3.147 \(\int \frac{\sinh (x)}{(1-\cosh (x))^3} \, dx\)

Optimal. Leaf size=12 \[ \frac{1}{2 (1-\cosh (x))^2} \]

[Out]

1/(2*(1 - Cosh[x])^2)

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Rubi [A]  time = 0.0211197, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2667, 32} \[ \frac{1}{2 (1-\cosh (x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]/(1 - Cosh[x])^3,x]

[Out]

1/(2*(1 - Cosh[x])^2)

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\sinh (x)}{(1-\cosh (x))^3} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{(1+x)^3} \, dx,x,-\cosh (x)\right )\\ &=\frac{1}{2 (1-\cosh (x))^2}\\ \end{align*}

Mathematica [A]  time = 0.0103202, size = 12, normalized size = 1. \[ \frac{1}{8} \text{csch}^4\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]/(1 - Cosh[x])^3,x]

[Out]

Csch[x/2]^4/8

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Maple [A]  time = 0.005, size = 11, normalized size = 0.9 \begin{align*}{\frac{1}{2\, \left ( 1-\cosh \left ( x \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(1-cosh(x))^2

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Maxima [A]  time = 1.06197, size = 11, normalized size = 0.92 \begin{align*} \frac{1}{2 \,{\left (\cosh \left (x\right ) - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(cosh(x) - 1)^2

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Fricas [B]  time = 2.13166, size = 196, normalized size = 16.33 \begin{align*} \frac{2 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}}{\cosh \left (x\right )^{3} +{\left (3 \, \cosh \left (x\right ) - 4\right )} \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} - 4 \, \cosh \left (x\right )^{2} +{\left (3 \, \cosh \left (x\right )^{2} - 8 \, \cosh \left (x\right ) + 5\right )} \sinh \left (x\right ) + 7 \, \cosh \left (x\right ) - 4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(cosh(x) + sinh(x))/(cosh(x)^3 + (3*cosh(x) - 4)*sinh(x)^2 + sinh(x)^3 - 4*cosh(x)^2 + (3*cosh(x)^2 - 8*cosh
(x) + 5)*sinh(x) + 7*cosh(x) - 4)

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Sympy [A]  time = 0.842427, size = 14, normalized size = 1.17 \begin{align*} \frac{1}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(2*cosh(x)**2 - 4*cosh(x) + 2)

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Giac [A]  time = 1.14395, size = 16, normalized size = 1.33 \begin{align*} \frac{2 \, e^{\left (2 \, x\right )}}{{\left (e^{x} - 1\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e^(2*x)/(e^x - 1)^4