3.275 \(\int e^x \text {sech}(2 x) \, dx\)

Optimal. Leaf size=92 \[ \frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} e^x+1\right )}{\sqrt {2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2282, 12, 297, 1162, 617, 204, 1165, 628} \[ \frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} e^x+1\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sech[2*x],x]

[Out]

-(ArcTan[1 - Sqrt[2]*E^x]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*E^x]/Sqrt[2] + Log[1 - Sqrt[2]*E^x + E^(2*x)]/(2*Sqrt[
2]) - Log[1 + Sqrt[2]*E^x + E^(2*x)]/(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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[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, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 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, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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

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Mathematica [C]  time = 0.01, size = 24, normalized size = 0.26 \[ \frac {2}{3} e^{3 x} \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-e^{4 x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sech[2*x],x]

[Out]

(2*E^(3*x)*Hypergeometric2F1[3/4, 1, 7/4, -E^(4*x)])/3

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fricas [A]  time = 0.59, size = 113, normalized size = 1.23 \[ -\sqrt {2} \arctan \left (-\sqrt {2} e^{x} + \sqrt {2} \sqrt {\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) - \sqrt {2} \arctan \left (-\sqrt {2} e^{x} + \frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(2*x),x, algorithm="fricas")

[Out]

-sqrt(2)*arctan(-sqrt(2)*e^x + sqrt(2)*sqrt(sqrt(2)*e^x + e^(2*x) + 1) - 1) - sqrt(2)*arctan(-sqrt(2)*e^x + 1/
2*sqrt(2)*sqrt(-4*sqrt(2)*e^x + 4*e^(2*x) + 4) + 1) - 1/4*sqrt(2)*log(4*sqrt(2)*e^x + 4*e^(2*x) + 4) + 1/4*sqr
t(2)*log(-4*sqrt(2)*e^x + 4*e^(2*x) + 4)

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giac [A]  time = 0.12, size = 76, normalized size = 0.83 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(2*x),x, algorithm="giac")

[Out]

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 1/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/4*s
qrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/4*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1)

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maple [C]  time = 0.14, size = 25, normalized size = 0.27 \[ 2 \left (\munderset {\textit {\_R} =\RootOf \left (256 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (64 \textit {\_R}^{3}+{\mathrm e}^{x}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*sech(2*x),x)

[Out]

2*sum(_R*ln(64*_R^3+exp(x)),_R=RootOf(256*_Z^4+1))

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maxima [A]  time = 0.42, size = 76, normalized size = 0.83 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(2*x),x, algorithm="maxima")

[Out]

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 1/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/4*s
qrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/4*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1)

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mupad [B]  time = 1.11, size = 77, normalized size = 0.84 \[ \sqrt {2}\,\ln \left (4+\sqrt {2}\,{\mathrm {e}}^x\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (4+\sqrt {2}\,{\mathrm {e}}^x\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (4+\sqrt {2}\,{\mathrm {e}}^x\,\left (2-2{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (4+\sqrt {2}\,{\mathrm {e}}^x\,\left (2+2{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/cosh(2*x),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \operatorname {sech}{\left (2 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(2*x),x)

[Out]

Integral(exp(x)*sech(2*x), x)

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