\(\int \frac {-2 x+(-11+2 x) \log (3)}{\log (3)} \, dx\) [2675]

Optimal result

Integrand size = 17, antiderivative size = 18 \[ \int \frac {-2 x+(-11+2 x) \log (3)}{\log (3)} \, dx=(-5+x)^2-x-\frac {x^2}{\log (3)} \]

[Out]

(-5+x)^2-x^2/ln(3)-x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12} \[ \int \frac {-2 x+(-11+2 x) \log (3)}{\log (3)} \, dx=\frac {1}{4} (11-2 x)^2-\frac {x^2}{\log (3)} \]

[In]

Int[(-2*x + (-11 + 2*x)*Log[3])/Log[3],x]

[Out]

(11 - 2*x)^2/4 - x^2/Log[3]

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 (-2 x+(-11+2 x) \log (3)) \, dx}{\log (3)} \\ & = \frac {1}{4} (11-2 x)^2-\frac {x^2}{\log (3)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {-2 x+(-11+2 x) \log (3)}{\log (3)} \, dx=-11 x+\frac {x^2 (-2+\log (9))}{2 \log (3)} \]

[In]

Integrate[(-2*x + (-11 + 2*x)*Log[3])/Log[3],x]

[Out]

-11*x + (x^2*(-2 + Log[9]))/(2*Log[3])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
norman	\(\frac {\left (\ln \left (3\right )-1\right ) x^{2}}{\ln \left (3\right )}-11 x\)	\(17\)
risch	\(x^{2}-11 x -\frac {x^{2}}{\ln \left (3\right )}\)	\(17\)
gosper	\(\frac {x \left (x \ln \left (3\right )-11 \ln \left (3\right )-x \right )}{\ln \left (3\right )}\)	\(19\)
default	\(\frac {x^{2} \ln \left (3\right )-11 x \ln \left (3\right )-x^{2}}{\ln \left (3\right )}\)	\(23\)
parallelrisch	\(\frac {x^{2} \ln \left (3\right )-11 x \ln \left (3\right )-x^{2}}{\ln \left (3\right )}\)	\(23\)

[In]

int(((2*x-11)*ln(3)-2*x)/ln(3),x,method=_RETURNVERBOSE)

[Out]

(ln(3)-1)/ln(3)*x^2-11*x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {-2 x+(-11+2 x) \log (3)}{\log (3)} \, dx=-\frac {x^{2} - {\left (x^{2} - 11 \, x\right )} \log \left (3\right )}{\log \left (3\right )} \]

[In]

integrate(((2*x-11)*log(3)-2*x)/log(3),x, algorithm="fricas")

[Out]

-(x^2 - (x^2 - 11*x)*log(3))/log(3)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-2 x+(-11+2 x) \log (3)}{\log (3)} \, dx=\frac {x^{2} \left (-1 + \log {\left (3 \right )}\right )}{\log {\left (3 \right )}} - 11 x \]

[In]

integrate(((2*x-11)*ln(3)-2*x)/ln(3),x)

[Out]

x**2*(-1 + log(3))/log(3) - 11*x

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {-2 x+(-11+2 x) \log (3)}{\log (3)} \, dx=-\frac {x^{2} - {\left (x^{2} - 11 \, x\right )} \log \left (3\right )}{\log \left (3\right )} \]

[In]

integrate(((2*x-11)*log(3)-2*x)/log(3),x, algorithm="maxima")

[Out]

-(x^2 - (x^2 - 11*x)*log(3))/log(3)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {-2 x+(-11+2 x) \log (3)}{\log (3)} \, dx=-\frac {x^{2} - {\left (x^{2} - 11 \, x\right )} \log \left (3\right )}{\log \left (3\right )} \]

[In]

integrate(((2*x-11)*log(3)-2*x)/log(3),x, algorithm="giac")

[Out]

-(x^2 - (x^2 - 11*x)*log(3))/log(3)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-2 x+(-11+2 x) \log (3)}{\log (3)} \, dx=\frac {x^2\,\left (\ln \left (3\right )-1\right )}{\ln \left (3\right )}-11\,x \]

[In]

int(-(2*x - log(3)*(2*x - 11))/log(3),x)

[Out]

(x^2*(log(3) - 1))/log(3) - 11*x