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 = 0.02, antiderivative size = 10, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 1.20 \[ -\frac {1}{8} \text {sech}^4\left (\frac {x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]/(1 + Cosh[x])^3,x]

[Out]

-1/8*Sech[x/2]^4

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fricas [B]  time = 0.82, size = 55, normalized size = 5.50 \[ -\frac {2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}}{\cosh \relax (x)^{3} + {\left (3 \, \cosh \relax (x) + 4\right )} \sinh \relax (x)^{2} + \sinh \relax (x)^{3} + 4 \, \cosh \relax (x)^{2} + {\left (3 \, \cosh \relax (x)^{2} + 8 \, \cosh \relax (x) + 5\right )} \sinh \relax (x) + 7 \, \cosh \relax (x) + 4} \]

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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giac [A]  time = 0.12, size = 12, normalized size = 1.20 \[ -\frac {2 \, e^{\left (2 \, x\right )}}{{\left (e^{x} + 1\right )}^{4}} \]

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

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maple [A]  time = 0.03, size = 9, normalized size = 0.90 \[ -\frac {1}{2 \left (1+\cosh \relax (x )\right )^{2}} \]

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 = 0.29, size = 8, normalized size = 0.80 \[ -\frac {1}{2 \, {\left (\cosh \relax (x) + 1\right )}^{2}} \]

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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mupad [B]  time = 0.92, size = 8, normalized size = 0.80 \[ -\frac {1}{2\,{\left (\mathrm {cosh}\relax (x)+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(cosh(x) + 1)^3,x)

[Out]

-1/(2*(cosh(x) + 1)^2)

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sympy [A]  time = 0.63, size = 15, normalized size = 1.50 \[ - \frac {1}{2 \cosh ^{2}{\relax (x )} + 4 \cosh {\relax (x )} + 2} \]

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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