\(\int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx\) [6757]

Optimal result

Integrand size = 16, antiderivative size = 13 \[ \int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx=-12 x \left (x+\frac {e^2}{\log (3)}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {9} \[ \int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx=-\frac {3 \left (2 x \log (3)+e^2\right )^2}{\log ^2(3)} \]

[In]

Int[(-12*E^2 - 24*x*Log[3])/Log[3],x]

[Out]

(-3*(E^2 + 2*x*Log[3])^2)/Log[3]^2

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[a*((b + c*x)^2/(2*c)), x] /; FreeQ[{a, b, c}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \left (e^2+2 x \log (3)\right )^2}{\log ^2(3)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.62 \[ \int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx=-\frac {12 \left (e^2 x+\frac {1}{2} x^2 \log (9)\right )}{\log (3)} \]

[In]

Integrate[(-12*E^2 - 24*x*Log[3])/Log[3],x]

[Out]

(-12*(E^2*x + (x^2*Log[9])/2))/Log[3]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15

method	result	size
gosper	\(-\frac {12 x \left (x \ln \left (3\right )+{\mathrm e}^{2}\right )}{\ln \left (3\right )}\)	\(15\)
norman	\(-12 x^{2}-\frac {12 \,{\mathrm e}^{2} x}{\ln \left (3\right )}\)	\(16\)
risch	\(-12 x^{2}-\frac {12 \,{\mathrm e}^{2} x}{\ln \left (3\right )}\)	\(16\)
parallelrisch	\(\frac {-12 x^{2} \ln \left (3\right )-12 \,{\mathrm e}^{2} x}{\ln \left (3\right )}\)	\(19\)
default	\(\frac {-12 x^{2} \ln \left (3\right )-12 \,{\mathrm e}^{2} x}{\ln \left (3\right )}\)	\(20\)

[In]

int((-24*x*ln(3)-12*exp(2))/ln(3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx=-\frac {12 \, {\left (x^{2} \log \left (3\right ) + x e^{2}\right )}}{\log \left (3\right )} \]

[In]

integrate((-24*x*log(3)-12*exp(2))/log(3),x, algorithm="fricas")

[Out]

-12*(x^2*log(3) + x*e^2)/log(3)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx=- 12 x^{2} - \frac {12 x e^{2}}{\log {\left (3 \right )}} \]

[In]

integrate((-24*x*ln(3)-12*exp(2))/ln(3),x)

[Out]

-12*x**2 - 12*x*exp(2)/log(3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx=-\frac {12 \, {\left (x^{2} \log \left (3\right ) + x e^{2}\right )}}{\log \left (3\right )} \]

[In]

integrate((-24*x*log(3)-12*exp(2))/log(3),x, algorithm="maxima")

[Out]

-12*(x^2*log(3) + x*e^2)/log(3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx=-\frac {12 \, {\left (x^{2} \log \left (3\right ) + x e^{2}\right )}}{\log \left (3\right )} \]

[In]

integrate((-24*x*log(3)-12*exp(2))/log(3),x, algorithm="giac")

[Out]

-12*(x^2*log(3) + x*e^2)/log(3)

Mupad [B] (verification not implemented)

Time = 12.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.38 \[ \int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx=-\frac {{\left (12\,{\mathrm {e}}^2+24\,x\,\ln \left (3\right )\right )}^2}{48\,{\ln \left (3\right )}^2} \]

[In]

int(-(12*exp(2) + 24*x*log(3))/log(3),x)

[Out]

-(12*exp(2) + 24*x*log(3))^2/(48*log(3)^2)