3.240 \(\int \frac{\coth (x)}{(a+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=32 \[ -\frac{\log (a+b \sinh (x))}{a^2}+\frac{\log (\sinh (x))}{a^2}+\frac{1}{a (a+b \sinh (x))} \]

[Out]

Log[Sinh[x]]/a^2 - Log[a + b*Sinh[x]]/a^2 + 1/(a*(a + b*Sinh[x]))

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Rubi [A]  time = 0.0530725, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2721, 44} \[ -\frac{\log (a+b \sinh (x))}{a^2}+\frac{\log (\sinh (x))}{a^2}+\frac{1}{a (a+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/(a + b*Sinh[x])^2,x]

[Out]

Log[Sinh[x]]/a^2 - Log[a + b*Sinh[x]]/a^2 + 1/(a*(a + b*Sinh[x]))

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\coth (x)}{(a+b \sinh (x))^2} \, dx &=\operatorname{Subst}\left (\int \frac{1}{x (a+x)^2} \, dx,x,b \sinh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{1}{a^2 x}-\frac{1}{a (a+x)^2}-\frac{1}{a^2 (a+x)}\right ) \, dx,x,b \sinh (x)\right )\\ &=\frac{\log (\sinh (x))}{a^2}-\frac{\log (a+b \sinh (x))}{a^2}+\frac{1}{a (a+b \sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.0467081, size = 27, normalized size = 0.84 \[ \frac{\frac{a}{a+b \sinh (x)}-\log (a+b \sinh (x))+\log (\sinh (x))}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]/(a + b*Sinh[x])^2,x]

[Out]

(Log[Sinh[x]] - Log[a + b*Sinh[x]] + a/(a + b*Sinh[x]))/a^2

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Maple [A]  time = 0.033, size = 33, normalized size = 1. \begin{align*}{\frac{\ln \left ( \sinh \left ( x \right ) \right ) }{{a}^{2}}}-{\frac{\ln \left ( a+b\sinh \left ( x \right ) \right ) }{{a}^{2}}}+{\frac{1}{a \left ( a+b\sinh \left ( x \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)/(a+b*sinh(x))^2,x)

[Out]

ln(sinh(x))/a^2-ln(a+b*sinh(x))/a^2+1/a/(a+b*sinh(x))

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Maxima [B]  time = 1.02619, size = 101, normalized size = 3.16 \begin{align*} \frac{2 \, e^{\left (-x\right )}}{2 \, a^{2} e^{\left (-x\right )} - a b e^{\left (-2 \, x\right )} + a b} - \frac{\log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{2}} + \frac{\log \left (e^{\left (-x\right )} + 1\right )}{a^{2}} + \frac{\log \left (e^{\left (-x\right )} - 1\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

2*e^(-x)/(2*a^2*e^(-x) - a*b*e^(-2*x) + a*b) - log(-2*a*e^(-x) + b*e^(-2*x) - b)/a^2 + log(e^(-x) + 1)/a^2 + l
og(e^(-x) - 1)/a^2

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Fricas [B]  time = 2.1266, size = 477, normalized size = 14.91 \begin{align*} \frac{2 \, a \cosh \left (x\right ) -{\left (b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b\right )} \log \left (\frac{2 \,{\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) +{\left (b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b\right )} \log \left (\frac{2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \, a \sinh \left (x\right )}{a^{2} b \cosh \left (x\right )^{2} + a^{2} b \sinh \left (x\right )^{2} + 2 \, a^{3} \cosh \left (x\right ) - a^{2} b + 2 \,{\left (a^{2} b \cosh \left (x\right ) + a^{3}\right )} \sinh \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+b*sinh(x))^2,x, algorithm="fricas")

[Out]

(2*a*cosh(x) - (b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)*log(2*(b*sinh(x) + a)
/(cosh(x) - sinh(x))) + (b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)*log(2*sinh(x
)/(cosh(x) - sinh(x))) + 2*a*sinh(x))/(a^2*b*cosh(x)^2 + a^2*b*sinh(x)^2 + 2*a^3*cosh(x) - a^2*b + 2*(a^2*b*co
sh(x) + a^3)*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth{\left (x \right )}}{\left (a + b \sinh{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+b*sinh(x))**2,x)

[Out]

Integral(coth(x)/(a + b*sinh(x))**2, x)

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Giac [B]  time = 1.13401, size = 101, normalized size = 3.16 \begin{align*} -\frac{\log \left ({\left | -b{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{2}} + \frac{\log \left ({\left | -e^{\left (-x\right )} + e^{x} \right |}\right )}{a^{2}} + \frac{b{\left (e^{\left (-x\right )} - e^{x}\right )} - 4 \, a}{{\left (b{\left (e^{\left (-x\right )} - e^{x}\right )} - 2 \, a\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+b*sinh(x))^2,x, algorithm="giac")

[Out]

-log(abs(-b*(e^(-x) - e^x) + 2*a))/a^2 + log(abs(-e^(-x) + e^x))/a^2 + (b*(e^(-x) - e^x) - 4*a)/((b*(e^(-x) -
e^x) - 2*a)*a^2)