3.15.68 \(\int \frac {1}{5} (6-5 e^{-1+e^{-1-x}-x}-e^x-2 e^{2 x}) \, dx\)

Optimal. Leaf size=30 \[ 6+e^{e^{-1-x}}+x+\frac {1}{5} \left (-e^x-e^{2 x}+x\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {12, 2282, 2194} \begin {gather*} \frac {6 x}{5}+e^{e^{-x-1}}-\frac {e^x}{5}-\frac {e^{2 x}}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 - 5*E^(-1 + E^(-1 - x) - x) - E^x - 2*E^(2*x))/5,x]

[Out]

E^E^(-1 - x) - E^x/5 - E^(2*x)/5 + (6*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 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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (6-5 e^{-1+e^{-1-x}-x}-e^x-2 e^{2 x}\right ) \, dx\\ &=\frac {6 x}{5}-\frac {\int e^x \, dx}{5}-\frac {2}{5} \int e^{2 x} \, dx-\int e^{-1+e^{-1-x}-x} \, dx\\ &=-\frac {e^x}{5}-\frac {e^{2 x}}{5}+\frac {6 x}{5}+\operatorname {Subst}\left (\int e^{-1+\frac {x}{e}} \, dx,x,e^{-x}\right )\\ &=e^{e^{-1-x}}-\frac {e^x}{5}-\frac {e^{2 x}}{5}+\frac {6 x}{5}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 31, normalized size = 1.03 \begin {gather*} \frac {1}{5} \left (5 e^{e^{-1-x}}-e^x-e^{2 x}+6 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 - 5*E^(-1 + E^(-1 - x) - x) - E^x - 2*E^(2*x))/5,x]

[Out]

(5*E^E^(-1 - x) - E^x - E^(2*x) + 6*x)/5

________________________________________________________________________________________

fricas [A]  time = 0.92, size = 36, normalized size = 1.20 \begin {gather*} \frac {6}{5} \, x + e^{\left (-{\left ({\left (x + 1\right )} e^{\left (x + 1\right )} - 1\right )} e^{\left (-x - 1\right )} + x + 1\right )} - \frac {1}{5} \, e^{\left (2 \, x\right )} - \frac {1}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-x-1)*exp(exp(-x-1))-2/5*exp(2*x)-1/5*exp(x)+6/5,x, algorithm="fricas")

[Out]

6/5*x + e^(-((x + 1)*e^(x + 1) - 1)*e^(-x - 1) + x + 1) - 1/5*e^(2*x) - 1/5*e^x

________________________________________________________________________________________

giac [A]  time = 0.29, size = 21, normalized size = 0.70 \begin {gather*} \frac {6}{5} \, x - \frac {1}{5} \, e^{\left (2 \, x\right )} - \frac {1}{5} \, e^{x} + e^{\left (e^{\left (-x - 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-x-1)*exp(exp(-x-1))-2/5*exp(2*x)-1/5*exp(x)+6/5,x, algorithm="giac")

[Out]

6/5*x - 1/5*e^(2*x) - 1/5*e^x + e^(e^(-x - 1))

________________________________________________________________________________________

maple [A]  time = 0.04, size = 22, normalized size = 0.73




method result size



default \(\frac {6 x}{5}+{\mathrm e}^{{\mathrm e}^{-x -1}}-\frac {{\mathrm e}^{x}}{5}-\frac {{\mathrm e}^{2 x}}{5}\) \(22\)
risch \(\frac {6 x}{5}+{\mathrm e}^{{\mathrm e}^{-x -1}}-\frac {{\mathrm e}^{x}}{5}-\frac {{\mathrm e}^{2 x}}{5}\) \(22\)
norman \(\left ({\mathrm e}^{-1} {\mathrm e} \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{-1} {\mathrm e}^{-x}}-\frac {{\mathrm e}^{2 x}}{5}-\frac {{\mathrm e}^{3 x}}{5}+\frac {6 \,{\mathrm e}^{x} x}{5}\right ) {\mathrm e}^{-x}\) \(43\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-x-1)*exp(exp(-x-1))-2/5*exp(2*x)-1/5*exp(x)+6/5,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.66, size = 21, normalized size = 0.70 \begin {gather*} \frac {6}{5} \, x - \frac {1}{5} \, e^{\left (2 \, x\right )} - \frac {1}{5} \, e^{x} + e^{\left (e^{\left (-x - 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-x-1)*exp(exp(-x-1))-2/5*exp(2*x)-1/5*exp(x)+6/5,x, algorithm="maxima")

[Out]

6/5*x - 1/5*e^(2*x) - 1/5*e^x + e^(e^(-x - 1))

________________________________________________________________________________________

mupad [B]  time = 1.04, size = 22, normalized size = 0.73 \begin {gather*} \frac {6\,x}{5}-\frac {{\mathrm {e}}^{2\,x}}{5}+{\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-1}}-\frac {{\mathrm {e}}^x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(6/5 - exp(x)/5 - exp(exp(- x - 1))*exp(- x - 1) - (2*exp(2*x))/5,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.21, size = 26, normalized size = 0.87 \begin {gather*} \frac {6 x}{5} - \frac {e^{2 x}}{5} - \frac {e^{x}}{5} + e^{\frac {e^{- x}}{e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-x-1)*exp(exp(-x-1))-2/5*exp(2*x)-1/5*exp(x)+6/5,x)

[Out]

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

________________________________________________________________________________________