3.282 \(\int \frac{\text{sech}^3(x)}{a+a \sinh ^2(x)} \, dx\)

Optimal. Leaf size=35 \[ \frac{3 \tan ^{-1}(\sinh (x))}{8 a}+\frac{\tanh (x) \text{sech}^3(x)}{4 a}+\frac{3 \tanh (x) \text{sech}(x)}{8 a} \]

[Out]

(3*ArcTan[Sinh[x]])/(8*a) + (3*Sech[x]*Tanh[x])/(8*a) + (Sech[x]^3*Tanh[x])/(4*a)

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Rubi [A]  time = 0.0598158, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3175, 3768, 3770} \[ \frac{3 \tan ^{-1}(\sinh (x))}{8 a}+\frac{\tanh (x) \text{sech}^3(x)}{4 a}+\frac{3 \tanh (x) \text{sech}(x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + a*Sinh[x]^2),x]

[Out]

(3*ArcTan[Sinh[x]])/(8*a) + (3*Sech[x]*Tanh[x])/(8*a) + (Sech[x]^3*Tanh[x])/(4*a)

Rule 3175

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

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]

Rubi steps

\begin{align*} \int \frac{\text{sech}^3(x)}{a+a \sinh ^2(x)} \, dx &=\frac{\int \text{sech}^5(x) \, dx}{a}\\ &=\frac{\text{sech}^3(x) \tanh (x)}{4 a}+\frac{3 \int \text{sech}^3(x) \, dx}{4 a}\\ &=\frac{3 \text{sech}(x) \tanh (x)}{8 a}+\frac{\text{sech}^3(x) \tanh (x)}{4 a}+\frac{3 \int \text{sech}(x) \, dx}{8 a}\\ &=\frac{3 \tan ^{-1}(\sinh (x))}{8 a}+\frac{3 \text{sech}(x) \tanh (x)}{8 a}+\frac{\text{sech}^3(x) \tanh (x)}{4 a}\\ \end{align*}

Mathematica [A]  time = 0.0044584, size = 34, normalized size = 0.97 \[ \frac{\frac{3}{4} \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+\frac{1}{4} \tanh (x) \text{sech}^3(x)+\frac{3}{8} \tanh (x) \text{sech}(x)}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(a + a*Sinh[x]^2),x]

[Out]

((3*ArcTan[Tanh[x/2]])/4 + (3*Sech[x]*Tanh[x])/8 + (Sech[x]^3*Tanh[x])/4)/a

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Maple [B]  time = 0.026, size = 94, normalized size = 2.7 \begin{align*} -{\frac{5}{4\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{7} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-4}}+{\frac{3}{4\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-4}}-{\frac{3}{4\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-4}}+{\frac{5}{4\,a}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-4}}+{\frac{3}{4\,a}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^3/(a+a*sinh(x)^2),x)

[Out]

-5/4/a/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^7+3/4/a/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^5-3/4/a/(tanh(1/2*x)^2+1)^4*tan
h(1/2*x)^3+5/4/a/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)+3/4/a*arctan(tanh(1/2*x))

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Maxima [B]  time = 1.51705, size = 93, normalized size = 2.66 \begin{align*} \frac{3 \, e^{\left (-x\right )} + 11 \, e^{\left (-3 \, x\right )} - 11 \, e^{\left (-5 \, x\right )} - 3 \, e^{\left (-7 \, x\right )}}{4 \,{\left (4 \, a e^{\left (-2 \, x\right )} + 6 \, a e^{\left (-4 \, x\right )} + 4 \, a e^{\left (-6 \, x\right )} + a e^{\left (-8 \, x\right )} + a\right )}} - \frac{3 \, \arctan \left (e^{\left (-x\right )}\right )}{4 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+a*sinh(x)^2),x, algorithm="maxima")

[Out]

1/4*(3*e^(-x) + 11*e^(-3*x) - 11*e^(-5*x) - 3*e^(-7*x))/(4*a*e^(-2*x) + 6*a*e^(-4*x) + 4*a*e^(-6*x) + a*e^(-8*
x) + a) - 3/4*arctan(e^(-x))/a

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Fricas [B]  time = 1.49142, size = 1607, normalized size = 45.91 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+a*sinh(x)^2),x, algorithm="fricas")

[Out]

1/4*(3*cosh(x)^7 + 21*cosh(x)*sinh(x)^6 + 3*sinh(x)^7 + (63*cosh(x)^2 + 11)*sinh(x)^5 + 11*cosh(x)^5 + 5*(21*c
osh(x)^3 + 11*cosh(x))*sinh(x)^4 + (105*cosh(x)^4 + 110*cosh(x)^2 - 11)*sinh(x)^3 - 11*cosh(x)^3 + (63*cosh(x)
^5 + 110*cosh(x)^3 - 33*cosh(x))*sinh(x)^2 + 3*(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 +
1)*sinh(x)^6 + 4*cosh(x)^6 + 8*(7*cosh(x)^3 + 3*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 + 30*cosh(x)^2 + 3)*sinh
(x)^4 + 6*cosh(x)^4 + 8*(7*cosh(x)^5 + 10*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 + 15*cosh(x)^4 + 9
*cosh(x)^2 + 1)*sinh(x)^2 + 4*cosh(x)^2 + 8*(cosh(x)^7 + 3*cosh(x)^5 + 3*cosh(x)^3 + cosh(x))*sinh(x) + 1)*arc
tan(cosh(x) + sinh(x)) + (21*cosh(x)^6 + 55*cosh(x)^4 - 33*cosh(x)^2 - 3)*sinh(x) - 3*cosh(x))/(a*cosh(x)^8 +
8*a*cosh(x)*sinh(x)^7 + a*sinh(x)^8 + 4*a*cosh(x)^6 + 4*(7*a*cosh(x)^2 + a)*sinh(x)^6 + 8*(7*a*cosh(x)^3 + 3*a
*cosh(x))*sinh(x)^5 + 6*a*cosh(x)^4 + 2*(35*a*cosh(x)^4 + 30*a*cosh(x)^2 + 3*a)*sinh(x)^4 + 8*(7*a*cosh(x)^5 +
10*a*cosh(x)^3 + 3*a*cosh(x))*sinh(x)^3 + 4*a*cosh(x)^2 + 4*(7*a*cosh(x)^6 + 15*a*cosh(x)^4 + 9*a*cosh(x)^2 +
a)*sinh(x)^2 + 8*(a*cosh(x)^7 + 3*a*cosh(x)^5 + 3*a*cosh(x)^3 + a*cosh(x))*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 )}}{\sinh ^{2}{\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*sinh(x)**2),x)

[Out]

Integral(sech(x)**3/(sinh(x)**2 + 1), x)/a

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Giac [B]  time = 1.15806, size = 90, normalized size = 2.57 \begin{align*} \frac{3 \,{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )}}{16 \, a} - \frac{3 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 20 \, e^{\left (-x\right )} - 20 \, e^{x}}{4 \,{\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}^{2} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+a*sinh(x)^2),x, algorithm="giac")

[Out]

3/16*(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x)))/a - 1/4*(3*(e^(-x) - e^x)^3 + 20*e^(-x) - 20*e^x)/(((e^(-x) - e
^x)^2 + 4)^2*a)