log(2))/log(2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {(-12+2 x) \log (3)+\log (4)}{\log (4)} \, dx=\frac {{\left (x^{2} - 12 \, x\right )} \log \left (3\right ) + 2 \, x \log \left (2\right )}{2 \, \log \left (2\right )} \]

[In]

integrate(1/2*(2*log(2)+(2*x-12)*log(3))/log(2),x, algorithm="giac")

[Out]

1/2*((x^2 - 12*x)*log(3) + 2*x*log(2))/log(2)

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {(-12+2 x) \log (3)+\log (4)}{\log (4)} \, dx=\frac {{\left (\ln \left (2\right )+\frac {\ln \left (3\right )\,\left (2\,x-12\right )}{2}\right )}^2}{2\,\ln \left (2\right )\,\ln \left (3\right )} \]

[In]

int((log(2) + (log(3)*(2*x - 12))/2)/log(2),x)

[Out]

(log(2) + (log(3)*(2*x - 12))/2)^2/(2*log(2)*log(3))