\(\int e^{-2 x} \tan (e^{-2 x}) \, dx\) [755]

Optimal result

Integrand size = 12, antiderivative size = 11 \[ \int e^{-2 x} \tan \left (e^{-2 x}\right ) \, dx=\frac {1}{2} \log \left (\cos \left (e^{-2 x}\right )\right ) \]

[Out]

1/2*ln(cos(exp(-2*x)))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, 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, 3556} \[ \int e^{-2 x} \tan \left (e^{-2 x}\right ) \, dx=\frac {1}{2} \log \left (\cos \left (e^{-2 x}\right )\right ) \]

[In]

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

[Out]

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

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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int e^{-2 x} \tan \left (e^{-2 x}\right ) \, dx=\frac {1}{2} \log \left (\cos \left (e^{-2 x}\right )\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\ln \left (\cos \left ({\mathrm e}^{-2 x}\right )\right )}{2}\)	\(9\)
norman	\(-\frac {\ln \left (1+\tan \left ({\mathrm e}^{-2 x}\right )^{2}\right )}{4}\)	\(13\)
risch	\(-\frac {i {\mathrm e}^{-2 x}}{2}+\frac {\ln \left ({\mathrm e}^{2 i {\mathrm e}^{-2 x}}+1\right )}{2}\)	\(22\)

[In]

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

[Out]

1/2*ln(cos(exp(-2*x)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int e^{-2 x} \tan \left (e^{-2 x}\right ) \, dx=\frac {1}{4} \, \log \left (\frac {1}{\tan \left (e^{\left (-2 \, x\right )}\right )^{2} + 1}\right ) \]

[In]

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

[Out]

1/4*log(1/(tan(e^(-2*x))^2 + 1))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.36 \[ \int e^{-2 x} \tan \left (e^{-2 x}\right ) \, dx=- \frac {\log {\left (\tan ^{2}{\left (e^{- 2 x} \right )} + 1 \right )}}{4} \]

[In]

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

[Out]

-log(tan(exp(-2*x))**2 + 1)/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int e^{-2 x} \tan \left (e^{-2 x}\right ) \, dx=-\frac {1}{2} \, \log \left (\sec \left (e^{\left (-2 \, x\right )}\right )\right ) \]

[In]

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

[Out]

-1/2*log(sec(e^(-2*x)))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int e^{-2 x} \tan \left (e^{-2 x}\right ) \, dx=\frac {1}{2} \, \log \left ({\left | \cos \left (e^{\left (-2 \, x\right )}\right ) \right |}\right ) \]

[In]

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

[Out]

1/2*log(abs(cos(e^(-2*x))))

Mupad [B] (verification not implemented)

Time = 27.40 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09 \[ \int e^{-2 x} \tan \left (e^{-2 x}\right ) \, dx=-\frac {\ln \left ({\mathrm {tan}\left ({\mathrm {e}}^{-2\,x}\right )}^2+1\right )}{4} \]

[In]

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

[Out]

-log(tan(exp(-2*x))^2 + 1)/4