3.525 \(\int \coth (x) \sqrt{a+b \sinh ^n(x)} \, dx\)

Optimal. Leaf size=47 \[ \frac{2 \sqrt{a+b \sinh ^n(x)}}{n}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^n(x)}}{\sqrt{a}}\right )}{n} \]

[Out]

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

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Rubi [A]  time = 0.0919908, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3230, 266, 50, 63, 208} \[ \frac{2 \sqrt{a+b \sinh ^n(x)}}{n}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^n(x)}}{\sqrt{a}}\right )}{n} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]*Sqrt[a + b*Sinh[x]^n],x]

[Out]

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

Rule 3230

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + b*(c*ff*x)^n)^p)/(1 - ff^2*x^2)^(
(m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 208

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

Rubi steps

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

Mathematica [A]  time = 0.0208877, size = 45, normalized size = 0.96 \[ \frac{2 \sqrt{a+b \sinh ^n(x)}-2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^n(x)}}{\sqrt{a}}\right )}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 38, normalized size = 0.8 \begin{align*}{\frac{1}{n} \left ( 2\,\sqrt{a+b \left ( \sinh \left ( x \right ) \right ) ^{n}}-2\,\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{a+b \left ( \sinh \left ( x \right ) \right ) ^{n}}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.81846, size = 552, normalized size = 11.74 \begin{align*} \left [\frac{\sqrt{a} \log \left (\frac{b \cosh \left (n \log \left (\sinh \left (x\right )\right )\right ) + b \sinh \left (n \log \left (\sinh \left (x\right )\right )\right ) - 2 \, \sqrt{b \cosh \left (n \log \left (\sinh \left (x\right )\right )\right ) + b \sinh \left (n \log \left (\sinh \left (x\right )\right )\right ) + a} \sqrt{a} + 2 \, a}{\cosh \left (n \log \left (\sinh \left (x\right )\right )\right ) + \sinh \left (n \log \left (\sinh \left (x\right )\right )\right )}\right ) + 2 \, \sqrt{b \cosh \left (n \log \left (\sinh \left (x\right )\right )\right ) + b \sinh \left (n \log \left (\sinh \left (x\right )\right )\right ) + a}}{n}, \frac{2 \,{\left (\sqrt{-a} \arctan \left (\frac{\sqrt{b \cosh \left (n \log \left (\sinh \left (x\right )\right )\right ) + b \sinh \left (n \log \left (\sinh \left (x\right )\right )\right ) + a} \sqrt{-a}}{a}\right ) + \sqrt{b \cosh \left (n \log \left (\sinh \left (x\right )\right )\right ) + b \sinh \left (n \log \left (\sinh \left (x\right )\right )\right ) + a}\right )}}{n}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(a)*log((b*cosh(n*log(sinh(x))) + b*sinh(n*log(sinh(x))) - 2*sqrt(b*cosh(n*log(sinh(x))) + b*sinh(n*log(
sinh(x))) + a)*sqrt(a) + 2*a)/(cosh(n*log(sinh(x))) + sinh(n*log(sinh(x))))) + 2*sqrt(b*cosh(n*log(sinh(x))) +
b*sinh(n*log(sinh(x))) + a))/n, 2*(sqrt(-a)*arctan(sqrt(b*cosh(n*log(sinh(x))) + b*sinh(n*log(sinh(x))) + a)*
sqrt(-a)/a) + sqrt(b*cosh(n*log(sinh(x))) + b*sinh(n*log(sinh(x))) + a))/n]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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