3.245 \(\int \frac{\coth (x)}{\sqrt{a+b \sinh (x)}} \, dx\)

Optimal. Leaf size=24 \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh (x)}}{\sqrt{a}}\right )}{\sqrt{a}} \]

[Out]

(-2*ArcTanh[Sqrt[a + b*Sinh[x]]/Sqrt[a]])/Sqrt[a]

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Rubi [A]  time = 0.0589528, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2721, 63, 207} \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh (x)}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/Sqrt[a + b*Sinh[x]],x]

[Out]

(-2*ArcTanh[Sqrt[a + b*Sinh[x]]/Sqrt[a]])/Sqrt[a]

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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0126825, size = 24, normalized size = 1. \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh (x)}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]/Sqrt[a + b*Sinh[x]],x]

[Out]

(-2*ArcTanh[Sqrt[a + b*Sinh[x]]/Sqrt[a]])/Sqrt[a]

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Maple [A]  time = 0.016, size = 19, normalized size = 0.8 \begin{align*} -2\,{\frac{1}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{a+b\sinh \left ( x \right ) }}{\sqrt{a}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)/(a+b*sinh(x))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+b*sinh(x))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.58261, size = 1099, normalized size = 45.79 \begin{align*} \left [\frac{\log \left (\frac{b^{2} \cosh \left (x\right )^{4} + b^{2} \sinh \left (x\right )^{4} + 16 \, a b \cosh \left (x\right )^{3} + 4 \,{\left (b^{2} \cosh \left (x\right ) + 4 \, a b\right )} \sinh \left (x\right )^{3} - 16 \, a b \cosh \left (x\right ) + 2 \,{\left (16 \, a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (3 \, b^{2} \cosh \left (x\right )^{2} + 24 \, a b \cosh \left (x\right ) + 16 \, a^{2} - b^{2}\right )} \sinh \left (x\right )^{2} - 8 \,{\left (b \cosh \left (x\right )^{3} + b \sinh \left (x\right )^{3} + 4 \, a \cosh \left (x\right )^{2} +{\left (3 \, b \cosh \left (x\right ) + 4 \, a\right )} \sinh \left (x\right )^{2} - b \cosh \left (x\right ) +{\left (3 \, b \cosh \left (x\right )^{2} + 8 \, a \cosh \left (x\right ) - b\right )} \sinh \left (x\right )\right )} \sqrt{b \sinh \left (x\right ) + a} \sqrt{a} + b^{2} + 4 \,{\left (b^{2} \cosh \left (x\right )^{3} + 12 \, a b \cosh \left (x\right )^{2} - 4 \, a b +{\left (16 \, a^{2} - b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1}\right )}{2 \, \sqrt{a}}, \frac{\sqrt{-a} \arctan \left (\frac{{\left (b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 4 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + 2 \, a\right )} \sinh \left (x\right ) - b\right )} \sqrt{b \sinh \left (x\right ) + a} \sqrt{-a}}{2 \,{\left (a b \cosh \left (x\right )^{2} + a b \sinh \left (x\right )^{2} + 2 \, a^{2} \cosh \left (x\right ) - a b + 2 \,{\left (a b \cosh \left (x\right ) + a^{2}\right )} \sinh \left (x\right )\right )}}\right )}{a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+b*sinh(x))^(1/2),x, algorithm="fricas")

[Out]

[1/2*log((b^2*cosh(x)^4 + b^2*sinh(x)^4 + 16*a*b*cosh(x)^3 + 4*(b^2*cosh(x) + 4*a*b)*sinh(x)^3 - 16*a*b*cosh(x
) + 2*(16*a^2 - b^2)*cosh(x)^2 + 2*(3*b^2*cosh(x)^2 + 24*a*b*cosh(x) + 16*a^2 - b^2)*sinh(x)^2 - 8*(b*cosh(x)^
3 + b*sinh(x)^3 + 4*a*cosh(x)^2 + (3*b*cosh(x) + 4*a)*sinh(x)^2 - b*cosh(x) + (3*b*cosh(x)^2 + 8*a*cosh(x) - b
)*sinh(x))*sqrt(b*sinh(x) + a)*sqrt(a) + b^2 + 4*(b^2*cosh(x)^3 + 12*a*b*cosh(x)^2 - 4*a*b + (16*a^2 - b^2)*co
sh(x))*sinh(x))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 - 1)*sinh(x)^2 - 2*cosh(x)^2 + 4
*(cosh(x)^3 - cosh(x))*sinh(x) + 1))/sqrt(a), sqrt(-a)*arctan(1/2*(b*cosh(x)^2 + b*sinh(x)^2 + 4*a*cosh(x) + 2
*(b*cosh(x) + 2*a)*sinh(x) - b)*sqrt(b*sinh(x) + a)*sqrt(-a)/(a*b*cosh(x)^2 + a*b*sinh(x)^2 + 2*a^2*cosh(x) -
a*b + 2*(a*b*cosh(x) + a^2)*sinh(x)))/a]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+b*sinh(x))**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+b*sinh(x))^(1/2),x, algorithm="giac")

[Out]

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