3.283 \(\int e^x \text{sech}(4 x) \, dx\)

Optimal. Leaf size=371 \[ -\frac{\log \left (-\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\log \left (\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\log \left (-\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\log \left (\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}} \]

[Out]

ArcTan[(Sqrt[2 - Sqrt[2]] - 2*E^x)/Sqrt[2 + Sqrt[2]]]/(2*Sqrt[2*(2 + Sqrt[2])]) - ArcTan[(Sqrt[2 + Sqrt[2]] -
2*E^x)/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 - Sqrt[2])]) - ArcTan[(Sqrt[2 - Sqrt[2]] + 2*E^x)/Sqrt[2 + Sqrt[2]]]/(2
*Sqrt[2*(2 + Sqrt[2])]) + ArcTan[(Sqrt[2 + Sqrt[2]] + 2*E^x)/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 - Sqrt[2])]) - Lo
g[1 - Sqrt[2 - Sqrt[2]]*E^x + E^(2*x)]/(4*Sqrt[2*(2 - Sqrt[2])]) + Log[1 + Sqrt[2 - Sqrt[2]]*E^x + E^(2*x)]/(4
*Sqrt[2*(2 - Sqrt[2])]) + Log[1 - Sqrt[2 + Sqrt[2]]*E^x + E^(2*x)]/(4*Sqrt[2*(2 + Sqrt[2])]) - Log[1 + Sqrt[2
+ Sqrt[2]]*E^x + E^(2*x)]/(4*Sqrt[2*(2 + Sqrt[2])])

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Rubi [A]  time = 0.321078, antiderivative size = 371, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 9, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.125, Rules used = {2282, 12, 299, 1127, 1161, 618, 204, 1164, 628} \[ -\frac{\log \left (-\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\log \left (\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\log \left (-\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\log \left (\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sech[4*x],x]

[Out]

ArcTan[(Sqrt[2 - Sqrt[2]] - 2*E^x)/Sqrt[2 + Sqrt[2]]]/(2*Sqrt[2*(2 + Sqrt[2])]) - ArcTan[(Sqrt[2 + Sqrt[2]] -
2*E^x)/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 - Sqrt[2])]) - ArcTan[(Sqrt[2 - Sqrt[2]] + 2*E^x)/Sqrt[2 + Sqrt[2]]]/(2
*Sqrt[2*(2 + Sqrt[2])]) + ArcTan[(Sqrt[2 + Sqrt[2]] + 2*E^x)/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 - Sqrt[2])]) - Lo
g[1 - Sqrt[2 - Sqrt[2]]*E^x + E^(2*x)]/(4*Sqrt[2*(2 - Sqrt[2])]) + Log[1 + Sqrt[2 - Sqrt[2]]*E^x + E^(2*x)]/(4
*Sqrt[2*(2 - Sqrt[2])]) + Log[1 - Sqrt[2 + Sqrt[2]]*E^x + E^(2*x)]/(4*Sqrt[2*(2 + Sqrt[2])]) - Log[1 + Sqrt[2
+ Sqrt[2]]*E^x + E^(2*x)]/(4*Sqrt[2*(2 + Sqrt[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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 299

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[Rt[a/b, 4]], s = Denominator[Rt[a/b,
4]]}, Dist[s^3/(2*Sqrt[2]*b*r), Int[x^(m - n/4)/(r^2 - Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x] - Dist[s^3/
(2*Sqrt[2]*b*r), Int[x^(m - n/4)/(r^2 + Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x]] /; FreeQ[{a, b}, x] && IGt
Q[n/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && GtQ[a/b, 0]

Rule 1127

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, Dist[1/2, Int[(q + x^2)/(
a + b*x^2 + c*x^4), x], x] - Dist[1/2, Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && Lt
Q[b^2 - 4*a*c, 0] && PosQ[a*c]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1164

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int e^x \text{sech}(4 x) \, dx &=\operatorname{Subst}\left (\int \frac{2 x^4}{1+x^8} \, dx,x,e^x\right )\\ &=2 \operatorname{Subst}\left (\int \frac{x^4}{1+x^8} \, dx,x,e^x\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{1-\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{\sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{x^2}{1+\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{\sqrt{2}}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{1-\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{2 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{1-\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{2 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{1+\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{2 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{1+\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{2 \sqrt{2}}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{4 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{4 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{4 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{4 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2-\sqrt{2}}+2 x}{-1-\sqrt{2-\sqrt{2}} x-x^2} \, dx,x,e^x\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2-\sqrt{2}}-2 x}{-1+\sqrt{2-\sqrt{2}} x-x^2} \, dx,x,e^x\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2+\sqrt{2}}+2 x}{-1-\sqrt{2+\sqrt{2}} x-x^2} \, dx,x,e^x\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2+\sqrt{2}}-2 x}{-1+\sqrt{2+\sqrt{2}} x-x^2} \, dx,x,e^x\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}\\ &=-\frac{\log \left (1-\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\log \left (1+\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\log \left (1-\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\log \left (1+\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-2-\sqrt{2}-x^2} \, dx,x,-\sqrt{2-\sqrt{2}}+2 e^x\right )}{2 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-2-\sqrt{2}-x^2} \, dx,x,\sqrt{2-\sqrt{2}}+2 e^x\right )}{2 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-2+\sqrt{2}-x^2} \, dx,x,-\sqrt{2+\sqrt{2}}+2 e^x\right )}{2 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-2+\sqrt{2}-x^2} \, dx,x,\sqrt{2+\sqrt{2}}+2 e^x\right )}{2 \sqrt{2}}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}+2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}+2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\log \left (1-\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\log \left (1+\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )}{4 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\log \left (1-\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\log \left (1+\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )}{4 \sqrt{2 \left (2+\sqrt{2}\right )}}\\ \end{align*}

Mathematica [C]  time = 0.0101792, size = 24, normalized size = 0.06 \[ \frac{2}{5} e^{5 x} \, _2F_1\left (\frac{5}{8},1;\frac{13}{8};-e^{8 x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sech[4*x],x]

[Out]

(2*E^(5*x)*Hypergeometric2F1[5/8, 1, 13/8, -E^(8*x)])/5

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Maple [C]  time = 0.036, size = 25, normalized size = 0.1 \begin{align*} 2\,\sum _{{\it \_R}={\it RootOf} \left ( 16777216\,{{\it \_Z}}^{8}+1 \right ) }{\it \_R}\,\ln \left ( -32768\,{{\it \_R}}^{5}+{{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*sech(4*x),x)

[Out]

2*sum(_R*ln(-32768*_R^5+exp(x)),_R=RootOf(16777216*_Z^8+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{x} \operatorname{sech}\left (4 \, x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(4*x),x, algorithm="maxima")

[Out]

integrate(e^x*sech(4*x), x)

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Fricas [B]  time = 2.21957, size = 3322, normalized size = 8.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(4*x),x, algorithm="fricas")

[Out]

1/8*(sqrt(2)*sqrt(sqrt(2) + 2) - sqrt(2)*sqrt(-sqrt(2) + 2))*arctan(-(2*sqrt(2)*e^x - sqrt(2)*sqrt(2*sqrt(2)*s
qrt(sqrt(2) + 2)*e^x - 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) + sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) +
2))/(sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))) + 1/8*(sqrt(2)*sqrt(sqrt(2) + 2) - sqrt(2)*sqrt(-sqrt(2) + 2))*
arctan(-(2*sqrt(2)*e^x - sqrt(2)*sqrt(-2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x + 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*
e^(2*x) + 4) - sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))/(sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))) + 1/8*(sqrt(2
)*sqrt(sqrt(2) + 2) + sqrt(2)*sqrt(-sqrt(2) + 2))*arctan((2*sqrt(2)*e^x - sqrt(2)*sqrt(2*sqrt(2)*sqrt(sqrt(2)
+ 2)*e^x + 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) + sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))/(sqrt(s
qrt(2) + 2) - sqrt(-sqrt(2) + 2))) + 1/8*(sqrt(2)*sqrt(sqrt(2) + 2) + sqrt(2)*sqrt(-sqrt(2) + 2))*arctan((2*sq
rt(2)*e^x - sqrt(2)*sqrt(-2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x - 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4)
- sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))/(sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))) - 1/32*(sqrt(2)*sqrt(sqrt(
2) + 2) - sqrt(2)*sqrt(-sqrt(2) + 2))*log(2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x + 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x +
4*e^(2*x) + 4) + 1/32*(sqrt(2)*sqrt(sqrt(2) + 2) + sqrt(2)*sqrt(-sqrt(2) + 2))*log(2*sqrt(2)*sqrt(sqrt(2) + 2
)*e^x - 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) - 1/32*(sqrt(2)*sqrt(sqrt(2) + 2) + sqrt(2)*sqrt(-sq
rt(2) + 2))*log(-2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x + 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) + 1/32*(s
qrt(2)*sqrt(sqrt(2) + 2) - sqrt(2)*sqrt(-sqrt(2) + 2))*log(-2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x - 2*sqrt(2)*sqrt(-
sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) - 1/4*sqrt(sqrt(2) + 2)*arctan((2*sqrt(sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1)
- sqrt(sqrt(2) + 2) - 2*e^x)/sqrt(-sqrt(2) + 2)) - 1/4*sqrt(sqrt(2) + 2)*arctan((2*sqrt(-sqrt(sqrt(2) + 2)*e^x
+ e^(2*x) + 1) + sqrt(sqrt(2) + 2) - 2*e^x)/sqrt(-sqrt(2) + 2)) + 1/4*sqrt(-sqrt(2) + 2)*arctan((2*sqrt(sqrt(
-sqrt(2) + 2)*e^x + e^(2*x) + 1) - sqrt(-sqrt(2) + 2) - 2*e^x)/sqrt(sqrt(2) + 2)) + 1/4*sqrt(-sqrt(2) + 2)*arc
tan((2*sqrt(-sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) + sqrt(-sqrt(2) + 2) - 2*e^x)/sqrt(sqrt(2) + 2)) - 1/16*sqr
t(-sqrt(2) + 2)*log(sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1) + 1/16*sqrt(-sqrt(2) + 2)*log(-sqrt(sqrt(2) + 2)*e^x
+ e^(2*x) + 1) + 1/16*sqrt(sqrt(2) + 2)*log(sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) - 1/16*sqrt(sqrt(2) + 2)*log
(-sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{x} \operatorname{sech}{\left (4 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(4*x),x)

[Out]

Integral(exp(x)*sech(4*x), x)

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Giac [A]  time = 1.35742, size = 336, normalized size = 0.91 \begin{align*} \frac{1}{4} \, \sqrt{\sqrt{2} + 2} \arctan \left (\frac{\sqrt{\sqrt{2} + 2} + 2 \, e^{x}}{\sqrt{-\sqrt{2} + 2}}\right ) + \frac{1}{4} \, \sqrt{\sqrt{2} + 2} \arctan \left (-\frac{\sqrt{\sqrt{2} + 2} - 2 \, e^{x}}{\sqrt{-\sqrt{2} + 2}}\right ) - \frac{1}{4} \, \sqrt{-\sqrt{2} + 2} \arctan \left (\frac{\sqrt{-\sqrt{2} + 2} + 2 \, e^{x}}{\sqrt{\sqrt{2} + 2}}\right ) - \frac{1}{4} \, \sqrt{-\sqrt{2} + 2} \arctan \left (-\frac{\sqrt{-\sqrt{2} + 2} - 2 \, e^{x}}{\sqrt{\sqrt{2} + 2}}\right ) - \frac{1}{8} \, \sqrt{-\sqrt{2} + 2} \log \left (\sqrt{\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{8} \, \sqrt{-\sqrt{2} + 2} \log \left (-\sqrt{\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{8} \, \sqrt{\sqrt{2} + 2} \log \left (\sqrt{-\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{1}{8} \, \sqrt{\sqrt{2} + 2} \log \left (-\sqrt{-\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(4*x),x, algorithm="giac")

[Out]

1/4*sqrt(sqrt(2) + 2)*arctan((sqrt(sqrt(2) + 2) + 2*e^x)/sqrt(-sqrt(2) + 2)) + 1/4*sqrt(sqrt(2) + 2)*arctan(-(
sqrt(sqrt(2) + 2) - 2*e^x)/sqrt(-sqrt(2) + 2)) - 1/4*sqrt(-sqrt(2) + 2)*arctan((sqrt(-sqrt(2) + 2) + 2*e^x)/sq
rt(sqrt(2) + 2)) - 1/4*sqrt(-sqrt(2) + 2)*arctan(-(sqrt(-sqrt(2) + 2) - 2*e^x)/sqrt(sqrt(2) + 2)) - 1/8*sqrt(-
sqrt(2) + 2)*log(sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1) + 1/8*sqrt(-sqrt(2) + 2)*log(-sqrt(sqrt(2) + 2)*e^x + e^
(2*x) + 1) + 1/8*sqrt(sqrt(2) + 2)*log(sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) - 1/8*sqrt(sqrt(2) + 2)*log(-sqrt
(-sqrt(2) + 2)*e^x + e^(2*x) + 1)