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]

-1/10*arctan((5-2*5^(1/2))^(1/2)*tanh(x))*(10-2*5^(1/2))^(1/2)+1/10*arctan((5+2*5^(1/2))^(1/2)*tanh(x))*(10+2*
5^(1/2))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 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])

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]

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 = 0.11, 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])

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 173, normalized size = 2.31 \[ -\frac {1}{5} \, \sqrt {2} \sqrt {\sqrt {5} + 5} \arctan \left (\frac {1}{40} \, \sqrt {5} \sqrt {2} \sqrt {-32 \, {\left (\sqrt {5} + 1\right )} e^{\left (2 \, x\right )} + 64 \, e^{\left (4 \, x\right )} + 64} \sqrt {\sqrt {5} + 5} - \frac {1}{20} \, {\left (4 \, \sqrt {5} \sqrt {2} e^{\left (2 \, x\right )} - \sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} \sqrt {\sqrt {5} + 5}\right ) + \frac {1}{5} \, \sqrt {2} \sqrt {-\sqrt {5} + 5} \arctan \left (-\frac {1}{20} \, {\left (4 \, \sqrt {5} \sqrt {2} e^{\left (2 \, x\right )} - \sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {-\sqrt {5} + 5} + \frac {1}{5} \, \sqrt {5} \sqrt {{\left (\sqrt {5} - 1\right )} e^{\left (2 \, x\right )} + 2 \, e^{\left (4 \, x\right )} + 2} \sqrt {-\sqrt {5} + 5}\right ) \]

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))

________________________________________________________________________________________

giac [A]  time = 0.16, size = 68, normalized size = 0.91 \[ -\frac {1}{10} \, \sqrt {-2 \, \sqrt {5} + 10} \arctan \left (\frac {\sqrt {5} + 4 \, e^{\left (2 \, x\right )} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right ) + \frac {1}{10} \, \sqrt {2 \, \sqrt {5} + 10} \arctan \left (-\frac {\sqrt {5} - 4 \, e^{\left (2 \, x\right )} + 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right ) \]

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))

________________________________________________________________________________________

maple [C]  time = 0.28, size = 41, normalized size = 0.55 \[ 2 \left (\munderset {\textit {\_R} =\RootOf \left (32000 \textit {\_Z}^{4}+400 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-4000 \textit {\_R}^{3}+200 \textit {\_R}^{2}+{\mathrm e}^{2 x}-30 \textit {\_R} +1\right )\right ) \]

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))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\sqrt {5} \arctan \left (\frac {\sqrt {5} + 4 \, e^{\left (-2 \, x\right )} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right )}{5 \, \sqrt {2 \, \sqrt {5} + 10}} - \frac {\sqrt {5} \arctan \left (-\frac {\sqrt {5} - 4 \, e^{\left (-2 \, x\right )} + 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right )}{5 \, \sqrt {-2 \, \sqrt {5} + 10}} - \frac {\log \left (-{\left (\sqrt {5} + 1\right )} e^{\left (-2 \, x\right )} + 2 \, e^{\left (-4 \, x\right )} + 2\right )}{10 \, {\left (\sqrt {5} + 1\right )}} + \frac {\log \left ({\left (\sqrt {5} - 1\right )} e^{\left (-2 \, x\right )} + 2 \, e^{\left (-4 \, x\right )} + 2\right )}{10 \, {\left (\sqrt {5} - 1\right )}} - \frac {1}{5} \, \int \frac {{\left (e^{\left (7 \, x\right )} - 2 \, e^{\left (5 \, x\right )} - 2 \, e^{\left (3 \, x\right )} + e^{x}\right )} e^{x}}{e^{\left (8 \, x\right )} - e^{\left (6 \, x\right )} + e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )} + 1}\,{d x} + \frac {1}{10} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{10} \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

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)

________________________________________________________________________________________

mupad [B]  time = 4.15, size = 297, normalized size = 3.96 \[ \ln \left (1-\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (4\,{\mathrm {e}}^{2\,x}+\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (48\,{\mathrm {e}}^{2\,x}+\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (360\,{\mathrm {e}}^{2\,x}-360\right )-72\right )-8\right )-{\mathrm {e}}^{2\,x}\right )\,\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}-\ln \left (\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (4\,{\mathrm {e}}^{2\,x}+\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (360\,{\mathrm {e}}^{2\,x}-360\right )-48\,{\mathrm {e}}^{2\,x}+72\right )-8\right )-{\mathrm {e}}^{2\,x}+1\right )\,\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}-\ln \left (\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (4\,{\mathrm {e}}^{2\,x}+\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (360\,{\mathrm {e}}^{2\,x}-360\right )-48\,{\mathrm {e}}^{2\,x}+72\right )-8\right )-{\mathrm {e}}^{2\,x}+1\right )\,\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}+\ln \left (1-\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (4\,{\mathrm {e}}^{2\,x}+\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (48\,{\mathrm {e}}^{2\,x}+\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (360\,{\mathrm {e}}^{2\,x}-360\right )-72\right )-8\right )-{\mathrm {e}}^{2\,x}\right )\,\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1 - (- 5^(1/2)/200 - 1/40)^(1/2)*(4*exp(2*x) + (- 5^(1/2)/200 - 1/40)^(1/2)*(48*exp(2*x) + (- 5^(1/2)/200
- 1/40)^(1/2)*(360*exp(2*x) - 360) - 72) - 8) - exp(2*x))*(- 5^(1/2)/200 - 1/40)^(1/2) - log((5^(1/2)/200 - 1/
40)^(1/2)*(4*exp(2*x) + (5^(1/2)/200 - 1/40)^(1/2)*((5^(1/2)/200 - 1/40)^(1/2)*(360*exp(2*x) - 360) - 48*exp(2
*x) + 72) - 8) - exp(2*x) + 1)*(5^(1/2)/200 - 1/40)^(1/2) - log((- 5^(1/2)/200 - 1/40)^(1/2)*(4*exp(2*x) + (-
5^(1/2)/200 - 1/40)^(1/2)*((- 5^(1/2)/200 - 1/40)^(1/2)*(360*exp(2*x) - 360) - 48*exp(2*x) + 72) - 8) - exp(2*
x) + 1)*(- 5^(1/2)/200 - 1/40)^(1/2) + log(1 - (5^(1/2)/200 - 1/40)^(1/2)*(4*exp(2*x) + (5^(1/2)/200 - 1/40)^(
1/2)*(48*exp(2*x) + (5^(1/2)/200 - 1/40)^(1/2)*(360*exp(2*x) - 360) - 72) - 8) - exp(2*x))*(5^(1/2)/200 - 1/40
)^(1/2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cosh {\relax (x )} \operatorname {sech}{\left (5 x \right )}\, dx \]

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)

________________________________________________________________________________________