3.30 \(\int \frac{\text{sech}^3(x)}{a+a \cosh (x)} \, dx\)

Optimal. Leaf size=43 \[ -\frac{2 \tanh (x)}{a}+\frac{3 \tan ^{-1}(\sinh (x))}{2 a}+\frac{3 \tanh (x) \text{sech}(x)}{2 a}-\frac{\tanh (x) \text{sech}(x)}{a \cosh (x)+a} \]

[Out]

(3*ArcTan[Sinh[x]])/(2*a) - (2*Tanh[x])/a + (3*Sech[x]*Tanh[x])/(2*a) - (Sech[x]*Tanh[x])/(a + a*Cosh[x])

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Rubi [A]  time = 0.0744833, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2768, 2748, 3768, 3770, 3767, 8} \[ -\frac{2 \tanh (x)}{a}+\frac{3 \tan ^{-1}(\sinh (x))}{2 a}+\frac{3 \tanh (x) \text{sech}(x)}{2 a}-\frac{\tanh (x) \text{sech}(x)}{a \cosh (x)+a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + a*Cosh[x]),x]

[Out]

(3*ArcTan[Sinh[x]])/(2*a) - (2*Tanh[x])/a + (3*Sech[x]*Tanh[x])/(2*a) - (Sech[x]*Tanh[x])/(a + a*Cosh[x])

Rule 2768

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b
^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x])), x] + Dist[d/(a*(b*c - a*
d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\text{sech}^3(x)}{a+a \cosh (x)} \, dx &=-\frac{\text{sech}(x) \tanh (x)}{a+a \cosh (x)}-\frac{\int (-3 a+2 a \cosh (x)) \text{sech}^3(x) \, dx}{a^2}\\ &=-\frac{\text{sech}(x) \tanh (x)}{a+a \cosh (x)}-\frac{2 \int \text{sech}^2(x) \, dx}{a}+\frac{3 \int \text{sech}^3(x) \, dx}{a}\\ &=\frac{3 \text{sech}(x) \tanh (x)}{2 a}-\frac{\text{sech}(x) \tanh (x)}{a+a \cosh (x)}-\frac{(2 i) \operatorname{Subst}(\int 1 \, dx,x,-i \tanh (x))}{a}+\frac{3 \int \text{sech}(x) \, dx}{2 a}\\ &=\frac{3 \tan ^{-1}(\sinh (x))}{2 a}-\frac{2 \tanh (x)}{a}+\frac{3 \text{sech}(x) \tanh (x)}{2 a}-\frac{\text{sech}(x) \tanh (x)}{a+a \cosh (x)}\\ \end{align*}

Mathematica [A]  time = 0.0830855, size = 49, normalized size = 1.14 \[ \frac{\cosh \left (\frac{x}{2}\right ) \left (\cosh \left (\frac{x}{2}\right ) \left (6 \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+\tanh (x) (\text{sech}(x)-2)\right )-2 \sinh \left (\frac{x}{2}\right )\right )}{a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(a + a*Cosh[x]),x]

[Out]

(Cosh[x/2]*(-2*Sinh[x/2] + Cosh[x/2]*(6*ArcTan[Tanh[x/2]] + (-2 + Sech[x])*Tanh[x])))/(a*(1 + Cosh[x]))

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Maple [A]  time = 0.024, size = 61, normalized size = 1.4 \begin{align*} -{\frac{1}{a}\tanh \left ({\frac{x}{2}} \right ) }-3\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{3}}{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{1}{a}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+3\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^3/(a+a*cosh(x)),x)

[Out]

-1/a*tanh(1/2*x)-3/a/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3-1/a/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)+3/a*arctan(tanh(1/2
*x))

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Maxima [A]  time = 1.67315, size = 99, normalized size = 2.3 \begin{align*} -\frac{e^{\left (-x\right )} + 5 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + 4}{a e^{\left (-x\right )} + 2 \, a e^{\left (-2 \, x\right )} + 2 \, a e^{\left (-3 \, x\right )} + a e^{\left (-4 \, x\right )} + a e^{\left (-5 \, x\right )} + a} - \frac{3 \, \arctan \left (e^{\left (-x\right )}\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+a*cosh(x)),x, algorithm="maxima")

[Out]

-(e^(-x) + 5*e^(-2*x) + 3*e^(-3*x) + 3*e^(-4*x) + 4)/(a*e^(-x) + 2*a*e^(-2*x) + 2*a*e^(-3*x) + a*e^(-4*x) + a*
e^(-5*x) + a) - 3*arctan(e^(-x))/a

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Fricas [B]  time = 1.89154, size = 1099, normalized size = 25.56 \begin{align*} \frac{3 \, \cosh \left (x\right )^{4} + 3 \,{\left (4 \, \cosh \left (x\right ) + 1\right )} \sinh \left (x\right )^{3} + 3 \, \sinh \left (x\right )^{4} + 3 \, \cosh \left (x\right )^{3} +{\left (18 \, \cosh \left (x\right )^{2} + 9 \, \cosh \left (x\right ) + 5\right )} \sinh \left (x\right )^{2} + 3 \,{\left (\cosh \left (x\right )^{5} +{\left (5 \, \cosh \left (x\right ) + 1\right )} \sinh \left (x\right )^{4} + \sinh \left (x\right )^{5} + \cosh \left (x\right )^{4} + 2 \,{\left (5 \, \cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) + 1\right )} \sinh \left (x\right )^{3} + 2 \, \cosh \left (x\right )^{3} + 2 \,{\left (5 \, \cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right )^{2} + 3 \, \cosh \left (x\right ) + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} +{\left (5 \, \cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} + 6 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) + 1\right )} \sinh \left (x\right ) + \cosh \left (x\right ) + 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + 5 \, \cosh \left (x\right )^{2} +{\left (12 \, \cosh \left (x\right )^{3} + 9 \, \cosh \left (x\right )^{2} + 10 \, \cosh \left (x\right ) + 1\right )} \sinh \left (x\right ) + \cosh \left (x\right ) + 4}{a \cosh \left (x\right )^{5} + a \sinh \left (x\right )^{5} + a \cosh \left (x\right )^{4} +{\left (5 \, a \cosh \left (x\right ) + a\right )} \sinh \left (x\right )^{4} + 2 \, a \cosh \left (x\right )^{3} + 2 \,{\left (5 \, a \cosh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + a\right )} \sinh \left (x\right )^{3} + 2 \, a \cosh \left (x\right )^{2} + 2 \,{\left (5 \, a \cosh \left (x\right )^{3} + 3 \, a \cosh \left (x\right )^{2} + 3 \, a \cosh \left (x\right ) + a\right )} \sinh \left (x\right )^{2} + a \cosh \left (x\right ) +{\left (5 \, a \cosh \left (x\right )^{4} + 4 \, a \cosh \left (x\right )^{3} + 6 \, a \cosh \left (x\right )^{2} + 4 \, a \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+a*cosh(x)),x, algorithm="fricas")

[Out]

(3*cosh(x)^4 + 3*(4*cosh(x) + 1)*sinh(x)^3 + 3*sinh(x)^4 + 3*cosh(x)^3 + (18*cosh(x)^2 + 9*cosh(x) + 5)*sinh(x
)^2 + 3*(cosh(x)^5 + (5*cosh(x) + 1)*sinh(x)^4 + sinh(x)^5 + cosh(x)^4 + 2*(5*cosh(x)^2 + 2*cosh(x) + 1)*sinh(
x)^3 + 2*cosh(x)^3 + 2*(5*cosh(x)^3 + 3*cosh(x)^2 + 3*cosh(x) + 1)*sinh(x)^2 + 2*cosh(x)^2 + (5*cosh(x)^4 + 4*
cosh(x)^3 + 6*cosh(x)^2 + 4*cosh(x) + 1)*sinh(x) + cosh(x) + 1)*arctan(cosh(x) + sinh(x)) + 5*cosh(x)^2 + (12*
cosh(x)^3 + 9*cosh(x)^2 + 10*cosh(x) + 1)*sinh(x) + cosh(x) + 4)/(a*cosh(x)^5 + a*sinh(x)^5 + a*cosh(x)^4 + (5
*a*cosh(x) + a)*sinh(x)^4 + 2*a*cosh(x)^3 + 2*(5*a*cosh(x)^2 + 2*a*cosh(x) + a)*sinh(x)^3 + 2*a*cosh(x)^2 + 2*
(5*a*cosh(x)^3 + 3*a*cosh(x)^2 + 3*a*cosh(x) + a)*sinh(x)^2 + a*cosh(x) + (5*a*cosh(x)^4 + 4*a*cosh(x)^3 + 6*a
*cosh(x)^2 + 4*a*cosh(x) + a)*sinh(x) + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{sech}^{3}{\left (x \right )}}{\cosh{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**3/(a+a*cosh(x)),x)

[Out]

Integral(sech(x)**3/(cosh(x) + 1), x)/a

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Giac [A]  time = 1.14482, size = 65, normalized size = 1.51 \begin{align*} \frac{3 \, \arctan \left (e^{x}\right )}{a} + \frac{e^{\left (3 \, x\right )} + 2 \, e^{\left (2 \, x\right )} - e^{x} + 2}{a{\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} + \frac{2}{a{\left (e^{x} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+a*cosh(x)),x, algorithm="giac")

[Out]

3*arctan(e^x)/a + (e^(3*x) + 2*e^(2*x) - e^x + 2)/(a*(e^(2*x) + 1)^2) + 2/(a*(e^x + 1))