\(\int e^{-4 x} (e^x+e^{2 x}+e^{3 x}) \, dx\) [675]

Optimal result

Integrand size = 20, antiderivative size = 26 \[ \int e^{-4 x} \left (e^x+e^{2 x}+e^{3 x}\right ) \, dx=-\frac {1}{3} e^{-3 x}-\frac {e^{-2 x}}{2}-e^{-x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2320, 14} \[ \int e^{-4 x} \left (e^x+e^{2 x}+e^{3 x}\right ) \, dx=-\frac {1}{3} e^{-3 x}-\frac {e^{-2 x}}{2}-e^{-x} \]

[In]

Int[(E^x + E^(2*x) + E^(3*x))/E^(4*x),x]

[Out]

-1/3*1/E^(3*x) - 1/(2*E^(2*x)) - E^(-x)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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}& = \text {Subst}\left (\int \frac {1+x+x^2}{x^4} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \left (\frac {1}{x^4}+\frac {1}{x^3}+\frac {1}{x^2}\right ) \, dx,x,e^x\right ) \\ & = -\frac {1}{3} e^{-3 x}-\frac {e^{-2 x}}{2}-e^{-x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int e^{-4 x} \left (e^x+e^{2 x}+e^{3 x}\right ) \, dx=\frac {1}{6} e^{-3 x} \left (-2-3 e^x-6 e^{2 x}\right ) \]

[In]

Integrate[(E^x + E^(2*x) + E^(3*x))/E^(4*x),x]

[Out]

(-2 - 3*E^x - 6*E^(2*x))/(6*E^(3*x))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {{\mathrm e}^{-3 x}}{3}-\frac {{\mathrm e}^{-2 x}}{2}-{\mathrm e}^{-x}\)	\(20\)
risch	\(-\frac {{\mathrm e}^{-3 x}}{3}-\frac {{\mathrm e}^{-2 x}}{2}-{\mathrm e}^{-x}\)	\(20\)
parts	\(-\frac {{\mathrm e}^{-3 x}}{3}-\frac {{\mathrm e}^{-2 x}}{2}-{\mathrm e}^{-x}\)	\(20\)
meijerg	\(\frac {11}{6}-\frac {{\mathrm e}^{-3 x}}{3}-\frac {{\mathrm e}^{-2 x}}{2}-{\mathrm e}^{-x}\)	\(21\)
norman	\(\left (-\frac {{\mathrm e}^{2 x}}{2}-{\mathrm e}^{3 x}-\frac {{\mathrm e}^{x}}{3}\right ) {\mathrm e}^{-4 x}\)	\(23\)
parallelrisch	\(\frac {\left (-2 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{2 x}-6 \,{\mathrm e}^{3 x}\right ) {\mathrm e}^{-4 x}}{6}\)	\(26\)

[In]

int((exp(x)+exp(2*x)+exp(3*x))/exp(4*x),x,method=_RETURNVERBOSE)

[Out]

-1/3/exp(x)^3-1/2/exp(x)^2-1/exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int e^{-4 x} \left (e^x+e^{2 x}+e^{3 x}\right ) \, dx=-\frac {1}{6} \, {\left (6 \, e^{\left (2 \, x\right )} + 3 \, e^{x} + 2\right )} e^{\left (-3 \, x\right )} \]

[In]

integrate((exp(x)+exp(2*x)+exp(3*x))/exp(4*x),x, algorithm="fricas")

[Out]

-1/6*(6*e^(2*x) + 3*e^x + 2)*e^(-3*x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int e^{-4 x} \left (e^x+e^{2 x}+e^{3 x}\right ) \, dx=- e^{- x} - \frac {e^{- 2 x}}{2} - \frac {e^{- 3 x}}{3} \]

[In]

integrate((exp(x)+exp(2*x)+exp(3*x))/exp(4*x),x)

[Out]

-exp(-x) - exp(-2*x)/2 - exp(-3*x)/3

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int e^{-4 x} \left (e^x+e^{2 x}+e^{3 x}\right ) \, dx=-e^{\left (-x\right )} - \frac {1}{2} \, e^{\left (-2 \, x\right )} - \frac {1}{3} \, e^{\left (-3 \, x\right )} \]

[In]

integrate((exp(x)+exp(2*x)+exp(3*x))/exp(4*x),x, algorithm="maxima")

[Out]

-e^(-x) - 1/2*e^(-2*x) - 1/3*e^(-3*x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int e^{-4 x} \left (e^x+e^{2 x}+e^{3 x}\right ) \, dx=-\frac {1}{6} \, {\left (6 \, e^{\left (2 \, x\right )} + 3 \, e^{x} + 2\right )} e^{\left (-3 \, x\right )} \]

[In]

integrate((exp(x)+exp(2*x)+exp(3*x))/exp(4*x),x, algorithm="giac")

[Out]

-1/6*(6*e^(2*x) + 3*e^x + 2)*e^(-3*x)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int e^{-4 x} \left (e^x+e^{2 x}+e^{3 x}\right ) \, dx=-\frac {{\mathrm {e}}^{-3\,x}\,\left (6\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^x+2\right )}{6} \]

[In]

int(exp(-4*x)*(exp(2*x) + exp(3*x) + exp(x)),x)

[Out]

-(exp(-3*x)*(6*exp(2*x) + 3*exp(x) + 2))/6