3.243 \(\int \cosh (x) \text{sech}(5 x) \, dx\)

Optimal. Leaf size=75 \[ \frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{5+2 \sqrt{5}} \tanh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{5-2 \sqrt{5}} \tanh (x)\right ) \]

[Out]

-(Sqrt[(5 - Sqrt[5])/2]*ArcTan[Sqrt[5 - 2*Sqrt[5]]*Tanh[x]])/5 + (Sqrt[(5 + Sqrt[5])/2]*ArcTan[Sqrt[5 + 2*Sqrt
[5]]*Tanh[x]])/5

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Rubi [A]  time = 0.125132, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {1166, 203} \[ \frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{5+2 \sqrt{5}} \tanh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{5-2 \sqrt{5}} \tanh (x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]*Sech[5*x],x]

[Out]

-(Sqrt[(5 - Sqrt[5])/2]*ArcTan[Sqrt[5 - 2*Sqrt[5]]*Tanh[x]])/5 + (Sqrt[(5 + Sqrt[5])/2]*ArcTan[Sqrt[5 + 2*Sqrt
[5]]*Tanh[x]])/5

Rule 1166

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

Rule 203

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

Rubi steps

\begin{align*} \int \cosh (x) \text{sech}(5 x) \, dx &=\operatorname{Subst}\left (\int \frac{1-x^2}{1+10 x^2+5 x^4} \, dx,x,\tanh (x)\right )\\ &=\frac{1}{2} \left (-1-\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{5+2 \sqrt{5}+5 x^2} \, dx,x,\tanh (x)\right )+\frac{1}{2} \left (-1+\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{5-2 \sqrt{5}+5 x^2} \, dx,x,\tanh (x)\right )\\ &=-\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{5-2 \sqrt{5}} \tanh (x)\right )+\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tan ^{-1}\left (\sqrt{5+2 \sqrt{5}} \tanh (x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0997007, size = 84, normalized size = 1.12 \[ \frac{\sqrt{5+\sqrt{5}} \tan ^{-1}\left (\frac{\left (5+\sqrt{5}\right ) \tanh (x)}{\sqrt{10-2 \sqrt{5}}}\right )+\sqrt{5-\sqrt{5}} \tan ^{-1}\left (\frac{\left (\sqrt{5}-5\right ) \tanh (x)}{\sqrt{2 \left (5+\sqrt{5}\right )}}\right )}{5 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]*Sech[5*x],x]

[Out]

(Sqrt[5 + Sqrt[5]]*ArcTan[((5 + Sqrt[5])*Tanh[x])/Sqrt[10 - 2*Sqrt[5]]] + Sqrt[5 - Sqrt[5]]*ArcTan[((-5 + Sqrt
[5])*Tanh[x])/Sqrt[2*(5 + Sqrt[5])]])/(5*Sqrt[2])

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Maple [C]  time = 0.037, size = 41, normalized size = 0.6 \begin{align*} 2\,\sum _{{\it \_R}={\it RootOf} \left ( 32000\,{{\it \_Z}}^{4}+400\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ( -4000\,{{\it \_R}}^{3}+200\,{{\it \_R}}^{2}+{{\rm e}^{2\,x}}-30\,{\it \_R}+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)*sech(5*x),x)

[Out]

2*sum(_R*ln(-4000*_R^3+200*_R^2+exp(2*x)-30*_R+1),_R=RootOf(32000*_Z^4+400*_Z^2+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{5} \arctan \left (\frac{\sqrt{5} + 4 \, e^{\left (-2 \, x\right )} - 1}{\sqrt{2 \, \sqrt{5} + 10}}\right )}{5 \, \sqrt{2 \, \sqrt{5} + 10}} - \frac{\sqrt{5} \arctan \left (-\frac{\sqrt{5} - 4 \, e^{\left (-2 \, x\right )} + 1}{\sqrt{-2 \, \sqrt{5} + 10}}\right )}{5 \, \sqrt{-2 \, \sqrt{5} + 10}} - \frac{\log \left (-{\left (\sqrt{5} + 1\right )} e^{\left (-2 \, x\right )} + 2 \, e^{\left (-4 \, x\right )} + 2\right )}{10 \,{\left (\sqrt{5} + 1\right )}} + \frac{\log \left ({\left (\sqrt{5} - 1\right )} e^{\left (-2 \, x\right )} + 2 \, e^{\left (-4 \, x\right )} + 2\right )}{10 \,{\left (\sqrt{5} - 1\right )}} - \frac{1}{5} \, \int \frac{{\left (e^{\left (7 \, x\right )} - 2 \, e^{\left (5 \, x\right )} - 2 \, e^{\left (3 \, x\right )} + e^{x}\right )} e^{x}}{e^{\left (8 \, x\right )} - e^{\left (6 \, x\right )} + e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )} + 1}\,{d x} + \frac{1}{10} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) - \frac{1}{10} \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sech(5*x),x, algorithm="maxima")

[Out]

1/5*sqrt(5)*arctan((sqrt(5) + 4*e^(-2*x) - 1)/sqrt(2*sqrt(5) + 10))/sqrt(2*sqrt(5) + 10) - 1/5*sqrt(5)*arctan(
-(sqrt(5) - 4*e^(-2*x) + 1)/sqrt(-2*sqrt(5) + 10))/sqrt(-2*sqrt(5) + 10) - 1/10*log(-(sqrt(5) + 1)*e^(-2*x) +
2*e^(-4*x) + 2)/(sqrt(5) + 1) + 1/10*log((sqrt(5) - 1)*e^(-2*x) + 2*e^(-4*x) + 2)/(sqrt(5) - 1) - 1/5*integrat
e((e^(7*x) - 2*e^(5*x) - 2*e^(3*x) + e^x)*e^x/(e^(8*x) - e^(6*x) + e^(4*x) - e^(2*x) + 1), x) + 1/10*log(e^(2*
x) + 1) - 1/10*log(e^(-2*x) + 1)

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Fricas [B]  time = 2.30668, size = 558, normalized size = 7.44 \begin{align*} -\frac{1}{5} \, \sqrt{2} \sqrt{\sqrt{5} + 5} \arctan \left (\frac{1}{40} \, \sqrt{5} \sqrt{2} \sqrt{-32 \,{\left (\sqrt{5} + 1\right )} e^{\left (2 \, x\right )} + 64 \, e^{\left (4 \, x\right )} + 64} \sqrt{\sqrt{5} + 5} - \frac{1}{20} \,{\left (4 \, \sqrt{5} \sqrt{2} e^{\left (2 \, x\right )} - \sqrt{5} \sqrt{2} - 5 \, \sqrt{2}\right )} \sqrt{\sqrt{5} + 5}\right ) + \frac{1}{5} \, \sqrt{2} \sqrt{-\sqrt{5} + 5} \arctan \left (-\frac{1}{20} \,{\left (4 \, \sqrt{5} \sqrt{2} e^{\left (2 \, x\right )} - \sqrt{5} \sqrt{2} + 5 \, \sqrt{2}\right )} \sqrt{-\sqrt{5} + 5} + \frac{1}{5} \, \sqrt{5} \sqrt{{\left (\sqrt{5} - 1\right )} e^{\left (2 \, x\right )} + 2 \, e^{\left (4 \, x\right )} + 2} \sqrt{-\sqrt{5} + 5}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sech(5*x),x, algorithm="fricas")

[Out]

-1/5*sqrt(2)*sqrt(sqrt(5) + 5)*arctan(1/40*sqrt(5)*sqrt(2)*sqrt(-32*(sqrt(5) + 1)*e^(2*x) + 64*e^(4*x) + 64)*s
qrt(sqrt(5) + 5) - 1/20*(4*sqrt(5)*sqrt(2)*e^(2*x) - sqrt(5)*sqrt(2) - 5*sqrt(2))*sqrt(sqrt(5) + 5)) + 1/5*sqr
t(2)*sqrt(-sqrt(5) + 5)*arctan(-1/20*(4*sqrt(5)*sqrt(2)*e^(2*x) - sqrt(5)*sqrt(2) + 5*sqrt(2))*sqrt(-sqrt(5) +
5) + 1/5*sqrt(5)*sqrt((sqrt(5) - 1)*e^(2*x) + 2*e^(4*x) + 2)*sqrt(-sqrt(5) + 5))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh{\left (x \right )} \operatorname{sech}{\left (5 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sech(5*x),x)

[Out]

Integral(cosh(x)*sech(5*x), x)

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Giac [A]  time = 1.17706, size = 92, normalized size = 1.23 \begin{align*} -\frac{1}{10} \, \sqrt{-2 \, \sqrt{5} + 10} \arctan \left (\frac{\sqrt{5} + 4 \, e^{\left (2 \, x\right )} - 1}{\sqrt{2 \, \sqrt{5} + 10}}\right ) + \frac{1}{10} \, \sqrt{2 \, \sqrt{5} + 10} \arctan \left (-\frac{\sqrt{5} - 4 \, e^{\left (2 \, x\right )} + 1}{\sqrt{-2 \, \sqrt{5} + 10}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sech(5*x),x, algorithm="giac")

[Out]

-1/10*sqrt(-2*sqrt(5) + 10)*arctan((sqrt(5) + 4*e^(2*x) - 1)/sqrt(2*sqrt(5) + 10)) + 1/10*sqrt(2*sqrt(5) + 10)
*arctan(-(sqrt(5) - 4*e^(2*x) + 1)/sqrt(-2*sqrt(5) + 10))