3.143 \(\int \frac {\sinh ^2(x)}{(1-\cosh (x))^2} \, dx\)

Optimal. Leaf size=14 \[ x+\frac {2 \sinh (x)}{1-\cosh (x)} \]

[Out]

x+2*sinh(x)/(1-cosh(x))

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Rubi [A]  time = 0.03, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2680, 8} \[ x+\frac {2 \sinh (x)}{1-\cosh (x)} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(1 - Cosh[x])^2,x]

[Out]

x + (2*Sinh[x])/(1 - Cosh[x])

Rule 8

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

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 24, normalized size = 1.71 \[ -2 \coth \left (\frac {x}{2}\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\tanh ^2\left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^2/(1 - Cosh[x])^2,x]

[Out]

-2*Coth[x/2]*Hypergeometric2F1[-1/2, 1, 1/2, Tanh[x/2]^2]

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fricas [A]  time = 2.15, size = 22, normalized size = 1.57 \[ \frac {x \cosh \relax (x) + x \sinh \relax (x) - x - 4}{\cosh \relax (x) + \sinh \relax (x) - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*cosh(x) + x*sinh(x) - x - 4)/(cosh(x) + sinh(x) - 1)

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giac [A]  time = 0.16, size = 10, normalized size = 0.71 \[ x - \frac {4}{e^{x} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 4/(e^x - 1)

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maple [A]  time = 0.07, size = 26, normalized size = 1.86 \[ -\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\frac {2}{\tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(tanh(1/2*x)-1)+ln(tanh(1/2*x)+1)-2/tanh(1/2*x)

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maxima [A]  time = 0.31, size = 12, normalized size = 0.86 \[ x + \frac {4}{e^{\left (-x\right )} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 4/(e^(-x) - 1)

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mupad [B]  time = 0.05, size = 10, normalized size = 0.71 \[ x-\frac {4}{{\mathrm {e}}^x-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(cosh(x) - 1)^2,x)

[Out]

x - 4/(exp(x) - 1)

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sympy [A]  time = 1.04, size = 7, normalized size = 0.50 \[ x - \frac {2}{\tanh {\left (\frac {x}{2} \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 2/tanh(x/2)

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