\(\int e^x \sec (e^x) \tan (e^x) \, dx\) [71]

Optimal result

Integrand size = 12, antiderivative size = 4 \[ \int e^x \sec \left (e^x\right ) \tan \left (e^x\right ) \, dx=\sec \left (e^x\right ) \]

[Out]

sec(exp(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2320, 2686, 8} \[ \int e^x \sec \left (e^x\right ) \tan \left (e^x\right ) \, dx=\sec \left (e^x\right ) \]

[In]

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

[Out]

Sec[E^x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int e^x \sec \left (e^x\right ) \tan \left (e^x\right ) \, dx=\sec \left (e^x\right ) \]

[In]

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

[Out]

Sec[E^x]

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\sec \left ({\mathrm e}^{x}\right )\)	\(4\)
default	\(\sec \left ({\mathrm e}^{x}\right )\)	\(4\)
risch	\(\frac {2 \,{\mathrm e}^{i {\mathrm e}^{x}}}{{\mathrm e}^{2 i {\mathrm e}^{x}}+1}\)	\(19\)

[In]

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

[Out]

sec(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.25 \[ \int e^x \sec \left (e^x\right ) \tan \left (e^x\right ) \, dx=\frac {1}{\cos \left (e^{x}\right )} \]

[In]

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

[Out]

1/cos(e^x)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int e^x \sec \left (e^x\right ) \tan \left (e^x\right ) \, dx=\sec {\left (e^{x} \right )} \]

[In]

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

[Out]

sec(exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.25 \[ \int e^x \sec \left (e^x\right ) \tan \left (e^x\right ) \, dx=\frac {1}{\cos \left (e^{x}\right )} \]

[In]

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

[Out]

1/cos(e^x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.25 \[ \int e^x \sec \left (e^x\right ) \tan \left (e^x\right ) \, dx=\frac {1}{\cos \left (e^{x}\right )} \]

[In]

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

[Out]

1/cos(e^x)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.25 \[ \int e^x \sec \left (e^x\right ) \tan \left (e^x\right ) \, dx=\frac {1}{\cos \left ({\mathrm {e}}^x\right )} \]

[In]

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

[Out]

1/cos(exp(x))