3.33.85 \(\int \frac {1}{2} (-4+e^{e^3-x}+8 e^x) \, dx\)

Optimal. Leaf size=22 \[ -\frac {1}{2} e^{e^3-x}+4 e^x-2 x \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2194} \begin {gather*} -2 x-\frac {e^{e^3-x}}{2}+4 e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + E^(E^3 - x) + 8*E^x)/2,x]

[Out]

-1/2*E^(E^3 - x) + 4*E^x - 2*x

Rule 12

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

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} &=\frac {1}{2} \int \left (-4+e^{e^3-x}+8 e^x\right ) \, dx\\ &=-2 x+\frac {1}{2} \int e^{e^3-x} \, dx+4 \int e^x \, dx\\ &=-\frac {1}{2} e^{e^3-x}+4 e^x-2 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.09 \begin {gather*} \frac {1}{2} \left (-e^{e^3-x}+8 e^x-4 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + E^(E^3 - x) + 8*E^x)/2,x]

[Out]

(-E^(E^3 - x) + 8*E^x - 4*x)/2

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fricas [A]  time = 0.68, size = 34, normalized size = 1.55 \begin {gather*} -\frac {1}{2} \, {\left (4 \, x e^{\left (-x + e^{3}\right )} + e^{\left (-2 \, x + 2 \, e^{3}\right )} - 8 \, e^{\left (e^{3}\right )}\right )} e^{\left (x - e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*x*e^(-x + e^3) + e^(-2*x + 2*e^3) - 8*e^(e^3))*e^(x - e^3)

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giac [A]  time = 0.19, size = 17, normalized size = 0.77 \begin {gather*} -2 \, x + 4 \, e^{x} - \frac {1}{2} \, e^{\left (-x + e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + 4*e^x - 1/2*e^(-x + e^3)

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maple [A]  time = 0.02, size = 18, normalized size = 0.82

method	result	size
default	\(-\frac {{\mathrm e}^{-x +{\mathrm e}^{3}}}{2}-2 x +4 \,{\mathrm e}^{x}\)	\(18\)
risch	\(-\frac {{\mathrm e}^{-x +{\mathrm e}^{3}}}{2}-2 x +4 \,{\mathrm e}^{x}\)	\(18\)
norman	\(\left (4 \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x -\frac {{\mathrm e}^{{\mathrm e}^{3}}}{2}\right ) {\mathrm e}^{-x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*exp(-x+exp(3))-2*x+4*exp(x)

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maxima [A]  time = 0.88, size = 17, normalized size = 0.77 \begin {gather*} -2 \, x + 4 \, e^{x} - \frac {1}{2} \, e^{\left (-x + e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + 4*e^x - 1/2*e^(-x + e^3)

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mupad [B]  time = 1.90, size = 17, normalized size = 0.77 \begin {gather*} 4\,{\mathrm {e}}^x-2\,x-\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^3}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(exp(3) - x)/2 + 4*exp(x) - 2,x)

[Out]

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

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sympy [A]  time = 0.14, size = 17, normalized size = 0.77 \begin {gather*} - 2 x + 4 e^{x} - \frac {e^{- x} e^{e^{3}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*exp(-x+exp(3))+4*exp(x)-2,x)

[Out]

-2*x + 4*exp(x) - exp(-x)*exp(exp(3))/2

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