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

Optimal. Leaf size=26 \[ x+\log (6)-\log \left (\frac {1}{2} e^{4+e^{1-\frac {x}{5}} x}\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {12, 6688, 2176, 2194} \begin {gather*} e^{1-\frac {x}{5}} (5-x)-5 e^{1-\frac {x}{5}}+x \end {gather*}

Antiderivative was successfully verified.

[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 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]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.54 \begin {gather*} x-e^{1-\frac {x}{5}} x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - E^(1 - x/5)*x

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fricas [A]  time = 1.07, size = 19, normalized size = 0.73 \begin {gather*} {\left (x e^{\left (\frac {1}{5} \, x - 1\right )} - x\right )} e^{\left (-\frac {1}{5} \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 11, normalized size = 0.42 \begin {gather*} -x e^{\left (-\frac {1}{5} \, x + 1\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 12, normalized size = 0.46

method	result	size
risch	\(x -x \,{\mathrm e}^{1-\frac {x}{5}}\)	\(12\)
norman	\(\left (x \,{\mathrm e}^{\frac {x}{5}-1}-x \right ) {\mathrm e}^{1-\frac {x}{5}}\)	\(22\)
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	\(x -5-5 \,{\mathrm e}^{1-\frac {x}{5}} \left (\frac {x}{5}-1\right )-5 \,{\mathrm e}^{1-\frac {x}{5}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.51, size = 25, normalized size = 0.96 \begin {gather*} -{\left (x e + 5 \, e\right )} e^{\left (-\frac {1}{5} \, x\right )} + x + 5 \, e^{\left (-\frac {1}{5} \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.06, size = 11, normalized size = 0.42 \begin {gather*} -x\,\left ({\mathrm {e}}^{1-\frac {x}{5}}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*(exp(1 - x/5) - 1)

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sympy [A]  time = 0.10, size = 8, normalized size = 0.31 \begin {gather*} - x e^{1 - \frac {x}{5}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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