3.453 \(\int \frac{\coth ^2(e+f x)}{(a+a \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=64 \[ -\frac{\coth (e+f x)}{a f \sqrt{a \cosh ^2(e+f x)}}-\frac{\cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{a f \sqrt{a \cosh ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.131732, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3176, 3207, 2621, 321, 207} \[ -\frac{\coth (e+f x)}{a f \sqrt{a \cosh ^2(e+f x)}}-\frac{\cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{a f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[e + f*x]^2/(a + a*Sinh[e + f*x]^2)^(3/2),x]

[Out]

-((ArcTan[Sinh[e + f*x]]*Cosh[e + f*x])/(a*f*Sqrt[a*Cosh[e + f*x]^2])) - Coth[e + f*x]/(a*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 2621

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
 c, n, m, p, 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 ^2(e+f x)}{\left (a+a \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\int \frac{\coth ^2(e+f x)}{\left (a \cosh ^2(e+f x)\right )^{3/2}} \, dx\\ &=\frac{\cosh (e+f x) \int \text{csch}^2(e+f x) \text{sech}(e+f x) \, dx}{a \sqrt{a \cosh ^2(e+f x)}}\\ &=-\frac{(i \cosh (e+f x)) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,-i \text{csch}(e+f x)\right )}{a f \sqrt{a \cosh ^2(e+f x)}}\\ &=-\frac{\coth (e+f x)}{a f \sqrt{a \cosh ^2(e+f x)}}-\frac{(i \cosh (e+f x)) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,-i \text{csch}(e+f x)\right )}{a f \sqrt{a \cosh ^2(e+f x)}}\\ &=-\frac{\tan ^{-1}(\sinh (e+f x)) \cosh (e+f x)}{a f \sqrt{a \cosh ^2(e+f x)}}-\frac{\coth (e+f x)}{a f \sqrt{a \cosh ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.0787, size = 46, normalized size = 0.72 \[ -\frac{\coth (e+f x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\sinh ^2(e+f x)\right )}{a f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[e + f*x]^2/(a + a*Sinh[e + f*x]^2)^(3/2),x]

[Out]

-((Coth[e + f*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Sinh[e + f*x]^2])/(a*f*Sqrt[a*Cosh[e + f*x]^2]))

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Maple [A]  time = 0.093, size = 51, normalized size = 0.8 \begin{align*} -{\frac{\cosh \left ( fx+e \right ) \left ( \arctan \left ( \sinh \left ( fx+e \right ) \right ) \sinh \left ( fx+e \right ) +1 \right ) }{a\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)^(3/2),x)

[Out]

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

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Maxima [B]  time = 1.71757, size = 433, normalized size = 6.77 \begin{align*} -\frac{\frac{3 \, \sqrt{a} e^{\left (-f x - e\right )} + 2 \, \sqrt{a} e^{\left (-3 \, f x - 3 \, e\right )} + 3 \, \sqrt{a} e^{\left (-5 \, f x - 5 \, e\right )}}{a^{2} e^{\left (-2 \, f x - 2 \, e\right )} - a^{2} e^{\left (-4 \, f x - 4 \, e\right )} - a^{2} e^{\left (-6 \, f x - 6 \, e\right )} + a^{2}} - \frac{3 \, \arctan \left (e^{\left (-f x - e\right )}\right )}{a^{\frac{3}{2}}}}{2 \, f} - \frac{5 \, \sqrt{a} e^{\left (-f x - e\right )} + 6 \, \sqrt{a} e^{\left (-3 \, f x - 3 \, e\right )} - 3 \, \sqrt{a} e^{\left (-5 \, f x - 5 \, e\right )}}{4 \,{\left (a^{2} e^{\left (-2 \, f x - 2 \, e\right )} - a^{2} e^{\left (-4 \, f x - 4 \, e\right )} - a^{2} e^{\left (-6 \, f x - 6 \, e\right )} + a^{2}\right )} f} + \frac{3 \, \sqrt{a} e^{\left (-f x - e\right )} - 6 \, \sqrt{a} e^{\left (-3 \, f x - 3 \, e\right )} - 5 \, \sqrt{a} e^{\left (-5 \, f x - 5 \, e\right )}}{4 \,{\left (a^{2} e^{\left (-2 \, f x - 2 \, e\right )} - a^{2} e^{\left (-4 \, f x - 4 \, e\right )} - a^{2} e^{\left (-6 \, f x - 6 \, e\right )} + a^{2}\right )} f} + \frac{\arctan \left (e^{\left (-f x - e\right )}\right )}{2 \, a^{\frac{3}{2}} 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)^(3/2),x, algorithm="maxima")

[Out]

-1/2*((3*sqrt(a)*e^(-f*x - e) + 2*sqrt(a)*e^(-3*f*x - 3*e) + 3*sqrt(a)*e^(-5*f*x - 5*e))/(a^2*e^(-2*f*x - 2*e)
 - a^2*e^(-4*f*x - 4*e) - a^2*e^(-6*f*x - 6*e) + a^2) - 3*arctan(e^(-f*x - e))/a^(3/2))/f - 1/4*(5*sqrt(a)*e^(
-f*x - e) + 6*sqrt(a)*e^(-3*f*x - 3*e) - 3*sqrt(a)*e^(-5*f*x - 5*e))/((a^2*e^(-2*f*x - 2*e) - a^2*e^(-4*f*x -
4*e) - a^2*e^(-6*f*x - 6*e) + a^2)*f) + 1/4*(3*sqrt(a)*e^(-f*x - e) - 6*sqrt(a)*e^(-3*f*x - 3*e) - 5*sqrt(a)*e
^(-5*f*x - 5*e))/((a^2*e^(-2*f*x - 2*e) - a^2*e^(-4*f*x - 4*e) - a^2*e^(-6*f*x - 6*e) + a^2)*f) + 1/2*arctan(e
^(-f*x - e))/(a^(3/2)*f)

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Fricas [B]  time = 1.80605, size = 652, normalized size = 10.19 \begin{align*} -\frac{2 \,{\left ({\left (2 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right ) + e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} +{\left (\cosh \left (f x + e\right )^{2} - 1\right )} e^{\left (f x + e\right )}\right )} \arctan \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right ) + \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 )} \sqrt{a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} e^{\left (-f x - e\right )}}{a^{2} f \cosh \left (f x + e\right )^{2} - a^{2} f +{\left (a^{2} f e^{\left (2 \, f x + 2 \, e\right )} + a^{2} f\right )} \sinh \left (f x + e\right )^{2} +{\left (a^{2} f \cosh \left (f x + e\right )^{2} - a^{2} f\right )} e^{\left (2 \, f x + 2 \, e\right )} + 2 \,{\left (a^{2} f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + a^{2} 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)^(3/2),x, algorithm="fricas")

[Out]

-2*((2*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e) + e^(f*x + e)*sinh(f*x + e)^2 + (cosh(f*x + e)^2 - 1)*e^(f*x +
e))*arctan(cosh(f*x + e) + sinh(f*x + e)) + cosh(f*x + e)*e^(f*x + e) + e^(f*x + e)*sinh(f*x + e))*sqrt(a*e^(4
*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*e^(-f*x - e)/(a^2*f*cosh(f*x + e)^2 - a^2*f + (a^2*f*e^(2*f*x + 2*e) +
a^2*f)*sinh(f*x + e)^2 + (a^2*f*cosh(f*x + e)^2 - a^2*f)*e^(2*f*x + 2*e) + 2*(a^2*f*cosh(f*x + e)*e^(2*f*x + 2
*e) + a^2*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 )}}{\left (a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, 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)**(3/2),x)

[Out]

Integral(coth(e + f*x)**2/(a*(sinh(e + f*x)**2 + 1))**(3/2), x)

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Giac [A]  time = 1.42971, size = 58, normalized size = 0.91 \begin{align*} -\frac{2 \,{\left (\frac{\arctan \left (e^{\left (f x + e\right )}\right )}{a^{\frac{3}{2}}} + \frac{e^{\left (f x + e\right )}}{a^{\frac{3}{2}}{\left (e^{\left (2 \, f x + 2 \, e\right )} - 1\right )}}\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)^(3/2),x, algorithm="giac")

[Out]

-2*(arctan(e^(f*x + e))/a^(3/2) + e^(f*x + e)/(a^(3/2)*(e^(2*f*x + 2*e) - 1)))/f