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

Optimal. Leaf size=66 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{2 \sqrt{a} f}-\frac{\text{csch}^2(e+f x) \sqrt{a \cosh ^2(e+f x)}}{2 a f} \]

[Out]

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

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Rubi [A]  time = 0.129395, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3176, 3205, 16, 47, 63, 206} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{2 \sqrt{a} f}-\frac{\text{csch}^2(e+f x) \sqrt{a \cosh ^2(e+f x)}}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Sqrt[a*Cosh[e + f*x]^2]/Sqrt[a]]/(2*Sqrt[a]*f) - (Sqrt[a*Cosh[e + f*x]^2]*Csch[e + f*x]^2)/(2*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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 ^3(e+f x)}{\sqrt{a+a \sinh ^2(e+f x)}} \, dx &=\int \frac{\coth ^3(e+f x)}{\sqrt{a \cosh ^2(e+f x)}} \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{(1-x)^2 \sqrt{a x}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a x}}{(1-x)^2} \, dx,x,\cosh ^2(e+f x)\right )}{2 a f}\\ &=-\frac{\sqrt{a \cosh ^2(e+f x)} \text{csch}^2(e+f x)}{2 a f}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a x}} \, dx,x,\cosh ^2(e+f x)\right )}{4 f}\\ &=-\frac{\sqrt{a \cosh ^2(e+f x)} \text{csch}^2(e+f x)}{2 a f}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-\frac{x^2}{a}} \, dx,x,\sqrt{a \cosh ^2(e+f x)}\right )}{2 a f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{2 \sqrt{a} f}-\frac{\sqrt{a \cosh ^2(e+f x)} \text{csch}^2(e+f x)}{2 a f}\\ \end{align*}

Mathematica [A]  time = 0.122417, size = 65, normalized size = 0.98 \[ -\frac{\cosh (e+f x) \left (\text{csch}^2\left (\frac{1}{2} (e+f x)\right )+\text{sech}^2\left (\frac{1}{2} (e+f x)\right )-4 \log \left (\tanh \left (\frac{1}{2} (e+f x)\right )\right )\right )}{8 f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cosh[e + f*x]*(Csch[(e + f*x)/2]^2 - 4*Log[Tanh[(e + f*x)/2]] + Sech[(e + f*x)/2]^2))/(8*f*Sqrt[a*Cosh[e + f
*x]^2])

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Maple [C]  time = 0.086, size = 42, normalized size = 0.6 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({( \left ( \sinh \left ( fx+e \right ) \right ) ^{-1}+ \left ( \sinh \left ( fx+e \right ) \right ) ^{-3}){\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)^3/(a+a*sinh(f*x+e)^2)^(1/2),x)

[Out]

`int/indef0`((1/sinh(f*x+e)+1/sinh(f*x+e)^3)/(a*cosh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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Maxima [A]  time = 1.90444, size = 135, normalized size = 2.05 \begin{align*} -\frac{\log \left (e^{\left (-f x - e\right )} + 1\right )}{2 \, \sqrt{a} f} + \frac{\log \left (e^{\left (-f x - e\right )} - 1\right )}{2 \, \sqrt{a} f} + \frac{e^{\left (-f x - e\right )} + e^{\left (-3 \, f x - 3 \, e\right )}}{{\left (2 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} - \sqrt{a} e^{\left (-4 \, f x - 4 \, e\right )} - \sqrt{a}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^3/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.94943, size = 1409, normalized size = 21.35 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^3/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*(6*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^2 + 2*e^(f*x + e)*sinh(f*x + e)^3 + 2*(3*cosh(f*x + e)^2 + 1)*
e^(f*x + e)*sinh(f*x + e) + 2*(cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e) - (4*cosh(f*x + e)*e^(f*x + e)*sin
h(f*x + e)^3 + e^(f*x + e)*sinh(f*x + e)^4 + 2*(3*cosh(f*x + e)^2 - 1)*e^(f*x + e)*sinh(f*x + e)^2 + 4*(cosh(f
*x + e)^3 - cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e) + (cosh(f*x + e)^4 - 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)))*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(
2*f*x + 2*e) + a)*e^(-f*x - e)/(a*f*cosh(f*x + e)^4 + (a*f*e^(2*f*x + 2*e) + a*f)*sinh(f*x + e)^4 - 2*a*f*cosh
(f*x + e)^2 + 4*(a*f*cosh(f*x + e)*e^(2*f*x + 2*e) + a*f*cosh(f*x + e))*sinh(f*x + e)^3 + 2*(3*a*f*cosh(f*x +
e)^2 - a*f + (3*a*f*cosh(f*x + e)^2 - a*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^2 + a*f + (a*f*cosh(f*x + e)^4 - 2*a
*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x + 2*e) + 4*(a*f*cosh(f*x + e)^3 - a*f*cosh(f*x + e) + (a*f*cosh(f*x + e)^3
- a*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{3}{\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)**3/(a+a*sinh(f*x+e)**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.36714, size = 109, normalized size = 1.65 \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}} + \frac{2 \,{\left (\sqrt{a} e^{\left (3 \, f x + 3 \, e\right )} + \sqrt{a} e^{\left (f x + e\right )}\right )}}{a{\left (e^{\left (2 \, f x + 2 \, e\right )} - 1\right )}^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^3/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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