\(\int \frac {1}{5} (e^2 (-18-4 x)+6 e^2 \log (3)) \, dx\) [10146]

Optimal result

Integrand size = 21, antiderivative size = 18 \[ \int \frac {1}{5} \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx=\frac {2}{5} e^2 x (-x+3 (-3+\log (3))) \]

[Out]

2/5*x*exp(2)*(3*ln(3)-9-x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12} \[ \int \frac {1}{5} \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx=\frac {6}{5} e^2 x \log (3)-\frac {1}{10} e^2 (2 x+9)^2 \]

[In]

Int[(E^2*(-18 - 4*x) + 6*E^2*Log[3])/5,x]

[Out]

-1/10*(E^2*(9 + 2*x)^2) + (6*E^2*x*Log[3])/5

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 {1}{5} \int \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx \\ & = -\frac {1}{10} e^2 (9+2 x)^2+\frac {6}{5} e^2 x \log (3) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1}{5} \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx=-\frac {2}{5} e^2 \left (9 x+x^2-3 x \log (3)\right ) \]

[In]

Integrate[(E^2*(-18 - 4*x) + 6*E^2*Log[3])/5,x]

[Out]

(-2*E^2*(9*x + x^2 - 3*x*Log[3]))/5

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {2 x \,{\mathrm e}^{2} \left (3 \ln \left (3\right )-9-x \right )}{5}\)	\(15\)
default	\(\frac {2 \,{\mathrm e}^{2} \left (3 x \ln \left (3\right )-x^{2}-9 x \right )}{5}\)	\(19\)
risch	\(\frac {6 x \,{\mathrm e}^{2} \ln \left (3\right )}{5}-\frac {2 x^{2} {\mathrm e}^{2}}{5}-\frac {18 \,{\mathrm e}^{2} x}{5}\)	\(21\)
parts	\(\frac {6 x \,{\mathrm e}^{2} \ln \left (3\right )}{5}-\frac {2 x^{2} {\mathrm e}^{2}}{5}-\frac {18 \,{\mathrm e}^{2} x}{5}\)	\(21\)
norman	\(\left (\frac {6 \,{\mathrm e}^{2} \ln \left (3\right )}{5}-\frac {18 \,{\mathrm e}^{2}}{5}\right ) x -\frac {2 x^{2} {\mathrm e}^{2}}{5}\)	\(22\)
parallelrisch	\(\frac {{\mathrm e}^{2} \left (-2 x^{2}-18 x \right )}{5}+\frac {6 x \,{\mathrm e}^{2} \ln \left (3\right )}{5}\)	\(22\)

[In]

int(6/5*exp(2)*ln(3)+1/5*(-4*x-18)*exp(2),x,method=_RETURNVERBOSE)

[Out]

2/5*x*exp(2)*(3*ln(3)-9-x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1}{5} \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx=\frac {6}{5} \, x e^{2} \log \left (3\right ) - \frac {2}{5} \, {\left (x^{2} + 9 \, x\right )} e^{2} \]

[In]

integrate(6/5*exp(2)*log(3)+1/5*(-4*x-18)*exp(2),x, algorithm="fricas")

[Out]

6/5*x*e^2*log(3) - 2/5*(x^2 + 9*x)*e^2

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {1}{5} \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx=- \frac {2 x^{2} e^{2}}{5} + x \left (- \frac {18 e^{2}}{5} + \frac {6 e^{2} \log {\left (3 \right )}}{5}\right ) \]

[In]

integrate(6/5*exp(2)*ln(3)+1/5*(-4*x-18)*exp(2),x)

[Out]

-2*x**2*exp(2)/5 + x*(-18*exp(2)/5 + 6*exp(2)*log(3)/5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1}{5} \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx=\frac {6}{5} \, x e^{2} \log \left (3\right ) - \frac {2}{5} \, {\left (x^{2} + 9 \, x\right )} e^{2} \]

[In]

integrate(6/5*exp(2)*log(3)+1/5*(-4*x-18)*exp(2),x, algorithm="maxima")

[Out]

6/5*x*e^2*log(3) - 2/5*(x^2 + 9*x)*e^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {1}{5} \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx=\frac {6}{5} \, x e^{2} \log \left (3\right ) - \frac {2}{5} \, {\left (x^{2} + 9 \, x\right )} e^{2} \]

[In]

integrate(6/5*exp(2)*log(3)+1/5*(-4*x-18)*exp(2),x, algorithm="giac")

[Out]

6/5*x*e^2*log(3) - 2/5*(x^2 + 9*x)*e^2

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx=-\frac {{\mathrm {e}}^2\,\left (4\,x+18\right )\,\left (4\,x-12\,\ln \left (3\right )+18\right )}{40} \]

[In]

int((6*exp(2)*log(3))/5 - (exp(2)*(4*x + 18))/5,x)

[Out]

-(exp(2)*(4*x + 18)*(4*x - 12*log(3) + 18))/40