3.2.42 \(\int \frac {\sinh ^2(x)}{(1+\cosh (x))^2} \, dx\) [142]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 12, normalized size of antiderivative = 1.00, 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 = {2759, 8} \begin {gather*} x-\frac {2 \sinh (x)}{\cosh (x)+1} \end {gather*}

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 2759

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 [A]
time = 0.01, size = 18, normalized size = 1.50 \begin {gather*} 2 \tanh ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) \end {gather*}

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 = 0.38, size = 24, normalized size = 2.00

method	result	size
risch	\(x +\frac {4}{{\mathrm e}^{x}+1}\)	\(11\)
default	\(-2 \tanh \left (\frac {x}{2}\right )-\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.26, size = 12, normalized size = 1.00 \begin {gather*} x - \frac {4}{e^{\left (-x\right )} + 1} \end {gather*}

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 = 0.41, size = 20, normalized size = 1.67 \begin {gather*} \frac {x \cosh \left (x\right ) + x \sinh \left (x\right ) + x + 4}{\cosh \left (x\right ) + \sinh \left (x\right ) + 1} \end {gather*}

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 = 0.20, size = 7, normalized size = 0.58 \begin {gather*} x - 2 \tanh {\left (\frac {x}{2} \right )} \end {gather*}

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 = 0.41, size = 10, normalized size = 0.83 \begin {gather*} x + \frac {4}{e^{x} + 1} \end {gather*}

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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Mupad [B]
time = 0.90, size = 10, normalized size = 0.83 \begin {gather*} x+\frac {4}{{\mathrm {e}}^x+1} \end {gather*}

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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