3.62.60 \(\int \frac {1}{5} e^x (-1-x) \, dx\)

Optimal. Leaf size=8 \[ -\frac {e^x x}{5} \]

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Rubi [B]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 2.25, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 2176, 2194} \begin {gather*} \frac {e^x}{5}-\frac {1}{5} e^x (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-1 - x))/5,x]

[Out]

E^x/5 - (E^x*(1 + x))/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 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} &=\frac {1}{5} \int e^x (-1-x) \, dx\\ &=-\frac {1}{5} e^x (1+x)+\frac {\int e^x \, dx}{5}\\ &=\frac {e^x}{5}-\frac {1}{5} e^x (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 8, normalized size = 1.00 \begin {gather*} -\frac {e^x x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-1 - x))/5,x]

[Out]

-1/5*(E^x*x)

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fricas [A]  time = 0.62, size = 5, normalized size = 0.62 \begin {gather*} -\frac {1}{5} \, x e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*x*e^x

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giac [A]  time = 0.18, size = 5, normalized size = 0.62 \begin {gather*} -\frac {1}{5} \, x e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*x*e^x

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maple [A]  time = 0.02, size = 6, normalized size = 0.75

method	result	size
gosper	\(-\frac {{\mathrm e}^{x} x}{5}\)	\(6\)
default	\(-\frac {{\mathrm e}^{x} x}{5}\)	\(6\)
norman	\(-\frac {{\mathrm e}^{x} x}{5}\)	\(6\)
risch	\(-\frac {{\mathrm e}^{x} x}{5}\)	\(6\)
meijerg	\(-\frac {{\mathrm e}^{x}}{5}+\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{10}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*exp(x)*x

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maxima [B]  time = 0.35, size = 12, normalized size = 1.50 \begin {gather*} -\frac {1}{5} \, {\left (x - 1\right )} e^{x} - \frac {1}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(x - 1)*e^x - 1/5*e^x

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mupad [B]  time = 0.02, size = 5, normalized size = 0.62 \begin {gather*} -\frac {x\,{\mathrm {e}}^x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(x + 1))/5,x)

[Out]

-(x*exp(x))/5

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sympy [A]  time = 0.08, size = 7, normalized size = 0.88 \begin {gather*} - \frac {x e^{x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*exp(x)/5

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