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]

2/3*cosh(x)-1/3*sinh(x)/(1+coth(x))

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Rubi [A]  time = 0.04, antiderivative size = 19, 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 = {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 2638

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

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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.39, size = 25, normalized size = 1.32 \[ \frac {\cosh \relax (x)^{2} + 4 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 3}{6 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}} \]

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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giac [A]  time = 0.13, size = 19, normalized size = 1.00 \[ \frac {1}{12} \, {\left (6 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-3 \, x\right )} + \frac {1}{4} \, e^{x} \]

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

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maple [B]  time = 0.09, size = 40, normalized size = 2.11 \[ -\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {2}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 \tanh \left (\frac {x}{2}\right )+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 17, normalized size = 0.89 \[ \frac {1}{2} \, e^{\left (-x\right )} - \frac {1}{12} \, e^{\left (-3 \, x\right )} + \frac {1}{4} \, e^{x} \]

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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mupad [B]  time = 1.24, size = 17, normalized size = 0.89 \[ \frac {{\mathrm {e}}^{-x}}{2}-\frac {{\mathrm {e}}^{-3\,x}}{12}+\frac {{\mathrm {e}}^x}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(coth(x) + 1),x)

[Out]

exp(-x)/2 - exp(-3*x)/12 + exp(x)/4

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\relax (x )}}{\coth {\relax (x )} + 1}\, dx \]

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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