\(\int e^x \tan (e^x) \, dx\) [69]

Optimal result

Integrand size = 8, antiderivative size = 7 \[ \int e^x \tan \left (e^x\right ) \, dx=-\log \left (\cos \left (e^x\right )\right ) \]

[Out]

-ln(cos(exp(x)))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2320, 3556} \[ \int e^x \tan \left (e^x\right ) \, dx=-\log \left (\cos \left (e^x\right )\right ) \]

[In]

Int[E^x*Tan[E^x],x]

[Out]

-Log[Cos[E^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 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}& = \text {Subst}\left (\int \tan (x) \, dx,x,e^x\right ) \\ & = -\log \left (\cos \left (e^x\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int e^x \tan \left (e^x\right ) \, dx=-\log \left (\cos \left (e^x\right )\right ) \]

[In]

Integrate[E^x*Tan[E^x],x]

[Out]

-Log[Cos[E^x]]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(-\ln \left (\cos \left ({\mathrm e}^{x}\right )\right )\)	\(7\)
default	\(-\ln \left (\cos \left ({\mathrm e}^{x}\right )\right )\)	\(7\)
norman	\(\frac {\ln \left (1+\tan \left ({\mathrm e}^{x}\right )^{2}\right )}{2}\)	\(11\)
parallelrisch	\(\frac {\ln \left (1+\tan \left ({\mathrm e}^{x}\right )^{2}\right )}{2}\)	\(11\)
risch	\(i {\mathrm e}^{x}-\ln \left ({\mathrm e}^{2 i {\mathrm e}^{x}}+1\right )\)	\(18\)

[In]

int(exp(x)*tan(exp(x)),x,method=_RETURNVERBOSE)

[Out]

-ln(cos(exp(x)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.71 \[ \int e^x \tan \left (e^x\right ) \, dx=-\frac {1}{2} \, \log \left (\frac {1}{\tan \left (e^{x}\right )^{2} + 1}\right ) \]

[In]

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

[Out]

-1/2*log(1/(tan(e^x)^2 + 1))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.43 \[ \int e^x \tan \left (e^x\right ) \, dx=\frac {\log {\left (\tan ^{2}{\left (e^{x} \right )} + 1 \right )}}{2} \]

[In]

integrate(exp(x)*tan(exp(x)),x)

[Out]

log(tan(exp(x))**2 + 1)/2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 4, normalized size of antiderivative = 0.57 \[ \int e^x \tan \left (e^x\right ) \, dx=\log \left (\sec \left (e^{x}\right )\right ) \]

[In]

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

[Out]

log(sec(e^x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int e^x \tan \left (e^x\right ) \, dx=-\log \left ({\left | \cos \left (e^{x}\right ) \right |}\right ) \]

[In]

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

[Out]

-log(abs(cos(e^x)))

Mupad [B] (verification not implemented)

Time = 27.55 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.43 \[ \int e^x \tan \left (e^x\right ) \, dx=\frac {\ln \left ({\mathrm {tan}\left ({\mathrm {e}}^x\right )}^2+1\right )}{2} \]

[In]

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

[Out]

log(tan(exp(x))^2 + 1)/2