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]

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

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Rubi [A]  time = 0.03, antiderivative size = 34, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 1.00 \[ -\frac {1}{2} \tanh ^{-1}\left (\sin \left (1-e^x\right )\right )-\frac {1}{2} \tan \left (1-e^x\right ) \sec \left (1-e^x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sec[1 - E^x]^3,x]

[Out]

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

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fricas [B]  time = 0.43, size = 52, normalized size = 1.53 \[ \frac {\cos \left (e^{x} - 1\right )^{2} \log \left (\sin \left (e^{x} - 1\right ) + 1\right ) - \cos \left (e^{x} - 1\right )^{2} \log \left (-\sin \left (e^{x} - 1\right ) + 1\right ) + 2 \, \sin \left (e^{x} - 1\right )}{4 \, \cos \left (e^{x} - 1\right )^{2}} \]

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

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giac [A]  time = 0.20, size = 41, normalized size = 1.21 \[ -\frac {\sin \left (e^{x} - 1\right )}{2 \, {\left (\sin \left (e^{x} - 1\right )^{2} - 1\right )}} + \frac {1}{4} \, \log \left (\sin \left (e^{x} - 1\right ) + 1\right ) - \frac {1}{4} \, \log \left (-\sin \left (e^{x} - 1\right ) + 1\right ) \]

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)

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maple [A]  time = 0.37, size = 28, normalized size = 0.82 \[ \frac {\sec \left ({\mathrm e}^{x}-1\right ) \tan \left ({\mathrm e}^{x}-1\right )}{2}+\frac {\ln \left (\sec \left ({\mathrm e}^{x}-1\right )+\tan \left ({\mathrm e}^{x}-1\right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.09, size = 39, normalized size = 1.15 \[ -\frac {\sin \left (e^{x} - 1\right )}{2 \, {\left (\sin \left (e^{x} - 1\right )^{2} - 1\right )}} + \frac {1}{4} \, \log \left (\sin \left (e^{x} - 1\right ) + 1\right ) - \frac {1}{4} \, \log \left (\sin \left (e^{x} - 1\right ) - 1\right ) \]

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)

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mupad [B]  time = 5.53, size = 78, normalized size = 2.29 \[ -\mathrm {atan}\left ({\mathrm {e}}^{-\mathrm {i}}\,{\mathrm {e}}^{{\mathrm {e}}^x\,1{}\mathrm {i}}\right )\,1{}\mathrm {i}-\frac {{\mathrm {e}}^{-\mathrm {i}}\,{\mathrm {e}}^{{\mathrm {e}}^x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{{\mathrm {e}}^{-2{}\mathrm {i}}\,{\mathrm {e}}^{{\mathrm {e}}^x\,2{}\mathrm {i}}+1}+\frac {{\mathrm {e}}^{-\mathrm {i}}\,{\mathrm {e}}^{{\mathrm {e}}^x\,1{}\mathrm {i}}\,2{}\mathrm {i}}{2\,{\mathrm {e}}^{-2{}\mathrm {i}}\,{\mathrm {e}}^{{\mathrm {e}}^x\,2{}\mathrm {i}}+{\mathrm {e}}^{-4{}\mathrm {i}}\,{\mathrm {e}}^{{\mathrm {e}}^x\,4{}\mathrm {i}}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/cos(exp(x) - 1)^3,x)

[Out]

(exp(-1i)*exp(exp(x)*1i)*2i)/(2*exp(-2i)*exp(exp(x)*2i) + exp(-4i)*exp(exp(x)*4i) + 1) - (exp(-1i)*exp(exp(x)*
1i)*1i)/(exp(-2i)*exp(exp(x)*2i) + 1) - atan(exp(-1i)*exp(exp(x)*1i))*1i

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \sec ^{3}{\left (e^{x} - 1 \right )}\, dx \]

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)

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