3.755 \(\int e^{-2 x} \tan (e^{-2 x}) \, dx\)

Optimal. Leaf size=11 \[ \frac{1}{2} \log \left (\cos \left (e^{-2 x}\right )\right ) \]

[Out]

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

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Rubi [A]  time = 0.0117125, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2282, 3475} \[ \frac{1}{2} \log \left (\cos \left (e^{-2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]]

Rule 3475

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*} \int e^{-2 x} \tan \left (e^{-2 x}\right ) \, dx &=-\left (\frac{1}{2} \operatorname{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]  time = 0.0080568, size = 11, normalized size = 1. \[ \frac{1}{2} \log \left (\cos \left (e^{-2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 9, normalized size = 0.8 \begin{align*}{\frac{\ln \left ( \cos \left ({{\rm e}^{-2\,x}} \right ) \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.958688, size = 11, normalized size = 1. \begin{align*} -\frac{1}{2} \, \log \left (\sec \left (e^{\left (-2 \, x\right )}\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.06475, size = 46, normalized size = 4.18 \begin{align*} \frac{1}{4} \, \log \left (\frac{1}{\tan \left (e^{\left (-2 \, x\right )}\right )^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(1/(tan(e^(-2*x))^2 + 1))

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Sympy [A]  time = 0.37184, size = 15, normalized size = 1.36 \begin{align*} - \frac{\log{\left (\tan ^{2}{\left (e^{- 2 x} \right )} + 1 \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(tan(exp(-2*x))**2 + 1)/4

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Giac [A]  time = 1.07857, size = 12, normalized size = 1.09 \begin{align*} \frac{1}{2} \, \log \left ({\left | \cos \left (e^{\left (-2 \, x\right )}\right ) \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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