3.92 \(\int \frac{\sinh (x)}{1+\coth (x)} \, dx\)

Optimal. Leaf size=19 \[ \frac{2 \cosh (x)}{3}-\frac{\sinh (x)}{3 (\coth (x)+1)} \]

[Out]

(2*Cosh[x])/3 - Sinh[x]/(3*(1 + Coth[x]))

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Rubi [A]  time = 0.0355857, antiderivative size = 19, 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 = {3502, 2638} \[ \frac{2 \cosh (x)}{3}-\frac{\sinh (x)}{3 (\coth (x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]/(1 + Coth[x]),x]

[Out]

(2*Cosh[x])/3 - Sinh[x]/(3*(1 + Coth[x]))

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sinh (x)}{1+\coth (x)} \, dx &=-\frac{\sinh (x)}{3 (1+\coth (x))}+\frac{2}{3} \int \sinh (x) \, dx\\ &=\frac{2 \cosh (x)}{3}-\frac{\sinh (x)}{3 (1+\coth (x))}\\ \end{align*}

Mathematica [A]  time = 0.0476313, size = 21, normalized size = 1.11 \[ \frac{1}{12} \left (4 \sinh ^3(x)+9 \cosh (x)-\cosh (3 x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]/(1 + Coth[x]),x]

[Out]

(9*Cosh[x] - Cosh[3*x] + 4*Sinh[x]^3)/12

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Maple [B]  time = 0.027, size = 40, normalized size = 2.1 \begin{align*} -{\frac{2}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}+{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(1+coth(x)),x)

[Out]

-2/3/(tanh(1/2*x)+1)^3+1/(tanh(1/2*x)+1)^2+1/2/(tanh(1/2*x)+1)-1/2/(tanh(1/2*x)-1)

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Maxima [A]  time = 1.10248, size = 23, normalized size = 1.21 \begin{align*} \frac{1}{2} \, e^{\left (-x\right )} - \frac{1}{12} \, e^{\left (-3 \, x\right )} + \frac{1}{4} \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(1+coth(x)),x, algorithm="maxima")

[Out]

1/2*e^(-x) - 1/12*e^(-3*x) + 1/4*e^x

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Fricas [A]  time = 2.47767, size = 99, normalized size = 5.21 \begin{align*} \frac{\cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 3}{6 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(1+coth(x)),x, algorithm="fricas")

[Out]

1/6*(cosh(x)^2 + 4*cosh(x)*sinh(x) + sinh(x)^2 + 3)/(cosh(x) + sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (x \right )}}{\coth{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(1+coth(x)),x)

[Out]

Integral(sinh(x)/(coth(x) + 1), x)

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Giac [A]  time = 1.12542, size = 26, normalized size = 1.37 \begin{align*} \frac{1}{12} \,{\left (6 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-3 \, x\right )} + \frac{1}{4} \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(1+coth(x)),x, algorithm="giac")

[Out]

1/12*(6*e^(2*x) - 1)*e^(-3*x) + 1/4*e^x