\(\int \frac {\coth ^3(x)}{\sqrt {a+b \text {sech}^2(x)}} \, dx\) [202]

Optimal result

Integrand size = 17, antiderivative size = 90 \[ \int \frac {\coth ^3(x)}{\sqrt {a+b \text {sech}^2(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{3/2}}-\frac {\coth ^2(x) \sqrt {a+b \text {sech}^2(x)}}{2 (a+b)} \] Output:

arctanh((a+b*sech(x)^2)^(1/2)/a^(1/2))/a^(1/2)-1/2*(2*a+3*b)*arctanh((a+b* 
sech(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)-coth(x)^2*(a+b*sech(x)^2)^(1/2)/ 
(2*a+2*b)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.77 \[ \int \frac {\coth ^3(x)}{\sqrt {a+b \text {sech}^2(x)}} \, dx=\frac {-\left ((a+2 b+a \cosh (2 x)) \text {csch}^2(x)\right )+\frac {\sqrt {2} \sqrt {a+2 b+a \cosh (2 x)} \left (-\sqrt {a} (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a+b} \cosh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right )+2 (a+b)^{3/2} \log \left (\sqrt {2} \sqrt {a} \cosh (x)+\sqrt {a+2 b+a \cosh (2 x)}\right )\right ) \text {sech}(x)}{\sqrt {a} \sqrt {a+b}}}{4 (a+b) \sqrt {a+b \text {sech}^2(x)}} \] Input:

Integrate[Coth[x]^3/Sqrt[a + b*Sech[x]^2],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {3042, 26, 4627, 354, 114, 27, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i}{\tan (i x)^3 \sqrt {a+b \sec (i x)^2}}dx\)

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2 (2 a+3 b) \int \frac {1}{\frac {a+b}{b}-\frac {\text {sech}^4(x)}{b}}d\sqrt {b \text {sech}^2(x)+a}}{b}+\frac {4 (a+b) \int \frac {1}{\frac {\text {sech}^4(x)}{b}-\frac {a}{b}}d\sqrt {b \text {sech}^2(x)+a}}{b}}{2 (a+b)}-\frac {\sqrt {a+b \text {sech}^2(x)}}{(a+b) \left (1-\text {sech}^2(x)\right )}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2 (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {4 (a+b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 (a+b)}-\frac {\sqrt {a+b \text {sech}^2(x)}}{(a+b) \left (1-\text {sech}^2(x)\right )}\right )\)

Input:

Int[Coth[x]^3/Sqrt[a + b*Sech[x]^2],x]
 

Output:

(-1/2*((-4*(a + b)*ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]])/Sqrt[a] + (2*(2 
*a + 3*b)*ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a + b]])/Sqrt[a + b])/(a + b) 
 - Sqrt[a + b*Sech[x]^2]/((a + b)*(1 - Sech[x]^2)))/2
 

Defintions of rubi rules used

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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

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])
 

Maple [F]

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

Input:

int(coth(x)^3/(a+sech(x)^2*b)^(1/2),x)
 

Output:

int(coth(x)^3/(a+sech(x)^2*b)^(1/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1154 vs. \(2 (72) = 144\).

Time = 0.30 (sec) , antiderivative size = 6357, normalized size of antiderivative = 70.63 \[ \int \frac {\coth ^3(x)}{\sqrt {a+b \text {sech}^2(x)}} \, dx=\text {Too large to display} \] Input:

integrate(coth(x)^3/(a+b*sech(x)^2)^(1/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate(coth(x)**3/(a+b*sech(x)**2)**(1/2),x)
 

Output:

Integral(coth(x)**3/sqrt(a + b*sech(x)**2), x)
 

Maxima [F]

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

integrate(coth(x)^3/(a+b*sech(x)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(coth(x)^3/sqrt(b*sech(x)^2 + a), x)
 

Giac [F(-2)]

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

integrate(coth(x)^3/(a+b*sech(x)^2)^(1/2),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
 

Mupad [F(-1)]

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

int(coth(x)^3/(a + b/cosh(x)^2)^(1/2),x)
 

Output:

int(coth(x)^3/(a + b/cosh(x)^2)^(1/2), x)
 

Reduce [F]

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

int(coth(x)^3/(a+b*sech(x)^2)^(1/2),x)
 

Output:

int((sqrt(sech(x)**2*b + a)*coth(x)**3)/(sech(x)**2*b + a),x)