3.684 \(\int e^x \sec ^3(1-e^x) \, dx\)

Optimal. Leaf size=34 \[ -\frac{1}{2} \tanh ^{-1}\left (\sin \left (1-e^x\right )\right )-\frac{1}{2} \tan \left (1-e^x\right ) \sec \left (1-e^x\right ) \]

[Out]

-ArcTanh[Sin[1 - E^x]]/2 - (Sec[1 - E^x]*Tan[1 - E^x])/2

________________________________________________________________________________________

Rubi [A]  time = 0.0298723, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2282, 3768, 3770} \[ -\frac{1}{2} \tanh ^{-1}\left (\sin \left (1-e^x\right )\right )-\frac{1}{2} \tan \left (1-e^x\right ) \sec \left (1-e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sec[1 - E^x]^3,x]

[Out]

-ArcTanh[Sin[1 - E^x]]/2 - (Sec[1 - E^x]*Tan[1 - E^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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin{align*} \int e^x \sec ^3\left (1-e^x\right ) \, dx &=\operatorname{Subst}\left (\int \sec ^3(1-x) \, dx,x,e^x\right )\\ &=-\frac{1}{2} \sec \left (1-e^x\right ) \tan \left (1-e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \sec (1-x) \, dx,x,e^x\right )\\ &=-\frac{1}{2} \tanh ^{-1}\left (\sin \left (1-e^x\right )\right )-\frac{1}{2} \sec \left (1-e^x\right ) \tan \left (1-e^x\right )\\ \end{align*}

Mathematica [A]  time = 0.0163438, size = 34, normalized size = 1. \[ -\frac{1}{2} \tanh ^{-1}\left (\sin \left (1-e^x\right )\right )-\frac{1}{2} \tan \left (1-e^x\right ) \sec \left (1-e^x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sec[1 - E^x]^3,x]

[Out]

-ArcTanh[Sin[1 - E^x]]/2 - (Sec[1 - E^x]*Tan[1 - E^x])/2

________________________________________________________________________________________

Maple [A]  time = 0.305, size = 28, normalized size = 0.8 \begin{align*}{\frac{\sec \left ( -1+{{\rm e}^{x}} \right ) \tan \left ( -1+{{\rm e}^{x}} \right ) }{2}}+{\frac{\ln \left ( \sec \left ( -1+{{\rm e}^{x}} \right ) +\tan \left ( -1+{{\rm e}^{x}} \right ) \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*sec(-1+exp(x))^3,x)

[Out]

1/2*sec(-1+exp(x))*tan(-1+exp(x))+1/2*ln(sec(-1+exp(x))+tan(-1+exp(x)))

________________________________________________________________________________________

Maxima [A]  time = 0.964896, size = 53, normalized size = 1.56 \begin{align*} -\frac{\sin \left (e^{x} - 1\right )}{2 \,{\left (\sin \left (e^{x} - 1\right )^{2} - 1\right )}} + \frac{1}{4} \, \log \left (\sin \left (e^{x} - 1\right ) + 1\right ) - \frac{1}{4} \, \log \left (\sin \left (e^{x} - 1\right ) - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sec(-1+exp(x))^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 0.735651, size = 157, normalized size = 4.62 \begin{align*} \frac{\cos \left (e^{x} - 1\right )^{2} \log \left (\sin \left (e^{x} - 1\right ) + 1\right ) - \cos \left (e^{x} - 1\right )^{2} \log \left (-\sin \left (e^{x} - 1\right ) + 1\right ) + 2 \, \sin \left (e^{x} - 1\right )}{4 \, \cos \left (e^{x} - 1\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sec(-1+exp(x))^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{x} \sec ^{3}{\left (e^{x} - 1 \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sec(-1+exp(x))**3,x)

[Out]

Integral(exp(x)*sec(exp(x) - 1)**3, x)

________________________________________________________________________________________

Giac [A]  time = 1.20622, size = 55, normalized size = 1.62 \begin{align*} -\frac{\sin \left (e^{x} - 1\right )}{2 \,{\left (\sin \left (e^{x} - 1\right )^{2} - 1\right )}} + \frac{1}{4} \, \log \left (\sin \left (e^{x} - 1\right ) + 1\right ) - \frac{1}{4} \, \log \left (-\sin \left (e^{x} - 1\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sec(-1+exp(x))^3,x, algorithm="giac")

[Out]

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