3.482 \(\int \frac{\coth (e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=33 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f} \]

[Out]

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

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Rubi [A]  time = 0.070531, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3194, 63, 208} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

Int[Coth[e + f*x]/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

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

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 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 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 \frac{\coth (e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^2(e+f x)}\right )}{b f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f}\\ \end{align*}

Mathematica [A]  time = 0.0353481, size = 33, normalized size = 1. \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[e + f*x]/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

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

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Maple [C]  time = 0.073, size = 35, normalized size = 1.1 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{1}{\sinh \left ( fx+e \right ) }{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}},\sinh \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

`int/indef0`(1/sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(coth(f*x + e)/sqrt(b*sinh(f*x + e)^2 + a), x)

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Fricas [B]  time = 2.22778, size = 1111, normalized size = 33.67 \begin{align*} \left [\frac{\log \left (\frac{b \cosh \left (f x + e\right )^{4} + 4 \, b \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + b \sinh \left (f x + e\right )^{4} + 2 \,{\left (4 \, a - b\right )} \cosh \left (f x + e\right )^{2} + 2 \,{\left (3 \, b \cosh \left (f x + e\right )^{2} + 4 \, a - b\right )} \sinh \left (f x + e\right )^{2} - 4 \, \sqrt{2} \sqrt{a} \sqrt{\frac{b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}{\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 4 \,{\left (b \cosh \left (f x + e\right )^{3} +{\left (4 \, a - b\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + b}{\cosh \left (f x + e\right )^{4} + 4 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} + \sinh \left (f x + e\right )^{4} + 2 \,{\left (3 \, \cosh \left (f x + e\right )^{2} - 1\right )} \sinh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right )^{2} + 4 \,{\left (\cosh \left (f x + e\right )^{3} - \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + 1}\right )}{2 \, \sqrt{a} f}, \frac{\sqrt{-a} \arctan \left (\frac{\sqrt{2} \sqrt{-a} \sqrt{\frac{b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{2 \,{\left (a \cosh \left (f x + e\right ) + a \sinh \left (f x + e\right )\right )}}\right )}{a f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*log((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(4*a - b)*cosh(f*x + e
)^2 + 2*(3*b*cosh(f*x + e)^2 + 4*a - b)*sinh(f*x + e)^2 - 4*sqrt(2)*sqrt(a)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f
*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))*(cosh(f*x + e) + sin
h(f*x + e)) + 4*(b*cosh(f*x + e)^3 + (4*a - b)*cosh(f*x + e))*sinh(f*x + e) + b)/(cosh(f*x + e)^4 + 4*cosh(f*x
+ e)*sinh(f*x + e)^3 + sinh(f*x + e)^4 + 2*(3*cosh(f*x + e)^2 - 1)*sinh(f*x + e)^2 - 2*cosh(f*x + e)^2 + 4*(c
osh(f*x + e)^3 - cosh(f*x + e))*sinh(f*x + e) + 1))/(sqrt(a)*f), sqrt(-a)*arctan(1/2*sqrt(2)*sqrt(-a)*sqrt((b*
cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e
)^2))/(a*cosh(f*x + e) + a*sinh(f*x + e)))/(a*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError