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

Optimal. Leaf size=66 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{3/2} f}-\frac{\text{csch}^2(e+f x) \sqrt{a \cosh ^2(e+f x)}}{2 a^2 f} \]

[Out]

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

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Rubi [A]  time = 0.139467, 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, 51, 63, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{3/2} f}-\frac{\text{csch}^2(e+f x) \sqrt{a \cosh ^2(e+f x)}}{2 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sqrt[a*Cosh[e + f*x]^2]/Sqrt[a]]/(2*a^(3/2)*f) - (Sqrt[a*Cosh[e + f*x]^2]*Csch[e + f*x]^2)/(2*a^2*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 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 ^3(e+f x)}{\left (a+a \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\int \frac{\coth ^3(e+f x)}{\left (a \cosh ^2(e+f x)\right )^{3/2}} \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{(1-x)^2 (a x)^{3/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x)^2 \sqrt{a x}} \, 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^2 f}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a x}} \, dx,x,\cosh ^2(e+f x)\right )}{4 a f}\\ &=-\frac{\sqrt{a \cosh ^2(e+f x)} \text{csch}^2(e+f x)}{2 a^2 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^2 f}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{3/2} f}-\frac{\sqrt{a \cosh ^2(e+f x)} \text{csch}^2(e+f x)}{2 a^2 f}\\ \end{align*}

Mathematica [A]  time = 0.131715, size = 67, normalized size = 1.02 \[ -\frac{\cosh ^3(e+f x) \left (\text{csch}^2\left (\frac{1}{2} (e+f x)\right )+\text{sech}^2\left (\frac{1}{2} (e+f x)\right )+4 \log \left (\tanh \left (\frac{1}{2} (e+f x)\right )\right )\right )}{8 f \left (a \cosh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cosh[e + f*x]^3*(Csch[(e + f*x)/2]^2 + 4*Log[Tanh[(e + f*x)/2]] + Sech[(e + f*x)/2]^2))/(8*f*(a*Cosh[e + f*x
]^2)^(3/2))

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Maple [C]  time = 0.069, size = 36, normalized size = 0.6 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{1}{ \left ( \sinh \left ( fx+e \right ) \right ) ^{3}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)^3/(a+a*sinh(f*x+e)^2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 2.11193, size = 135, normalized size = 2.05 \begin{align*} \frac{e^{\left (-f x - e\right )} + e^{\left (-3 \, f x - 3 \, e\right )}}{{\left (2 \, a^{\frac{3}{2}} e^{\left (-2 \, f x - 2 \, e\right )} - a^{\frac{3}{2}} e^{\left (-4 \, f x - 4 \, e\right )} - a^{\frac{3}{2}}\right )} f} + \frac{\log \left (e^{\left (-f x - e\right )} + 1\right )}{2 \, a^{\frac{3}{2}} f} - \frac{\log \left (e^{\left (-f x - e\right )} - 1\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)^3/(a+a*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.87461, size = 1458, normalized size = 22.09 \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)^(3/2),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^2*f*cosh(f*x + e)^4 - 2*a^2*f*cosh(f*x + e)^2 + (a^2*f*e^(2*f*x + 2*e) + a^2
*f)*sinh(f*x + e)^4 + 4*(a^2*f*cosh(f*x + e)*e^(2*f*x + 2*e) + a^2*f*cosh(f*x + e))*sinh(f*x + e)^3 + a^2*f +
2*(3*a^2*f*cosh(f*x + e)^2 - a^2*f + (3*a^2*f*cosh(f*x + e)^2 - a^2*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^2 + (a^2
*f*cosh(f*x + e)^4 - 2*a^2*f*cosh(f*x + e)^2 + a^2*f)*e^(2*f*x + 2*e) + 4*(a^2*f*cosh(f*x + e)^3 - a^2*f*cosh(
f*x + e) + (a^2*f*cosh(f*x + e)^3 - a^2*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 )}}{\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)**3/(a+a*sinh(f*x+e)**2)**(3/2),x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError