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

Optimal. Leaf size=31 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0856261, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3176, 3205, 63, 206} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/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 206

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

Rubi steps

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

Mathematica [A]  time = 0.053196, size = 49, normalized size = 1.58 \[ \frac{\cosh (e+f x) \left (\log \left (\sinh \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\cosh \left (\frac{1}{2} (e+f x)\right )\right )\right )}{f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.07, size = 33, 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 \left ( \cosh \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+a*sinh(f*x+e)^2)^(1/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.86608, size = 54, normalized size = 1.74 \begin{align*} -\frac{\log \left (e^{\left (-f x - e\right )} + 1\right )}{\sqrt{a} f} + \frac{\log \left (e^{\left (-f x - e\right )} - 1\right )}{\sqrt{a} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(e^(-f*x - e) + 1)/(sqrt(a)*f) + log(e^(-f*x - e) - 1)/(sqrt(a)*f)

________________________________________________________________________________________

Fricas [B]  time = 1.84154, size = 455, normalized size = 14.68 \begin{align*} \left [\frac{\sqrt{a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} \log \left (\frac{\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - 1}{\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + 1}\right )}{a f e^{\left (2 \, f x + 2 \, e\right )} + a f}, \frac{2 \, \sqrt{-a} \arctan \left (\frac{\sqrt{a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} \sqrt{-a}}{a \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + a \cosh \left (f x + e\right ) +{\left (a e^{\left (2 \, f x + 2 \, e\right )} + a\right )} \sinh \left (f x + e\right )}\right )}{a f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*log((cosh(f*x + e) + sinh(f*x + e) - 1)/(cosh(f*x + e) + si
nh(f*x + e) + 1))/(a*f*e^(2*f*x + 2*e) + a*f), 2*sqrt(-a)*arctan(sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e)
+ a)*sqrt(-a)/(a*cosh(f*x + e)*e^(2*f*x + 2*e) + a*cosh(f*x + e) + (a*e^(2*f*x + 2*e) + a)*sinh(f*x + e)))/(a*
f)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth{\left (e + f x \right )}}{\sqrt{a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(coth(e + f*x)/sqrt(a*(sinh(e + f*x)**2 + 1)), x)

________________________________________________________________________________________

Giac [A]  time = 1.32173, size = 49, normalized size = 1.58 \begin{align*} -\frac{\frac{\log \left (e^{\left (f x + e\right )} + 1\right )}{\sqrt{a}} - \frac{\log \left ({\left | e^{\left (f x + e\right )} - 1 \right |}\right )}{\sqrt{a}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(e^(f*x + e) + 1)/sqrt(a) - log(abs(e^(f*x + e) - 1))/sqrt(a))/f