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]

-1/2*arctan((-2*exp(x)+(2+2^(1/2))^(1/2))/(2-2^(1/2))^(1/2))/(4-2*2^(1/2))^(1/2)+1/2*arctan((2*exp(x)+(2+2^(1/
2))^(1/2))/(2-2^(1/2))^(1/2))/(4-2*2^(1/2))^(1/2)-1/4*ln(1+exp(2*x)-exp(x)*(2-2^(1/2))^(1/2))/(4-2*2^(1/2))^(1
/2)+1/4*ln(1+exp(2*x)+exp(x)*(2-2^(1/2))^(1/2))/(4-2*2^(1/2))^(1/2)+1/2*arctan((-2*exp(x)+(2-2^(1/2))^(1/2))/(
2+2^(1/2))^(1/2))/(4+2*2^(1/2))^(1/2)-1/2*arctan((2*exp(x)+(2-2^(1/2))^(1/2))/(2+2^(1/2))^(1/2))/(4+2*2^(1/2))
^(1/2)+1/4*ln(1+exp(2*x)-exp(x)*(2+2^(1/2))^(1/2))/(4+2*2^(1/2))^(1/2)-1/4*ln(1+exp(2*x)+exp(x)*(2+2^(1/2))^(1
/2))/(4+2*2^(1/2))^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 371, normalized size of antiderivative = 1.00, 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 12

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

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 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 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 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]

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 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 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]]]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [B]  time = 0.61, size = 1087, normalized size = 2.93 \[ \text {result too large to display} \]

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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giac [A]  time = 0.21, size = 249, normalized size = 0.67 \[ \frac {1}{4} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {\sqrt {\sqrt {2} + 2} + 2 \, e^{x}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{4} \, \sqrt {\sqrt {2} + 2} \arctan \left (-\frac {\sqrt {\sqrt {2} + 2} - 2 \, e^{x}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{4} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {\sqrt {-\sqrt {2} + 2} + 2 \, e^{x}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{4} \, \sqrt {-\sqrt {2} + 2} \arctan \left (-\frac {\sqrt {-\sqrt {2} + 2} - 2 \, e^{x}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \log \left (\sqrt {\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \log \left (-\sqrt {\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \log \left (\sqrt {-\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \log \left (-\sqrt {-\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) \]

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)

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maple [C]  time = 0.14, size = 25, normalized size = 0.07 \[ 2 \left (\munderset {\textit {\_R} =\RootOf \left (16777216 \textit {\_Z}^{8}+1\right )}{\sum }\textit {\_R} \ln \left (-32768 \textit {\_R}^{5}+{\mathrm e}^{x}\right )\right ) \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \operatorname {sech}\left (4 \, x\right )\,{d x} \]

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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mupad [B]  time = 4.56, size = 479, normalized size = 1.29 \[ -\ln \left (32768\,{\mathrm {e}}^x\,{\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )}^3-512\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )+\ln \left (32768\,{\mathrm {e}}^x\,{\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )}^3+512\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )-\ln \left (32768\,{\mathrm {e}}^x\,{\left (-\frac {\sqrt {2-\sqrt {2}}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )}^3-512\right )\,\left (-\frac {\sqrt {2-\sqrt {2}}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )+\ln \left (32768\,{\mathrm {e}}^x\,{\left (-\frac {\sqrt {2-\sqrt {2}}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )}^3+512\right )\,\left (-\frac {\sqrt {2-\sqrt {2}}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )+\sqrt {2}\,\ln \left (-512+\sqrt {2}\,{\mathrm {e}}^x\,{\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )}^3\,\left (16384-16384{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (512+\sqrt {2}\,{\mathrm {e}}^x\,{\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )}^3\,\left (16384-16384{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-512+\sqrt {2}\,{\mathrm {e}}^x\,{\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )}^3\,\left (16384+16384{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (512+\sqrt {2}\,{\mathrm {e}}^x\,{\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )}^3\,\left (16384+16384{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/cosh(4*x),x)

[Out]

log(32768*exp(x)*((2^(1/2) + 2)^(1/2)/8 + ((2 - 2^(1/2))^(1/2)*1i)/8)^3 + 512)*((2^(1/2) + 2)^(1/2)/8 + ((2 -
2^(1/2))^(1/2)*1i)/8) - log(32768*exp(x)*((2^(1/2) + 2)^(1/2)/8 + ((2 - 2^(1/2))^(1/2)*1i)/8)^3 - 512)*((2^(1/
2) + 2)^(1/2)/8 + ((2 - 2^(1/2))^(1/2)*1i)/8) - log(32768*exp(x)*(((2^(1/2) + 2)^(1/2)*1i)/8 - (2 - 2^(1/2))^(
1/2)/8)^3 - 512)*(((2^(1/2) + 2)^(1/2)*1i)/8 - (2 - 2^(1/2))^(1/2)/8) + log(32768*exp(x)*(((2^(1/2) + 2)^(1/2)
*1i)/8 - (2 - 2^(1/2))^(1/2)/8)^3 + 512)*(((2^(1/2) + 2)^(1/2)*1i)/8 - (2 - 2^(1/2))^(1/2)/8) + 2^(1/2)*log(2^
(1/2)*exp(x)*((2^(1/2) + 2)^(1/2)/8 + ((2 - 2^(1/2))^(1/2)*1i)/8)^3*(16384 - 16384i) - 512)*((2^(1/2) + 2)^(1/
2)/8 + ((2 - 2^(1/2))^(1/2)*1i)/8)*(1/2 + 1i/2) - 2^(1/2)*log(2^(1/2)*exp(x)*((2^(1/2) + 2)^(1/2)/8 + ((2 - 2^
(1/2))^(1/2)*1i)/8)^3*(16384 - 16384i) + 512)*((2^(1/2) + 2)^(1/2)/8 + ((2 - 2^(1/2))^(1/2)*1i)/8)*(1/2 + 1i/2
) + 2^(1/2)*log(2^(1/2)*exp(x)*((2^(1/2) + 2)^(1/2)/8 + ((2 - 2^(1/2))^(1/2)*1i)/8)^3*(16384 + 16384i) - 512)*
((2^(1/2) + 2)^(1/2)/8 + ((2 - 2^(1/2))^(1/2)*1i)/8)*(1/2 - 1i/2) - 2^(1/2)*log(2^(1/2)*exp(x)*((2^(1/2) + 2)^
(1/2)/8 + ((2 - 2^(1/2))^(1/2)*1i)/8)^3*(16384 + 16384i) + 512)*((2^(1/2) + 2)^(1/2)/8 + ((2 - 2^(1/2))^(1/2)*
1i)/8)*(1/2 - 1i/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \operatorname {sech}{\left (4 x \right )}\, dx \]

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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