3.279 \(\int e^x \text {sech}(3 x) \, dx\)

Optimal. Leaf size=55 \[ -\frac {1}{3} \log \left (e^{2 x}+1\right )+\frac {1}{6} \log \left (-e^{2 x}+e^{4 x}+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 e^{2 x}}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

-1/3*ln(exp(2*x)+1)+1/6*ln(1-exp(2*x)+exp(4*x))-1/3*arctan(1/3*(1-2*exp(2*x))*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {2282, 12, 275, 292, 31, 634, 618, 204, 628} \[ -\frac {1}{3} \log \left (e^{2 x}+1\right )+\frac {1}{6} \log \left (-e^{2 x}+e^{4 x}+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 e^{2 x}}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sech[3*x],x]

[Out]

-(ArcTan[(1 - 2*E^(2*x))/Sqrt[3]]/Sqrt[3]) - Log[1 + E^(2*x)]/3 + Log[1 - E^(2*x) + E^(4*x)]/6

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, 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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

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 634

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

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}(3 x) \, dx &=\operatorname {Subst}\left (\int \frac {2 x^3}{1+x^6} \, dx,x,e^x\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,e^{2 x}\right )\\ &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,e^{2 x}\right )\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,e^{2 x}\right )\\ &=-\frac {1}{3} \log \left (1+e^{2 x}\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,e^{2 x}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,e^{2 x}\right )\\ &=-\frac {1}{3} \log \left (1+e^{2 x}\right )+\frac {1}{6} \log \left (1-e^{2 x}+e^{4 x}\right )-\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 e^{2 x}\right )\\ &=\frac {\tan ^{-1}\left (\frac {-1+2 e^{2 x}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (1+e^{2 x}\right )+\frac {1}{6} \log \left (1-e^{2 x}+e^{4 x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 24, normalized size = 0.44 \[ \frac {1}{2} e^{4 x} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-e^{6 x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sech[3*x],x]

[Out]

(E^(4*x)*Hypergeometric2F1[2/3, 1, 5/3, -E^(6*x)])/2

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fricas [A]  time = 0.53, size = 83, normalized size = 1.51 \[ -\frac {1}{3} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} \cosh \relax (x) + 3 \, \sqrt {3} \sinh \relax (x)}{3 \, {\left (\cosh \relax (x) - \sinh \relax (x)\right )}}\right ) + \frac {1}{6} \, \log \left (\frac {2 \, \cosh \relax (x)^{2} + 2 \, \sinh \relax (x)^{2} - 1}{\cosh \relax (x)^{2} - 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2}}\right ) - \frac {1}{3} \, \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(3*x),x, algorithm="fricas")

[Out]

-1/3*sqrt(3)*arctan(-1/3*(sqrt(3)*cosh(x) + 3*sqrt(3)*sinh(x))/(cosh(x) - sinh(x))) + 1/6*log((2*cosh(x)^2 + 2
*sinh(x)^2 - 1)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) - 1/3*log(2*cosh(x)/(cosh(x) - sinh(x)))

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giac [A]  time = 0.11, size = 44, normalized size = 0.80 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (2 \, x\right )} - 1\right )}\right ) + \frac {1}{6} \, \log \left (e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{3} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(3*x),x, algorithm="giac")

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(2*x) - 1)) + 1/6*log(e^(4*x) - e^(2*x) + 1) - 1/3*log(e^(2*x) + 1)

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maple [C]  time = 0.14, size = 79, normalized size = 1.44 \[ -\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{3}+\frac {\ln \left ({\mathrm e}^{2 x}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{6}+\frac {i \ln \left ({\mathrm e}^{2 x}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{6}+\frac {\ln \left ({\mathrm e}^{2 x}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{6}-\frac {i \ln \left ({\mathrm e}^{2 x}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*sech(3*x),x)

[Out]

-1/3*ln(1+exp(2*x))+1/6*ln(exp(2*x)-1/2+1/2*I*3^(1/2))+1/6*I*ln(exp(2*x)-1/2+1/2*I*3^(1/2))*3^(1/2)+1/6*ln(exp
(2*x)-1/2-1/2*I*3^(1/2))-1/6*I*ln(exp(2*x)-1/2-1/2*I*3^(1/2))*3^(1/2)

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maxima [A]  time = 0.43, size = 71, normalized size = 1.29 \[ -\frac {1}{3} \, \sqrt {3} \arctan \left (\sqrt {3} + 2 \, e^{x}\right ) + \frac {1}{3} \, \sqrt {3} \arctan \left (-\sqrt {3} + 2 \, e^{x}\right ) + \frac {1}{6} \, \log \left (\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{6} \, \log \left (-\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{3} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(3*x),x, algorithm="maxima")

[Out]

-1/3*sqrt(3)*arctan(sqrt(3) + 2*e^x) + 1/3*sqrt(3)*arctan(-sqrt(3) + 2*e^x) + 1/6*log(sqrt(3)*e^x + e^(2*x) +
1) + 1/6*log(-sqrt(3)*e^x + e^(2*x) + 1) - 1/3*log(e^(2*x) + 1)

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mupad [B]  time = 1.04, size = 65, normalized size = 1.18 \[ -\frac {\ln \left (8\,{\mathrm {e}}^{2\,x}+8\right )}{3}-\ln \left (24\,{\mathrm {e}}^{2\,x}\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+8\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\ln \left (8-24\,{\mathrm {e}}^{2\,x}\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/cosh(3*x),x)

[Out]

log(8 - 24*exp(2*x)*((3^(1/2)*1i)/6 + 1/6))*((3^(1/2)*1i)/6 + 1/6) - log(24*exp(2*x)*((3^(1/2)*1i)/6 - 1/6) +
8)*((3^(1/2)*1i)/6 - 1/6) - log(8*exp(2*x) + 8)/3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(3*x),x)

[Out]

Integral(exp(x)*sech(3*x), x)

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