3.242 \(\int \cosh (x) \text{sech}(4 x) \, dx\)

Optimal. Leaf size=71 \[ \frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}} \]

[Out]

ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 - Sqrt[2])]) - ArcTan[(2*Sinh[x])/Sqrt[2 + Sqrt[2]]]/(2*Sqr
t[2*(2 + Sqrt[2])])

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Rubi [A]  time = 0.0389463, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4356, 1093, 203} \[ \frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]*Sech[4*x],x]

[Out]

ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 - Sqrt[2])]) - ArcTan[(2*Sinh[x])/Sqrt[2 + Sqrt[2]]]/(2*Sqr
t[2*(2 + Sqrt[2])])

Rule 4356

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist[d/(b
*c), Subst[Int[SubstFor[1, Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a +
b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cos] || EqQ[F, cos])

Rule 1093

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 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}(4 x) \, dx &=\operatorname{Subst}\left (\int \frac{1}{1+8 x^2+8 x^4} \, dx,x,\sinh (x)\right )\\ &=\sqrt{2} \operatorname{Subst}\left (\int \frac{1}{4-2 \sqrt{2}+8 x^2} \, dx,x,\sinh (x)\right )-\sqrt{2} \operatorname{Subst}\left (\int \frac{1}{4+2 \sqrt{2}+8 x^2} \, dx,x,\sinh (x)\right )\\ &=\frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}\\ \end{align*}

Mathematica [A]  time = 0.0967995, size = 67, normalized size = 0.94 \[ \frac{1}{4} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2-\sqrt{2}}}\right )-\frac{\tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]*Sech[4*x],x]

[Out]

(Sqrt[2 + Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[2]]])/4 - ArcTan[(2*Sinh[x])/Sqrt[2 + Sqrt[2]]]/(2*Sqrt[2*
(2 + Sqrt[2])])

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Maple [C]  time = 0.056, size = 40, normalized size = 0.6 \begin{align*} 2\,\sum _{{\it \_R}={\it RootOf} \left ( 32768\,{{\it \_Z}}^{4}+512\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ({{\rm e}^{2\,x}}+ \left ( -4096\,{{\it \_R}}^{3}-48\,{\it \_R} \right ){{\rm e}^{x}}-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)*sech(4*x),x)

[Out]

2*sum(_R*ln(exp(2*x)+(-4096*_R^3-48*_R)*exp(x)-1),_R=RootOf(32768*_Z^4+512*_Z^2+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh \left (x\right ) \operatorname{sech}\left (4 \, x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sech(4*x),x, algorithm="maxima")

[Out]

integrate(cosh(x)*sech(4*x), x)

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Fricas [B]  time = 2.23634, size = 427, normalized size = 6.01 \begin{align*} -\frac{1}{2} \, \sqrt{\sqrt{2} + 2} \arctan \left (-\frac{1}{2} \,{\left ({\left (\sqrt{2} e^{\left (2 \, x\right )} - \sqrt{2}\right )} \sqrt{\sqrt{2} + 2} - \sqrt{2} \sqrt{-\sqrt{2} e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )} + 1} \sqrt{\sqrt{2} + 2}\right )} e^{\left (-x\right )}\right ) + \frac{1}{2} \, \sqrt{-\sqrt{2} + 2} \arctan \left (-\frac{1}{2} \,{\left ({\left (\sqrt{2} e^{\left (2 \, x\right )} - \sqrt{2}\right )} \sqrt{-\sqrt{2} + 2} - \sqrt{2} \sqrt{\sqrt{2} e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )} + 1} \sqrt{-\sqrt{2} + 2}\right )} e^{\left (-x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sech(4*x),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh{\left (x \right )} \operatorname{sech}{\left (4 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sech(4*x),x)

[Out]

Integral(cosh(x)*sech(4*x), x)

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Giac [B]  time = 1.21649, size = 182, normalized size = 2.56 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(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))