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

Optimal. Leaf size=53 \[ \frac{1}{a f \sqrt{a \cosh ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f} \]

[Out]

-(ArcTanh[Sqrt[a*Cosh[e + f*x]^2]/Sqrt[a]]/(a^(3/2)*f)) + 1/(a*f*Sqrt[a*Cosh[e + f*x]^2])

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Rubi [A]  time = 0.107651, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3176, 3205, 51, 63, 206} \[ \frac{1}{a f \sqrt{a \cosh ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sqrt[a*Cosh[e + f*x]^2]/Sqrt[a]]/(a^(3/2)*f)) + 1/(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 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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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)}{\left (a+a \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\int \frac{\coth (e+f x)}{\left (a \cosh ^2(e+f x)\right )^{3/2}} \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x) (a x)^{3/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac{1}{a f \sqrt{a \cosh ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a x}} \, dx,x,\cosh ^2(e+f x)\right )}{2 a f}\\ &=\frac{1}{a f \sqrt{a \cosh ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-\frac{x^2}{a}} \, dx,x,\sqrt{a \cosh ^2(e+f x)}\right )}{a^2 f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{a^{3/2} f}+\frac{1}{a f \sqrt{a \cosh ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0621527, size = 41, normalized size = 0.77 \[ \frac{\cosh (e+f x) \log \left (\tanh \left (\frac{1}{2} (e+f x)\right )\right )+1}{a f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.066, size = 44, normalized size = 0.8 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{1}{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}\sinh \left ( fx+e \right ) a}{\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)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.79287, size = 103, normalized size = 1.94 \begin{align*} \frac{2 \, \sqrt{a} e^{\left (-f x - e\right )}}{{\left (a^{2} e^{\left (-2 \, f x - 2 \, e\right )} + a^{2}\right )} f} - \frac{\log \left (e^{\left (-f x - e\right )} + 1\right )}{a^{\frac{3}{2}} f} + \frac{\log \left (e^{\left (-f x - e\right )} - 1\right )}{a^{\frac{3}{2}} 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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.85301, size = 706, normalized size = 13.32 \begin{align*} \frac{\sqrt{a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a}{\left (2 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} +{\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 )} \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 ) + 2 \, e^{\left (f x + e\right )} \sinh \left (f x + e\right )\right )} 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)/(a+a*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*(2*cosh(f*x + e)*e^(f*x + e) + (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))*log((cosh(f*x + e) + sinh(f*x
+ e) - 1)/(cosh(f*x + e) + sinh(f*x + e) + 1)) + 2*e^(f*x + e)*sinh(f*x + e))*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{\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)/(a+a*sinh(f*x+e)**2)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.35105, size = 84, normalized size = 1.58 \begin{align*} -\frac{\frac{\log \left (e^{\left (f x + e\right )} + 1\right )}{a^{\frac{3}{2}}} - \frac{\log \left ({\left | e^{\left (f x + e\right )} - 1 \right |}\right )}{a^{\frac{3}{2}}} - \frac{2 \, e^{\left (f x + e\right )}}{a^{\frac{3}{2}}{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}}}{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)^(3/2),x, algorithm="giac")

[Out]

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