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

Optimal. Leaf size=25 \[ -\frac{\coth (e+f x)}{f \sqrt{a \cosh ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.104443, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3176, 3207, 2606, 8} \[ -\frac{\coth (e+f x)}{f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.0373885, size = 25, normalized size = 1. \[ -\frac{\coth (e+f x)}{f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.07, size = 32, normalized size = 1.3 \begin{align*} -{\frac{\cosh \left ( fx+e \right ) }{\sinh \left ( fx+e \right ) f}{\frac{1}{\sqrt{a \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cosh(f*x+e)/sinh(f*x+e)/(a*cosh(f*x+e)^2)^(1/2)/f

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Maxima [B]  time = 1.84918, size = 136, normalized size = 5.44 \begin{align*} \frac{\frac{\arctan \left (e^{\left (-f x - e\right )}\right )}{\sqrt{a}} + \frac{\sqrt{a} e^{\left (-f x - e\right )}}{a e^{\left (-2 \, f x - 2 \, e\right )} - a}}{f} - \frac{\arctan \left (e^{\left (-f x - e\right )}\right )}{\sqrt{a} f} + \frac{\sqrt{a} e^{\left (-f x - e\right )}}{{\left (a e^{\left (-2 \, f x - 2 \, e\right )} - a\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(arctan(e^(-f*x - e))/sqrt(a) + sqrt(a)*e^(-f*x - e)/(a*e^(-2*f*x - 2*e) - a))/f - arctan(e^(-f*x - e))/(sqrt(
a)*f) + sqrt(a)*e^(-f*x - e)/((a*e^(-2*f*x - 2*e) - a)*f)

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Fricas [B]  time = 1.87446, size = 428, normalized size = 17.12 \begin{align*} -\frac{2 \, \sqrt{a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a}{\left (\cosh \left (f x + e\right ) e^{\left (f x + e\right )} + e^{\left (f x + e\right )} \sinh \left (f x + e\right )\right )} e^{\left (-f x - e\right )}}{a f \cosh \left (f x + e\right )^{2} +{\left (a f e^{\left (2 \, f x + 2 \, e\right )} + a f\right )} \sinh \left (f x + e\right )^{2} - a f +{\left (a f \cosh \left (f x + e\right )^{2} - a f\right )} e^{\left (2 \, f x + 2 \, e\right )} + 2 \,{\left (a f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + a f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*(cosh(f*x + e)*e^(f*x + e) + e^(f*x + e)*sinh(f*x + e))*e
^(-f*x - e)/(a*f*cosh(f*x + e)^2 + (a*f*e^(2*f*x + 2*e) + a*f)*sinh(f*x + e)^2 - a*f + (a*f*cosh(f*x + e)^2 -
a*f)*e^(2*f*x + 2*e) + 2*(a*f*cosh(f*x + e)*e^(2*f*x + 2*e) + a*f*cosh(f*x + e))*sinh(f*x + e))

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

[Out]

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

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Giac [A]  time = 1.37947, size = 39, normalized size = 1.56 \begin{align*} -\frac{2 \, e^{\left (f x + e\right )}}{\sqrt{a} f{\left (e^{\left (2 \, f x + 2 \, e\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^(f*x + e)/(sqrt(a)*f*(e^(2*f*x + 2*e) - 1))