\(\int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\) [137]

Optimal result

Integrand size = 23, antiderivative size = 175 \[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {(4 a-5 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{3/2} d}-\frac {(4 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{3/2} d}-\frac {\coth ^2(c+d x) (a-b \text {sech}(c+d x)) \sqrt {a+b \text {sech}(c+d x)}}{2 \left (a^2-b^2\right ) d} \] Output:

2*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/2))/a^(1/2)/d-1/4*(4*a-5*b)*arctanh 
((a+b*sech(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(3/2)/d-1/4*(4*a+5*b)*arctanh( 
(a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)/d-1/2*coth(d*x+c)^2*(a-b* 
sech(d*x+c))*(a+b*sech(d*x+c))^(1/2)/(a^2-b^2)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.98 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.60 \[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=-\frac {-\frac {b^2 \arctan \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {-a+b}}\right )}{(-a+b)^{3/2}}-\frac {8 b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {4 b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {4 a \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+4 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )-\frac {b \sqrt {a+b \text {sech}(c+d x)}}{(a+b) (-1+\text {sech}(c+d x))}+\frac {b \sqrt {a+b \text {sech}(c+d x)}}{(a-b) (1+\text {sech}(c+d x))}}{4 b d} \] Input:

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

Output:

-1/4*(-((b^2*ArcTan[Sqrt[a + b*Sech[c + d*x]]/Sqrt[-a + b]])/(-a + b)^(3/2 
)) - (8*b*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/Sqrt[a] + (4*b*ArcTa 
nh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]])/Sqrt[a - b] + (b^2*ArcTanh[Sqrt 
[a + b*Sech[c + d*x]]/Sqrt[a + b]])/(a + b)^(3/2) - (4*a*ArcTanh[Sqrt[a + 
b*Sech[c + d*x]]/Sqrt[a + b]])/Sqrt[a + b] + 4*Sqrt[a + b]*ArcTanh[Sqrt[a 
+ b*Sech[c + d*x]]/Sqrt[a + b]] - (b*Sqrt[a + b*Sech[c + d*x]])/((a + b)*( 
-1 + Sech[c + d*x])) + (b*Sqrt[a + b*Sech[c + d*x]])/((a - b)*(1 + Sech[c 
+ d*x])))/(b*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.52 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.55, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4373, 561, 25, 1567, 2009}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i}{\cot \left (i c+i d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {1}{\cot \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^3 \sqrt {a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )}}dx\)

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1567

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 b^4 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} b^4}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 b^4 \sqrt {a-b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 b^4 \sqrt {a+b}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{8 b^3 (a-b)^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{8 b^3 (a+b)^{3/2}}-\frac {\text {sech}(c+d x)}{8 b^2 (a-b) \left (a-b^2 \text {sech}^2(c+d x)-b\right )}+\frac {\text {sech}(c+d x)}{8 b^2 (a+b) \left (a-b^2 \text {sech}^2(c+d x)+b\right )}\right )}{d}\)

Input:

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

Output:

(-2*b^4*(-(ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]]/(Sqrt[a]*b^4)) + Arc 
Tanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]]/(2*Sqrt[a - b]*b^4) - ArcTanh[ 
Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]]/(8*(a - b)^(3/2)*b^3) + ArcTanh[Sqr 
t[a + b*Sech[c + d*x]]/Sqrt[a + b]]/(8*b^3*(a + b)^(3/2)) + ArcTanh[Sqrt[a 
 + b*Sech[c + d*x]]/Sqrt[a + b]]/(2*b^4*Sqrt[a + b]) - Sech[c + d*x]/(8*(a 
 - b)*b^2*(a - b - b^2*Sech[c + d*x]^2)) + Sech[c + d*x]/(8*b^2*(a + b)*(a 
 + b - b^2*Sech[c + d*x]^2))))/d
 

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 561

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{k = Denominator[n]}, Simp[k/d   Subst[Int[x^(k*(n + 1) - 1)*(-c 
/d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), 
 x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac 
tionQ[n] && IntegerQ[p] && IntegerQ[m]
 

rule 1567

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x 
_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] 
 /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b^2 - 4*a*c, 0] && ((IntegerQ[p] 
 && IntegerQ[q]) || IGtQ[p, 0] || IGtQ[q, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4373

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1))   Subst[Int[(b^2 - x^ 
2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, 
 d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2009 vs. \(2 (150) = 300\).

Time = 4.99 (sec) , antiderivative size = 20851, normalized size of antiderivative = 119.15 \[ \int \frac {\coth ^3(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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