\(\int \frac {1}{5} (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}) \, dx\) [5412]

Optimal result

Integrand size = 26, antiderivative size = 18 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=-4+e^{\frac {1}{5} e^{5-3 x}}-2 x \]

[Out]

-4-2*x+exp(1/5*exp(-3*x+5))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {12, 2320, 2225} \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=e^{\frac {1}{5} e^{5-3 x}}-2 x \]

[In]

Int[(-10 - 3*E^(5 + E^(5 - 3*x)/5 - 3*x))/5,x]

[Out]

E^(E^(5 - 3*x)/5) - 2*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 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 2320

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{align*} \text {integral}& = \frac {1}{5} \int \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx \\ & = -2 x-\frac {3}{5} \int e^{5+\frac {1}{5} e^{5-3 x}-3 x} \, dx \\ & = -2 x+\frac {1}{5} \text {Subst}\left (\int e^{5+\frac {e^5 x}{5}} \, dx,x,e^{-3 x}\right ) \\ & = e^{\frac {1}{5} e^{5-3 x}}-2 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=e^{\frac {1}{5} e^{5-3 x}}-2 x \]

[In]

Integrate[(-10 - 3*E^(5 + E^(5 - 3*x)/5 - 3*x))/5,x]

[Out]

E^(E^(5 - 3*x)/5) - 2*x

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78

method	result	size
default	\(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\)	\(14\)
norman	\(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\)	\(14\)
risch	\(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\)	\(14\)
parallelrisch	\(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\)	\(14\)
parts	\(-2 x +{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\)	\(14\)
derivativedivides	\(\frac {2 \ln \left (\frac {{\mathrm e}^{-3 x +5}}{5}\right )}{3}+{\mathrm e}^{\frac {{\mathrm e}^{-3 x +5}}{5}}\)	\(22\)

[In]

int(-3/5*exp(-3*x+5)*exp(1/5*exp(-3*x+5))-2,x,method=_RETURNVERBOSE)

[Out]

-2*x+exp(1/5*exp(-3*x+5))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (14) = 28\).

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.89 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=-{\left (2 \, x e^{\left (-3 \, x + 5\right )} - e^{\left (-3 \, x + \frac {1}{5} \, e^{\left (-3 \, x + 5\right )} + 5\right )}\right )} e^{\left (3 \, x - 5\right )} \]

[In]

integrate(-3/5*exp(-3*x+5)*exp(1/5*exp(-3*x+5))-2,x, algorithm="fricas")

[Out]

-(2*x*e^(-3*x + 5) - e^(-3*x + 1/5*e^(-3*x + 5) + 5))*e^(3*x - 5)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=- 2 x + e^{\frac {e^{5 - 3 x}}{5}} \]

[In]

integrate(-3/5*exp(-3*x+5)*exp(1/5*exp(-3*x+5))-2,x)

[Out]

-2*x + exp(exp(5 - 3*x)/5)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=-2 \, x + e^{\left (\frac {1}{5} \, e^{\left (-3 \, x + 5\right )}\right )} \]

[In]

integrate(-3/5*exp(-3*x+5)*exp(1/5*exp(-3*x+5))-2,x, algorithm="maxima")

[Out]

-2*x + e^(1/5*e^(-3*x + 5))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx=-2 \, x + e^{\left (\frac {1}{5} \, e^{\left (-3 \, x + 5\right )}\right )} \]

[In]

integrate(-3/5*exp(-3*x+5)*exp(1/5*exp(-3*x+5))-2,x, algorithm="giac")

[Out]

-2*x + e^(1/5*e^(-3*x + 5))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{5} \left (-10-3 e^{5+\frac {1}{5} e^{5-3 x}-3 x}\right ) \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{5-3\,x}}{5}}-2\,x \]

[In]

int(- (3*exp(exp(5 - 3*x)/5)*exp(5 - 3*x))/5 - 2,x)

[Out]

exp(exp(5 - 3*x)/5) - 2*x