3.142 \(\int \frac{\sinh ^2(x)}{(1+\cosh (x))^2} \, dx\)

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

[Out]

x - (2*Sinh[x])/(1 + Cosh[x])

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Rubi [A]  time = 0.0314202, 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 = {2680, 8} \[ x-\frac{2 \sinh (x)}{\cosh (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(1 + Cosh[x])^2,x]

[Out]

x - (2*Sinh[x])/(1 + Cosh[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]

Rule 8

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

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*}

Mathematica [A]  time = 0.0057045, size = 18, normalized size = 1.5 \[ 2 \tanh ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )-2 \tanh \left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*ArcTanh[Tanh[x/2]] - 2*Tanh[x/2]

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Maple [A]  time = 0.014, size = 24, normalized size = 2. \begin{align*} -2\,\tanh \left ( x/2 \right ) -\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) +\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.08282, size = 16, normalized size = 1.33 \begin{align*} x - \frac{4}{e^{\left (-x\right )} + 1} \end{align*}

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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Fricas [A]  time = 2.19453, size = 77, normalized size = 6.42 \begin{align*} \frac{x \cosh \left (x\right ) + x \sinh \left (x\right ) + x + 4}{\cosh \left (x\right ) + \sinh \left (x\right ) + 1} \end{align*}

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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Sympy [A]  time = 0.733914, size = 7, normalized size = 0.58 \begin{align*} x - 2 \tanh{\left (\frac{x}{2} \right )} \end{align*}

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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Giac [A]  time = 1.22489, size = 14, normalized size = 1.17 \begin{align*} x + \frac{4}{e^{x} + 1} \end{align*}

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)