3.146 \(\int \frac{\sinh (x)}{(1+\cosh (x))^3} \, dx\)

Optimal. Leaf size=10 \[ -\frac{1}{2 (\cosh (x)+1)^2} \]

[Out]

-1/(2*(1 + Cosh[x])^2)

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Rubi [A]  time = 0.0205692, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2667, 32} \[ -\frac{1}{2 (\cosh (x)+1)^2} \]

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 = 0.0088983, size = 12, normalized size = 1.2 \[ -\frac{1}{8} \text{sech}^4\left (\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]/(1 + Cosh[x])^3,x]

[Out]

-Sech[x/2]^4/8

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Maple [A]  time = 0.004, size = 9, 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 = 1.06343, size = 11, normalized size = 1.1 \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 = 2.04528, size = 197, normalized size = 19.7 \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*cos
h(x) + 5)*sinh(x) + 7*cosh(x) + 4)

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Sympy [A]  time = 0.81847, size = 15, normalized size = 1.5 \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 = 1.18451, size = 16, normalized size = 1.6 \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