\(\int \frac {1}{5} e^{\frac {5-x}{5}} (-5+5 e^{\frac {1}{5} (-5+x)}+x) \, dx\) [3482]

Optimal result

Integrand size = 29, antiderivative size = 26 \[ \int \frac {1}{5} e^{\frac {5-x}{5}} \left (-5+5 e^{\frac {1}{5} (-5+x)}+x\right ) \, dx=x+\log (6)-\log \left (\frac {1}{2} e^{4+e^{1-\frac {x}{5}} x}\right ) \]

[Out]

ln(6)+x-ln(1/2*exp(x/exp(1/5*x-1)+4))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {12, 6820, 2207, 2225} \[ \int \frac {1}{5} e^{\frac {5-x}{5}} \left (-5+5 e^{\frac {1}{5} (-5+x)}+x\right ) \, dx=e^{1-\frac {x}{5}} (5-x)-5 e^{1-\frac {x}{5}}+x \]

[In]

Int[(E^((5 - x)/5)*(-5 + 5*E^((-5 + x)/5) + x))/5,x]

[Out]

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

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int e^{\frac {5-x}{5}} \left (-5+5 e^{\frac {1}{5} (-5+x)}+x\right ) \, dx \\ & = \frac {1}{5} \int \left (5+e^{1-\frac {x}{5}} (-5+x)\right ) \, dx \\ & = x+\frac {1}{5} \int e^{1-\frac {x}{5}} (-5+x) \, dx \\ & = e^{1-\frac {x}{5}} (5-x)+x+\int e^{1-\frac {x}{5}} \, dx \\ & = -5 e^{1-\frac {x}{5}}+e^{1-\frac {x}{5}} (5-x)+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.54 \[ \int \frac {1}{5} e^{\frac {5-x}{5}} \left (-5+5 e^{\frac {1}{5} (-5+x)}+x\right ) \, dx=x-e^{1-\frac {x}{5}} x \]

[In]

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

[Out]

x - E^(1 - x/5)*x

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.46

method	result	size
risch	\(x -x \,{\mathrm e}^{1-\frac {x}{5}}\)	\(12\)
norman	\(\left (x \,{\mathrm e}^{\frac {x}{5}-1}-x \right ) {\mathrm e}^{1-\frac {x}{5}}\)	\(22\)
parts	\(x -5 \,{\mathrm e}^{1-\frac {x}{5}} \left (\frac {x}{5}-1\right )-5 \,{\mathrm e}^{1-\frac {x}{5}}\)	\(28\)
derivativedivides	\(x -5-5 \,{\mathrm e}^{1-\frac {x}{5}} \left (\frac {x}{5}-1\right )-5 \,{\mathrm e}^{1-\frac {x}{5}}\)	\(29\)
default	\(x -5-5 \,{\mathrm e}^{1-\frac {x}{5}} \left (\frac {x}{5}-1\right )-5 \,{\mathrm e}^{1-\frac {x}{5}}\)	\(29\)
parallelrisch	\(-\frac {\left (-25 \ln \left ({\mathrm e}^{\frac {x}{5}-1}\right ) {\mathrm e}^{\frac {x}{5}-1}+5 x \right ) {\mathrm e}^{1-\frac {x}{5}}}{5}\)	\(30\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {1}{5} e^{\frac {5-x}{5}} \left (-5+5 e^{\frac {1}{5} (-5+x)}+x\right ) \, dx={\left (x e^{\left (\frac {1}{5} \, x - 1\right )} - x\right )} e^{\left (-\frac {1}{5} \, x + 1\right )} \]

[In]

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

[Out]

(x*e^(1/5*x - 1) - x)*e^(-1/5*x + 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.31 \[ \int \frac {1}{5} e^{\frac {5-x}{5}} \left (-5+5 e^{\frac {1}{5} (-5+x)}+x\right ) \, dx=- x e^{1 - \frac {x}{5}} + x \]

[In]

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

[Out]

-x*exp(1 - x/5) + x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {1}{5} e^{\frac {5-x}{5}} \left (-5+5 e^{\frac {1}{5} (-5+x)}+x\right ) \, dx=-{\left (x e + 5 \, e\right )} e^{\left (-\frac {1}{5} \, x\right )} + x + 5 \, e^{\left (-\frac {1}{5} \, x + 1\right )} \]

[In]

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

[Out]

-(x*e + 5*e)*e^(-1/5*x) + x + 5*e^(-1/5*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.46 \[ \int \frac {1}{5} e^{\frac {5-x}{5}} \left (-5+5 e^{\frac {1}{5} (-5+x)}+x\right ) \, dx=-x e^{\left (-\frac {1}{5} \, x + 1\right )} + x - 5 \]

[In]

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

[Out]

-x*e^(-1/5*x + 1) + x - 5

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.42 \[ \int \frac {1}{5} e^{\frac {5-x}{5}} \left (-5+5 e^{\frac {1}{5} (-5+x)}+x\right ) \, dx=-x\,\left ({\mathrm {e}}^{1-\frac {x}{5}}-1\right ) \]

[In]

int(exp(1 - x/5)*(x/5 + exp(x/5 - 1) - 1),x)

[Out]

-x*(exp(1 - x/5) - 1)