\(\int (2+e^{\frac {1}{2} (3+e^2)+x} (-2-2 x)) \, dx\) [5874]

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

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2207, 2225} \[ \int \left (2+e^{\frac {1}{2} \left (3+e^2\right )+x} (-2-2 x)\right ) \, dx=2 x+2 e^{x+\frac {1}{2} \left (3+e^2\right )}-2 e^{x+\frac {1}{2} \left (3+e^2\right )} (x+1) \]

[In]

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

[Out]

2*E^((3 + E^2)/2 + x) + 2*x - 2*E^((3 + E^2)/2 + x)*(1 + x)

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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}& = 2 x+\int e^{\frac {1}{2} \left (3+e^2\right )+x} (-2-2 x) \, dx \\ & = 2 x-2 e^{\frac {1}{2} \left (3+e^2\right )+x} (1+x)+2 \int e^{\frac {1}{2} \left (3+e^2\right )+x} \, dx \\ & = 2 e^{\frac {1}{2} \left (3+e^2\right )+x}+2 x-2 e^{\frac {1}{2} \left (3+e^2\right )+x} (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \left (2+e^{\frac {1}{2} \left (3+e^2\right )+x} (-2-2 x)\right ) \, dx=2 x-2 e^{\frac {3}{2}+\frac {e^2}{2}+x} x \]

[In]

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

[Out]

2*x - 2*E^(3/2 + E^2/2 + x)*x

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76

method result size
risch \(-2 x \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{2}+\frac {3}{2}+x}+2 x\) \(16\)
norman \(-2 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x} x +2 x\) \(22\)
parallelrisch \(-2 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x} x +2 x\) \(22\)
default \(2 x -2 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x} \left ({\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x \right )+3 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x}+{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x} {\mathrm e}^{2}\) \(67\)
parts \(2 x -2 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x} \left ({\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x \right )+3 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x}+{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x} {\mathrm e}^{2}\) \(67\)
derivativedivides \(2 \,{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+2 x -2 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x} \left ({\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x \right )+3 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x}+{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{2}+3\right )-\ln \left (2\right )}+x} {\mathrm e}^{2}\) \(80\)

[In]

int((-2-2*x)*exp(exp(ln(exp(2)+3)-ln(2))+x)+2,x,method=_RETURNVERBOSE)

[Out]

-2*x*exp(1/2*exp(2)+3/2+x)+2*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \left (2+e^{\frac {1}{2} \left (3+e^2\right )+x} (-2-2 x)\right ) \, dx=-2 \, x e^{\left (x + \frac {1}{2} \, e^{2} + \frac {3}{2}\right )} + 2 \, x \]

[In]

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

[Out]

-2*x*e^(x + 1/2*e^2 + 3/2) + 2*x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (2+e^{\frac {1}{2} \left (3+e^2\right )+x} (-2-2 x)\right ) \, dx=- 2 x e^{x + \frac {3}{2} + \frac {e^{2}}{2}} + 2 x \]

[In]

integrate((-2-2*x)*exp(exp(ln(exp(2)+3)-ln(2))+x)+2,x)

[Out]

-2*x*exp(x + 3/2 + exp(2)/2) + 2*x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76 \[ \int \left (2+e^{\frac {1}{2} \left (3+e^2\right )+x} (-2-2 x)\right ) \, dx=-2 \, {\left (x e^{\left (\frac {1}{2} \, e^{2} + \frac {3}{2}\right )} - e^{\left (\frac {1}{2} \, e^{2} + \frac {3}{2}\right )}\right )} e^{x} + 2 \, x - 2 \, e^{\left (x + \frac {1}{2} \, e^{2} + \frac {3}{2}\right )} \]

[In]

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

[Out]

-2*(x*e^(1/2*e^2 + 3/2) - e^(1/2*e^2 + 3/2))*e^x + 2*x - 2*e^(x + 1/2*e^2 + 3/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \left (2+e^{\frac {1}{2} \left (3+e^2\right )+x} (-2-2 x)\right ) \, dx=-2 \, x e^{\left (x + e^{\left (-\log \left (2\right ) + \log \left (e^{2} + 3\right )\right )}\right )} + 2 \, x \]

[In]

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

[Out]

-2*x*e^(x + e^(-log(2) + log(e^2 + 3))) + 2*x

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \left (2+e^{\frac {1}{2} \left (3+e^2\right )+x} (-2-2 x)\right ) \, dx=-2\,x\,\left ({\mathrm {e}}^{\frac {{\mathrm {e}}^2}{2}}\,{\mathrm {e}}^{3/2}\,{\mathrm {e}}^x-1\right ) \]

[In]

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

[Out]

-2*x*(exp(exp(2)/2)*exp(3/2)*exp(x) - 1)