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

Optimal. Leaf size=21 \[ -2 \left (-x+e^{\frac {1}{2} \left (3+e^2\right )+x} x\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.76, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2176, 2194} \begin {gather*} 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) \end {gather*}

Antiderivative was successfully verified.

[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 2176

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] &&  !$UseGamma === True

Rule 2194

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.00 \begin {gather*} 2 x-2 e^{\frac {3}{2}+\frac {e^2}{2}+x} x \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.60, size = 15, normalized size = 0.71 \begin {gather*} -2 \, x e^{\left (x + \frac {1}{2} \, e^{2} + \frac {3}{2}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-2)*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

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giac [A]  time = 0.15, size = 21, normalized size = 1.00 \begin {gather*} -2 \, x e^{\left (x + e^{\left (-\log \relax (2) + \log \left (e^{2} + 3\right )\right )}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-2)*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

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maple [A]  time = 0.06, size = 16, normalized size = 0.76

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x-2)*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

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maxima [A]  time = 0.35, size = 37, normalized size = 1.76 \begin {gather*} -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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-2)*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)

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mupad [B]  time = 0.09, size = 15, normalized size = 0.71 \begin {gather*} -2\,x\,\left ({\mathrm {e}}^{\frac {{\mathrm {e}}^2}{2}}\,{\mathrm {e}}^{3/2}\,{\mathrm {e}}^x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.08, size = 17, normalized size = 0.81 \begin {gather*} - 2 x e^{x + \frac {3}{2} + \frac {e^{2}}{2}} + 2 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*exp(x + 3/2 + exp(2)/2) + 2*x

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