3.80.63 \(\int e^{-5+x} (-2+e^5 (-1+x)) \, dx\)

Optimal. Leaf size=22 \[ 3-\frac {2 \left (1+e^5\right ) \left (3+e^x\right )}{e^5}+e^x x \]

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2187, 2176, 2194} \begin {gather*} -e^{x-5} \left (-e^5 x+e^5+2\right )-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-5 + x)*(-2 + E^5*(-1 + x)),x]

[Out]

-E^x - E^(-5 + x)*(2 + E^5 - E^5*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 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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} &=\int e^{-5+x} \left (-2-e^5+e^5 x\right ) \, dx\\ &=-e^{-5+x} \left (2+e^5-e^5 x\right )-e^5 \int e^{-5+x} \, dx\\ &=-e^x-e^{-5+x} \left (2+e^5-e^5 x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 12, normalized size = 0.55 \begin {gather*} e^x \left (-2-\frac {2}{e^5}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-5 + x)*(-2 + E^5*(-1 + x)),x]

[Out]

E^x*(-2 - 2/E^5 + x)

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fricas [A]  time = 0.85, size = 13, normalized size = 0.59 \begin {gather*} {\left ({\left (x - 2\right )} e^{5} - 2\right )} e^{\left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x - 2)*e^5 - 2)*e^(x - 5)

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giac [A]  time = 0.17, size = 13, normalized size = 0.59 \begin {gather*} {\left (x - 2\right )} e^{x} - 2 \, e^{\left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x - 2)*e^x - 2*e^(x - 5)

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maple [A]  time = 0.03, size = 16, normalized size = 0.73

method	result	size
risch	\(\left (x \,{\mathrm e}^{5}-2 \,{\mathrm e}^{5}-2\right ) {\mathrm e}^{x -5}\)	\(16\)
gosper	\({\mathrm e}^{x} \left (x \,{\mathrm e}^{5}-2 \,{\mathrm e}^{5}-2\right ) {\mathrm e}^{-5}\)	\(18\)
norman	\({\mathrm e}^{x} x -2 \left ({\mathrm e}^{5}+1\right ) {\mathrm e}^{-5} {\mathrm e}^{x}\)	\(18\)
meijerg	\(2 \,{\mathrm e}^{-5} \left (1-{\mathrm e}^{x}\right )+2-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}-{\mathrm e}^{x}\)	\(26\)
default	\({\mathrm e}^{-5} \left ({\mathrm e}^{5} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )-{\mathrm e}^{5} {\mathrm e}^{x}-2 \,{\mathrm e}^{x}\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-1)*exp(5)-2)*exp(x)/exp(5),x,method=_RETURNVERBOSE)

[Out]

(x*exp(5)-2*exp(5)-2)*exp(x-5)

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maxima [A]  time = 0.36, size = 17, normalized size = 0.77 \begin {gather*} {\left (x - 1\right )} e^{x} - 2 \, e^{\left (x - 5\right )} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x - 1)*e^x - 2*e^(x - 5) - e^x

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mupad [B]  time = 4.79, size = 17, normalized size = 0.77 \begin {gather*} -{\mathrm {e}}^{x-5}\,\left (2\,{\mathrm {e}}^5-x\,{\mathrm {e}}^5+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-5)*exp(x)*(exp(5)*(x - 1) - 2),x)

[Out]

-exp(x - 5)*(2*exp(5) - x*exp(5) + 2)

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sympy [A]  time = 0.10, size = 17, normalized size = 0.77 \begin {gather*} \frac {\left (x e^{5} - 2 e^{5} - 2\right ) e^{x}}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-1)*exp(5)-2)*exp(x)/exp(5),x)

[Out]

(x*exp(5) - 2*exp(5) - 2)*exp(-5)*exp(x)

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