3.40 \(\int \frac{1}{(1-\cosh ^2(x))^3} \, dx\)

Optimal. Leaf size=19 \[ \frac{\coth ^5(x)}{5}-\frac{2 \coth ^3(x)}{3}+\coth (x) \]

[Out]

Coth[x] - (2*Coth[x]^3)/3 + Coth[x]^5/5

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Rubi [A]  time = 0.0200164, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3175, 3767} \[ \frac{\coth ^5(x)}{5}-\frac{2 \coth ^3(x)}{3}+\coth (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cosh[x]^2)^(-3),x]

[Out]

Coth[x] - (2*Coth[x]^3)/3 + Coth[x]^5/5

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 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]

Rubi steps

\begin{align*} \int \frac{1}{\left (1-\cosh ^2(x)\right )^3} \, dx &=-\int \text{csch}^6(x) \, dx\\ &=i \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-i \coth (x)\right )\\ &=\coth (x)-\frac{2 \coth ^3(x)}{3}+\frac{\coth ^5(x)}{5}\\ \end{align*}

Mathematica [A]  time = 0.0038613, size = 27, normalized size = 1.42 \[ \frac{8 \coth (x)}{15}+\frac{1}{5} \coth (x) \text{csch}^4(x)-\frac{4}{15} \coth (x) \text{csch}^2(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cosh[x]^2)^(-3),x]

[Out]

(8*Coth[x])/15 - (4*Coth[x]*Csch[x]^2)/15 + (Coth[x]*Csch[x]^4)/5

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Maple [B]  time = 0.016, size = 48, normalized size = 2.5 \begin{align*}{\frac{1}{160} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5}}-{\frac{5}{96} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{5}{16}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{5}{16} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{5}{96} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}+{\frac{1}{160} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cosh(x)^2)^3,x)

[Out]

1/160*tanh(1/2*x)^5-5/96*tanh(1/2*x)^3+5/16*tanh(1/2*x)+5/16/tanh(1/2*x)-5/96/tanh(1/2*x)^3+1/160/tanh(1/2*x)^
5

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Maxima [B]  time = 1.09252, size = 150, normalized size = 7.89 \begin{align*} \frac{16 \, e^{\left (-2 \, x\right )}}{3 \,{\left (5 \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} - 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} - 1\right )}} - \frac{32 \, e^{\left (-4 \, x\right )}}{3 \,{\left (5 \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} - 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} - 1\right )}} - \frac{16}{15 \,{\left (5 \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} - 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)^2)^3,x, algorithm="maxima")

[Out]

16/3*e^(-2*x)/(5*e^(-2*x) - 10*e^(-4*x) + 10*e^(-6*x) - 5*e^(-8*x) + e^(-10*x) - 1) - 32/3*e^(-4*x)/(5*e^(-2*x
) - 10*e^(-4*x) + 10*e^(-6*x) - 5*e^(-8*x) + e^(-10*x) - 1) - 16/15/(5*e^(-2*x) - 10*e^(-4*x) + 10*e^(-6*x) -
5*e^(-8*x) + e^(-10*x) - 1)

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Fricas [B]  time = 2.0004, size = 626, normalized size = 32.95 \begin{align*} \frac{16 \,{\left (11 \, \cosh \left (x\right )^{2} + 18 \, \cosh \left (x\right ) \sinh \left (x\right ) + 11 \, \sinh \left (x\right )^{2} - 5\right )}}{15 \,{\left (\cosh \left (x\right )^{8} + 8 \, \cosh \left (x\right ) \sinh \left (x\right )^{7} + \sinh \left (x\right )^{8} +{\left (28 \, \cosh \left (x\right )^{2} - 5\right )} \sinh \left (x\right )^{6} - 5 \, \cosh \left (x\right )^{6} + 2 \,{\left (28 \, \cosh \left (x\right )^{3} - 15 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{5} + 5 \,{\left (14 \, \cosh \left (x\right )^{4} - 15 \, \cosh \left (x\right )^{2} + 2\right )} \sinh \left (x\right )^{4} + 10 \, \cosh \left (x\right )^{4} + 4 \,{\left (14 \, \cosh \left (x\right )^{5} - 25 \, \cosh \left (x\right )^{3} + 10 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} +{\left (28 \, \cosh \left (x\right )^{6} - 75 \, \cosh \left (x\right )^{4} + 60 \, \cosh \left (x\right )^{2} - 11\right )} \sinh \left (x\right )^{2} - 11 \, \cosh \left (x\right )^{2} + 2 \,{\left (4 \, \cosh \left (x\right )^{7} - 15 \, \cosh \left (x\right )^{5} + 20 \, \cosh \left (x\right )^{3} - 9 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) + 5\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)^2)^3,x, algorithm="fricas")

[Out]

16/15*(11*cosh(x)^2 + 18*cosh(x)*sinh(x) + 11*sinh(x)^2 - 5)/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + (2
8*cosh(x)^2 - 5)*sinh(x)^6 - 5*cosh(x)^6 + 2*(28*cosh(x)^3 - 15*cosh(x))*sinh(x)^5 + 5*(14*cosh(x)^4 - 15*cosh
(x)^2 + 2)*sinh(x)^4 + 10*cosh(x)^4 + 4*(14*cosh(x)^5 - 25*cosh(x)^3 + 10*cosh(x))*sinh(x)^3 + (28*cosh(x)^6 -
75*cosh(x)^4 + 60*cosh(x)^2 - 11)*sinh(x)^2 - 11*cosh(x)^2 + 2*(4*cosh(x)^7 - 15*cosh(x)^5 + 20*cosh(x)^3 - 9
*cosh(x))*sinh(x) + 5)

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Sympy [B]  time = 7.84964, size = 54, normalized size = 2.84 \begin{align*} \frac{\tanh ^{5}{\left (\frac{x}{2} \right )}}{160} - \frac{5 \tanh ^{3}{\left (\frac{x}{2} \right )}}{96} + \frac{5 \tanh{\left (\frac{x}{2} \right )}}{16} + \frac{5}{16 \tanh{\left (\frac{x}{2} \right )}} - \frac{5}{96 \tanh ^{3}{\left (\frac{x}{2} \right )}} + \frac{1}{160 \tanh ^{5}{\left (\frac{x}{2} \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)**2)**3,x)

[Out]

tanh(x/2)**5/160 - 5*tanh(x/2)**3/96 + 5*tanh(x/2)/16 + 5/(16*tanh(x/2)) - 5/(96*tanh(x/2)**3) + 1/(160*tanh(x
/2)**5)

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Giac [A]  time = 1.32176, size = 32, normalized size = 1.68 \begin{align*} \frac{16 \,{\left (10 \, e^{\left (4 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + 1\right )}}{15 \,{\left (e^{\left (2 \, x\right )} - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)^2)^3,x, algorithm="giac")

[Out]

16/15*(10*e^(4*x) - 5*e^(2*x) + 1)/(e^(2*x) - 1)^5