\(\int \frac {1}{5} e^x (-1-x) \, dx\) [6160]

Optimal result

Integrand size = 12, antiderivative size = 8 \[ \int \frac {1}{5} e^x (-1-x) \, dx=-\frac {e^x x}{5} \]

[Out]

-1/5*exp(x)*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(18\) vs. \(2(8)=16\).

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 2.25, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 2207, 2225} \[ \int \frac {1}{5} e^x (-1-x) \, dx=\frac {e^x}{5}-\frac {1}{5} e^x (x+1) \]

[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 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}& = \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} e^x (-1-x) \, dx=-\frac {e^x x}{5} \]

[In]

Integrate[(E^x*(-1 - x))/5,x]

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 6, normalized size of antiderivative = 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	\(-\frac {{\mathrm e}^{x} x}{5}\)	\(6\)
parallelrisch	\(-\frac {{\mathrm e}^{x} x}{5}\)	\(6\)
parts	\(-\frac {{\mathrm e}^{x} x}{5}\)	\(6\)
meijerg	\(-\frac {{\mathrm e}^{x}}{5}+\frac {\left (2-2 x \right ) {\mathrm e}^{x}}{10}\)	\(15\)

[In]

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

[Out]

-1/5*exp(x)*x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5} e^x (-1-x) \, dx=-\frac {1}{5} \, x e^{x} \]

[In]

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

[Out]

-1/5*x*e^x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {1}{5} e^x (-1-x) \, dx=- \frac {x e^{x}}{5} \]

[In]

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

[Out]

-x*exp(x)/5

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12 vs. \(2 (5) = 10\).

Time = 0.17 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.50 \[ \int \frac {1}{5} e^x (-1-x) \, dx=-\frac {1}{5} \, {\left (x - 1\right )} e^{x} - \frac {1}{5} \, e^{x} \]

[In]

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

[Out]

-1/5*(x - 1)*e^x - 1/5*e^x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5} e^x (-1-x) \, dx=-\frac {1}{5} \, x e^{x} \]

[In]

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

[Out]

-1/5*x*e^x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5} e^x (-1-x) \, dx=-\frac {x\,{\mathrm {e}}^x}{5} \]

[In]

int(-(exp(x)*(x + 1))/5,x)

[Out]

-(x*exp(x))/5