\(\int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx\) [5507]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 18 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=-6+x-\frac {e^{2 (4+x)}}{3 \log (5)} \]

[Out]

x-1/3*exp(2*x+8)/ln(5)-6

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 2225} \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=x-\frac {e^{2 x+8}}{3 \log (5)} \]

[In]

Int[(-2*E^(8 + 2*x) + 3*Log[5])/(3*Log[5]),x]

[Out]

x - E^(8 + 2*x)/(3*Log[5])

Rule 12

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

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-2 e^{8+2 x}+3 \log (5)\right ) \, dx}{3 \log (5)} \\ & = x-\frac {2 \int e^{8+2 x} \, dx}{3 \log (5)} \\ & = x-\frac {e^{8+2 x}}{3 \log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=x-\frac {e^{8+2 x}}{\log (125)} \]

[In]

Integrate[(-2*E^(8 + 2*x) + 3*Log[5])/(3*Log[5]),x]

[Out]

x - E^(8 + 2*x)/Log[125]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method result size
norman \(x -\frac {{\mathrm e}^{2 x +8}}{3 \ln \left (5\right )}\) \(15\)
risch \(x -\frac {{\mathrm e}^{2 x +8}}{3 \ln \left (5\right )}\) \(15\)
parts \(x -\frac {{\mathrm e}^{2 x +8}}{3 \ln \left (5\right )}\) \(15\)
default \(\frac {-{\mathrm e}^{2 x +8}+3 x \ln \left (5\right )}{3 \ln \left (5\right )}\) \(21\)
parallelrisch \(\frac {-{\mathrm e}^{2 x +8}+3 x \ln \left (5\right )}{3 \ln \left (5\right )}\) \(21\)
derivativedivides \(\frac {-2 \,{\mathrm e}^{2 x +8}+3 \ln \left (5\right ) \ln \left ({\mathrm e}^{2 x +8}\right )}{6 \ln \left (5\right )}\) \(27\)

[In]

int(1/3*(-2*exp(2*x+8)+3*ln(5))/ln(5),x,method=_RETURNVERBOSE)

[Out]

x-1/3*exp(2*x+8)/ln(5)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=\frac {3 \, x \log \left (5\right ) - e^{\left (2 \, x + 8\right )}}{3 \, \log \left (5\right )} \]

[In]

integrate(1/3*(-2*exp(2*x+8)+3*log(5))/log(5),x, algorithm="fricas")

[Out]

1/3*(3*x*log(5) - e^(2*x + 8))/log(5)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=x - \frac {e^{2 x + 8}}{3 \log {\left (5 \right )}} \]

[In]

integrate(1/3*(-2*exp(2*x+8)+3*ln(5))/ln(5),x)

[Out]

x - exp(2*x + 8)/(3*log(5))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=\frac {3 \, x \log \left (5\right ) - e^{\left (2 \, x + 8\right )}}{3 \, \log \left (5\right )} \]

[In]

integrate(1/3*(-2*exp(2*x+8)+3*log(5))/log(5),x, algorithm="maxima")

[Out]

1/3*(3*x*log(5) - e^(2*x + 8))/log(5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=\frac {3 \, x \log \left (5\right ) - e^{\left (2 \, x + 8\right )}}{3 \, \log \left (5\right )} \]

[In]

integrate(1/3*(-2*exp(2*x+8)+3*log(5))/log(5),x, algorithm="giac")

[Out]

1/3*(3*x*log(5) - e^(2*x + 8))/log(5)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx=x-\frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^8}{3\,\ln \left (5\right )} \]

[In]

int((log(5) - (2*exp(2*x + 8))/3)/log(5),x)

[Out]

x - (exp(2*x)*exp(8))/(3*log(5))