3.38 \(\int \frac{1}{(1-\cosh (c+d x))^3} \, dx\)

Optimal. Leaf size=76 \[ -\frac{2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))}-\frac{2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}-\frac{\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3} \]

[Out]

-Sinh[c + d*x]/(5*d*(1 - Cosh[c + d*x])^3) - (2*Sinh[c + d*x])/(15*d*(1 - Cosh[c + d*x])^2) - (2*Sinh[c + d*x]
)/(15*d*(1 - Cosh[c + d*x]))

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Rubi [A]  time = 0.0403379, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2650, 2648} \[ -\frac{2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))}-\frac{2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}-\frac{\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cosh[c + d*x])^(-3),x]

[Out]

-Sinh[c + d*x]/(5*d*(1 - Cosh[c + d*x])^3) - (2*Sinh[c + d*x])/(15*d*(1 - Cosh[c + d*x])^2) - (2*Sinh[c + d*x]
)/(15*d*(1 - Cosh[c + d*x]))

Rule 2650

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{1}{(1-\cosh (c+d x))^3} \, dx &=-\frac{\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}+\frac{2}{5} \int \frac{1}{(1-\cosh (c+d x))^2} \, dx\\ &=-\frac{\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}-\frac{2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}+\frac{2}{15} \int \frac{1}{1-\cosh (c+d x)} \, dx\\ &=-\frac{\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}-\frac{2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}-\frac{2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.0506645, size = 41, normalized size = 0.54 \[ \frac{\sinh (c+d x) (-6 \cosh (c+d x)+\cosh (2 (c+d x))+8)}{15 d (\cosh (c+d x)-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cosh[c + d*x])^(-3),x]

[Out]

((8 - 6*Cosh[c + d*x] + Cosh[2*(c + d*x)])*Sinh[c + d*x])/(15*d*(-1 + Cosh[c + d*x])^3)

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Maple [A]  time = 0.013, size = 45, normalized size = 0.6 \begin{align*}{\frac{1}{d} \left ({\frac{1}{4} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{6} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{1}{20} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cosh(d*x+c))^3,x)

[Out]

1/d*(1/4/tanh(1/2*d*x+1/2*c)-1/6/tanh(1/2*d*x+1/2*c)^3+1/20/tanh(1/2*d*x+1/2*c)^5)

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Maxima [B]  time = 1.0811, size = 277, normalized size = 3.64 \begin{align*} \frac{4 \, e^{\left (-d x - c\right )}}{3 \, d{\left (5 \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} - \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{3 \, d{\left (5 \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} - \frac{4}{15 \, d{\left (5 \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(d*x+c))^3,x, algorithm="maxima")

[Out]

4/3*e^(-d*x - c)/(d*(5*e^(-d*x - c) - 10*e^(-2*d*x - 2*c) + 10*e^(-3*d*x - 3*c) - 5*e^(-4*d*x - 4*c) + e^(-5*d
*x - 5*c) - 1)) - 8/3*e^(-2*d*x - 2*c)/(d*(5*e^(-d*x - c) - 10*e^(-2*d*x - 2*c) + 10*e^(-3*d*x - 3*c) - 5*e^(-
4*d*x - 4*c) + e^(-5*d*x - 5*c) - 1)) - 4/15/(d*(5*e^(-d*x - c) - 10*e^(-2*d*x - 2*c) + 10*e^(-3*d*x - 3*c) -
5*e^(-4*d*x - 4*c) + e^(-5*d*x - 5*c) - 1))

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Fricas [B]  time = 1.79807, size = 485, normalized size = 6.38 \begin{align*} \frac{4 \,{\left (11 \, \cosh \left (d x + c\right ) + 9 \, \sinh \left (d x + c\right ) - 5\right )}}{15 \,{\left (d \cosh \left (d x + c\right )^{4} + d \sinh \left (d x + c\right )^{4} - 5 \, d \cosh \left (d x + c\right )^{3} +{\left (4 \, d \cosh \left (d x + c\right ) - 5 \, d\right )} \sinh \left (d x + c\right )^{3} + 10 \, d \cosh \left (d x + c\right )^{2} +{\left (6 \, d \cosh \left (d x + c\right )^{2} - 15 \, d \cosh \left (d x + c\right ) + 10 \, d\right )} \sinh \left (d x + c\right )^{2} - 11 \, d \cosh \left (d x + c\right ) +{\left (4 \, d \cosh \left (d x + c\right )^{3} - 15 \, d \cosh \left (d x + c\right )^{2} + 20 \, d \cosh \left (d x + c\right ) - 9 \, d\right )} \sinh \left (d x + c\right ) + 5 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(d*x+c))^3,x, algorithm="fricas")

[Out]

4/15*(11*cosh(d*x + c) + 9*sinh(d*x + c) - 5)/(d*cosh(d*x + c)^4 + d*sinh(d*x + c)^4 - 5*d*cosh(d*x + c)^3 + (
4*d*cosh(d*x + c) - 5*d)*sinh(d*x + c)^3 + 10*d*cosh(d*x + c)^2 + (6*d*cosh(d*x + c)^2 - 15*d*cosh(d*x + c) +
10*d)*sinh(d*x + c)^2 - 11*d*cosh(d*x + c) + (4*d*cosh(d*x + c)^3 - 15*d*cosh(d*x + c)^2 + 20*d*cosh(d*x + c)
- 9*d)*sinh(d*x + c) + 5*d)

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Sympy [A]  time = 3.94714, size = 65, normalized size = 0.86 \begin{align*} \begin{cases} \tilde{\infty } x & \text{for}\: c = 0 \wedge d = 0 \\\frac{x}{\left (1 - \cosh{\left (c \right )}\right )^{3}} & \text{for}\: d = 0 \\\tilde{\infty } x & \text{for}\: c = - d x \\\frac{1}{4 d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )}} - \frac{1}{6 d \tanh ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}} + \frac{1}{20 d \tanh ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(d*x+c))**3,x)

[Out]

Piecewise((zoo*x, Eq(c, 0) & Eq(d, 0)), (x/(1 - cosh(c))**3, Eq(d, 0)), (zoo*x, Eq(c, -d*x)), (1/(4*d*tanh(c/2
+ d*x/2)) - 1/(6*d*tanh(c/2 + d*x/2)**3) + 1/(20*d*tanh(c/2 + d*x/2)**5), True))

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Giac [A]  time = 1.16561, size = 49, normalized size = 0.64 \begin{align*} \frac{4 \,{\left (10 \, e^{\left (2 \, d x + 2 \, c\right )} - 5 \, e^{\left (d x + c\right )} + 1\right )}}{15 \, d{\left (e^{\left (d x + c\right )} - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(d*x+c))^3,x, algorithm="giac")

[Out]

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