\(\int e^{-2 x} \cos (e^{-2 x}) \, dx\) [66]

Optimal result

Integrand size = 12, antiderivative size = 10 \[ \int e^{-2 x} \cos \left (e^{-2 x}\right ) \, dx=-\frac {1}{2} \sin \left (e^{-2 x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2320, 2717} \[ \int e^{-2 x} \cos \left (e^{-2 x}\right ) \, dx=-\frac {1}{2} \sin \left (e^{-2 x}\right ) \]

[In]

Int[Cos[E^(-2*x)]/E^(2*x),x]

[Out]

-1/2*Sin[E^(-2*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]]]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{2} \text {Subst}\left (\int \cos (x) \, dx,x,e^{-2 x}\right )\right ) \\ & = -\frac {1}{2} \sin \left (e^{-2 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int e^{-2 x} \cos \left (e^{-2 x}\right ) \, dx=-\frac {1}{2} \sin \left (e^{-2 x}\right ) \]

[In]

Integrate[Cos[E^(-2*x)]/E^(2*x),x]

[Out]

-1/2*Sin[E^(-2*x)]

Maple [A] (verified)

Time = 1.50 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {\sin \left ({\mathrm e}^{-2 x}\right )}{2}\)	\(8\)
risch	\(-\frac {\sin \left ({\mathrm e}^{-2 x}\right )}{2}\)	\(8\)
norman	\(-\frac {\tan \left (\frac {{\mathrm e}^{-2 x}}{2}\right )}{1+\tan \left (\frac {{\mathrm e}^{-2 x}}{2}\right )^{2}}\)	\(23\)

[In]

int(cos(exp(-2*x))/exp(2*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int e^{-2 x} \cos \left (e^{-2 x}\right ) \, dx=-\frac {1}{2} \, \sin \left (e^{\left (-2 \, x\right )}\right ) \]

[In]

integrate(cos(exp(-2*x))/exp(2*x),x, algorithm="fricas")

[Out]

-1/2*sin(e^(-2*x))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int e^{-2 x} \cos \left (e^{-2 x}\right ) \, dx=- \frac {\sin {\left (e^{- 2 x} \right )}}{2} \]

[In]

integrate(cos(exp(-2*x))/exp(2*x),x)

[Out]

-sin(exp(-2*x))/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int e^{-2 x} \cos \left (e^{-2 x}\right ) \, dx=-\frac {1}{2} \, \sin \left (e^{\left (-2 \, x\right )}\right ) \]

[In]

integrate(cos(exp(-2*x))/exp(2*x),x, algorithm="maxima")

[Out]

-1/2*sin(e^(-2*x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int e^{-2 x} \cos \left (e^{-2 x}\right ) \, dx=-\frac {1}{2} \, \sin \left (e^{\left (-2 \, x\right )}\right ) \]

[In]

integrate(cos(exp(-2*x))/exp(2*x),x, algorithm="giac")

[Out]

-1/2*sin(e^(-2*x))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int e^{-2 x} \cos \left (e^{-2 x}\right ) \, dx=-\frac {\sin \left ({\mathrm {e}}^{-2\,x}\right )}{2} \]

[In]

int(exp(-2*x)*cos(exp(-2*x)),x)

[Out]

-sin(exp(-2*x))/2