3.3.9 \(\int \frac {\coth (x)}{(a+b \text {sech}^2(x))^{3/2}} \, dx\) [209]

3.3.9.1 Optimal result
3.3.9.2 Mathematica [A] (verified)
3.3.9.3 Rubi [A] (verified)
3.3.9.4 Maple [F]
3.3.9.5 Fricas [B] (verification not implemented)
3.3.9.6 Sympy [F]
3.3.9.7 Maxima [F]
3.3.9.8 Giac [F(-2)]
3.3.9.9 Mupad [F(-1)]

3.3.9.1 Optimal result

Integrand size = 15, antiderivative size = 79 \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {b}{a (a+b) \sqrt {a+b \text {sech}^2(x)}} \]

output
arctanh((a+b*sech(x)^2)^(1/2)/a^(1/2))/a^(3/2)-arctanh((a+b*sech(x)^2)^(1/ 
2)/(a+b)^(1/2))/(a+b)^(3/2)-b/a/(a+b)/(a+b*sech(x)^2)^(1/2)
 
3.3.9.2 Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.96 \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\frac {\text {sech}^2(x) \left (-2 b (a+2 b+a \cosh (2 x))+\frac {\sqrt {2} (a+2 b+a \cosh (2 x))^{3/2} \left (-a^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a+b} \cosh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )+(a+b)^{3/2} \log \left (\sqrt {2} \sqrt {a} \cosh (x)+\sqrt {a+2 b+a \cosh (2 x)}\right )\right ) \text {sech}(x)}{\sqrt {a} \sqrt {a+b}}\right )}{4 a (a+b) \left (a+b \text {sech}^2(x)\right )^{3/2}} \]

input
Integrate[Coth[x]/(a + b*Sech[x]^2)^(3/2),x]
 
output
(Sech[x]^2*(-2*b*(a + 2*b + a*Cosh[2*x]) + (Sqrt[2]*(a + 2*b + a*Cosh[2*x] 
)^(3/2)*(-(a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a + b]*Cosh[x])/Sqrt[a + 2*b + a* 
Cosh[2*x]]]) + (a + b)^(3/2)*Log[Sqrt[2]*Sqrt[a]*Cosh[x] + Sqrt[a + 2*b + 
a*Cosh[2*x]]])*Sech[x])/(Sqrt[a]*Sqrt[a + b])))/(4*a*(a + b)*(a + b*Sech[x 
]^2)^(3/2))
 
3.3.9.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.25, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 26, 4627, 25, 354, 96, 25, 174, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i}{\tan (i x) \left (a+b \sec (i x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {1}{\left (b \sec (i x)^2+a\right )^{3/2} \tan (i x)}dx\)

\(\Big \downarrow \) 4627

\(\displaystyle \int -\frac {\cosh (x)}{\left (1-\text {sech}^2(x)\right ) \left (a+b \text {sech}^2(x)\right )^{3/2}}d\text {sech}(x)\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\cosh (x)}{\left (1-\text {sech}^2(x)\right ) \left (b \text {sech}^2(x)+a\right )^{3/2}}d\text {sech}(x)\)

\(\Big \downarrow \) 354

\(\displaystyle -\frac {1}{2} \int \frac {\cosh (x)}{\left (1-\text {sech}^2(x)\right ) \left (b \text {sech}^2(x)+a\right )^{3/2}}d\text {sech}^2(x)\)

\(\Big \downarrow \) 96

\(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {\cosh (x) \left (-b \text {sech}^2(x)+a+b\right )}{\left (1-\text {sech}^2(x)\right ) \sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)}{a (a+b)}-\frac {2 b}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {\cosh (x) \left (-b \text {sech}^2(x)+a+b\right )}{\left (1-\text {sech}^2(x)\right ) \sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)}{a (a+b)}-\frac {2 b}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}\right )\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{2} \left (-\frac {a \int \frac {1}{\left (1-\text {sech}^2(x)\right ) \sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)+(a+b) \int \frac {\cosh (x)}{\sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)}{a (a+b)}-\frac {2 b}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2 a \int \frac {1}{\frac {a+b}{b}-\frac {\text {sech}^4(x)}{b}}d\sqrt {b \text {sech}^2(x)+a}}{b}+\frac {2 (a+b) \int \frac {1}{\frac {\text {sech}^4(x)}{b}-\frac {a}{b}}d\sqrt {b \text {sech}^2(x)+a}}{b}}{a (a+b)}-\frac {2 b}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2 a \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {2 (a+b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}}{a (a+b)}-\frac {2 b}{a (a+b) \sqrt {a+b \text {sech}^2(x)}}\right )\)

input
Int[Coth[x]/(a + b*Sech[x]^2)^(3/2),x]
 
output
(-(((-2*(a + b)*ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]])/Sqrt[a] + (2*a*Arc 
Tanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a + b]])/Sqrt[a + b])/(a*(a + b))) - (2*b) 
/(a*(a + b)*Sqrt[a + b*Sech[x]^2]))/2
 

3.3.9.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 26
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 96
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[f*((e + f*x)^(p + 1)/((p + 1)*(b*e - a*f)*(d*e - c*f))), x] + S 
imp[1/((b*e - a*f)*(d*e - c*f))   Int[(b*d*e - b*c*f - a*d*f - b*d*f*x)*((e 
 + f*x)^(p + 1)/((a + b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, 
 x] && LtQ[p, -1]
 

rule 174
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 354
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4627
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] 
, x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( 
m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 
3.3.9.4 Maple [F]

\[\int \frac {\coth \left (x \right )}{\left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {3}{2}}}d x\]

input
int(coth(x)/(a+sech(x)^2*b)^(3/2),x)
 
output
int(coth(x)/(a+sech(x)^2*b)^(3/2),x)
 
3.3.9.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1330 vs. \(2 (65) = 130\).

Time = 0.47 (sec) , antiderivative size = 6939, normalized size of antiderivative = 87.84 \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\text {Too large to display} \]

input
integrate(coth(x)/(a+b*sech(x)^2)^(3/2),x, algorithm="fricas")
 
output
Too large to include
 
3.3.9.6 Sympy [F]

\[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int \frac {\coth {\left (x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \]

input
integrate(coth(x)/(a+b*sech(x)**2)**(3/2),x)
 
output
Integral(coth(x)/(a + b*sech(x)**2)**(3/2), x)
 
3.3.9.7 Maxima [F]

\[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int { \frac {\coth \left (x\right )}{{\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

input
integrate(coth(x)/(a+b*sech(x)^2)^(3/2),x, algorithm="maxima")
 
output
integrate(coth(x)/(b*sech(x)^2 + a)^(3/2), x)
 
3.3.9.8 Giac [F(-2)]

Exception generated. \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

input
integrate(coth(x)/(a+b*sech(x)^2)^(3/2),x, algorithm="giac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 
3.3.9.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\coth (x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int \frac {\mathrm {coth}\left (x\right )}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{3/2}} \,d x \]

input
int(coth(x)/(a + b/cosh(x)^2)^(3/2),x)
 
output
int(coth(x)/(a + b/cosh(x)^2)^(3/2), x)