\(\int e^x \csc (e^x) \sec (e^x) \, dx\) [63]

Optimal result

Integrand size = 12, antiderivative size = 5 \[ \int e^x \csc \left (e^x\right ) \sec \left (e^x\right ) \, dx=\log \left (\tan \left (e^x\right )\right ) \]

[Out]

ln(tan(exp(x)))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 5, 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, 2700, 29} \[ \int e^x \csc \left (e^x\right ) \sec \left (e^x\right ) \, dx=\log \left (\tan \left (e^x\right )\right ) \]

[In]

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

[Out]

Log[Tan[E^x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(21\) vs. \(2(5)=10\).

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 4.20 \[ \int e^x \csc \left (e^x\right ) \sec \left (e^x\right ) \, dx=2 \left (-\frac {1}{2} \log \left (\cos \left (e^x\right )\right )+\frac {1}{2} \log \left (\sin \left (e^x\right )\right )\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\ln \left (\tan \left ({\mathrm e}^{x}\right )\right )\)	\(5\)
default	\(\ln \left (\tan \left ({\mathrm e}^{x}\right )\right )\)	\(5\)
risch	\(-\ln \left ({\mathrm e}^{2 i {\mathrm e}^{x}}+1\right )+\ln \left ({\mathrm e}^{2 i {\mathrm e}^{x}}-1\right )\)	\(22\)
norman	\(-\ln \left (\tan \left (\frac {{\mathrm e}^{x}}{2}\right )-1\right )-\ln \left (\tan \left (\frac {{\mathrm e}^{x}}{2}\right )+1\right )+\ln \left (\tan \left (\frac {{\mathrm e}^{x}}{2}\right )\right )\)	\(28\)
parallelrisch	\(-\ln \left (\tan \left (\frac {{\mathrm e}^{x}}{2}\right )-1\right )-\ln \left (\tan \left (\frac {{\mathrm e}^{x}}{2}\right )+1\right )+\ln \left (\tan \left (\frac {{\mathrm e}^{x}}{2}\right )\right )\)	\(28\)

[In]

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

[Out]

ln(tan(exp(x)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 21 vs. \(2 (4) = 8\).

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 4.20 \[ \int e^x \csc \left (e^x\right ) \sec \left (e^x\right ) \, dx=-\frac {1}{2} \, \log \left (\cos \left (e^{x}\right )^{2}\right ) + \frac {1}{2} \, \log \left (-\frac {1}{4} \, \cos \left (e^{x}\right )^{2} + \frac {1}{4}\right ) \]

[In]

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

[Out]

-1/2*log(cos(e^x)^2) + 1/2*log(-1/4*cos(e^x)^2 + 1/4)

Sympy [F]

\[ \int e^x \csc \left (e^x\right ) \sec \left (e^x\right ) \, dx=\int e^{x} \csc {\left (e^{x} \right )} \sec {\left (e^{x} \right )}\, dx \]

[In]

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

[Out]

Integral(exp(x)*csc(exp(x))*sec(exp(x)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 19 vs. \(2 (4) = 8\).

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 3.80 \[ \int e^x \csc \left (e^x\right ) \sec \left (e^x\right ) \, dx=-\frac {1}{2} \, \log \left (\sin \left (e^{x}\right )^{2} - 1\right ) + \frac {1}{2} \, \log \left (\sin \left (e^{x}\right )^{2}\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (4) = 8\).

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 3.40 \[ \int e^x \csc \left (e^x\right ) \sec \left (e^x\right ) \, dx=-\frac {1}{2} \, \log \left ({\left | \sin \left (e^{x}\right )^{2} - 1 \right |}\right ) + \log \left ({\left | \sin \left (e^{x}\right ) \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 43, normalized size of antiderivative = 8.60 \[ \int e^x \csc \left (e^x\right ) \sec \left (e^x\right ) \, dx=-\ln \left (-16\,{\mathrm {e}}^{2\,x}-16\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{{\mathrm {e}}^x\,2{}\mathrm {i}}\right )+\ln \left (16\,{\mathrm {e}}^{2\,x}-16\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{{\mathrm {e}}^x\,2{}\mathrm {i}}\right ) \]

[In]

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

[Out]

log(16*exp(2*x) - 16*exp(2*x)*exp(exp(x)*2i)) - log(- 16*exp(2*x) - 16*exp(2*x)*exp(exp(x)*2i))