\(\int \frac {1}{(a+b \text {csch}^2(c+d x))^{3/2}} \, dx\) [13]

Optimal result

Integrand size = 16, antiderivative size = 82 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{a^{3/2} d}+\frac {b \coth (c+d x)}{a (a-b) d \sqrt {a-b+b \coth ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.80 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \, dx=\frac {\text {csch}^2(c+d x) \left (\frac {2 \sqrt {a} b (-a+2 b+a \cosh (2 (c+d x))) \coth (c+d x)}{a-b}+\sqrt {2} (-a+2 b+a \cosh (2 (c+d x)))^{3/2} \text {csch}(c+d x) \log \left (\sqrt {2} \sqrt {a} \cosh (c+d x)+\sqrt {-a+2 b+a \cosh (2 (c+d x))}\right )\right )}{4 a^{3/2} d \left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \] Input:

Integrate[(a + b*Csch[c + d*x]^2)^(-3/2),x]
 

Output:

(Csch[c + d*x]^2*((2*Sqrt[a]*b*(-a + 2*b + a*Cosh[2*(c + d*x)])*Coth[c + d 
*x])/(a - b) + Sqrt[2]*(-a + 2*b + a*Cosh[2*(c + d*x)])^(3/2)*Csch[c + d*x 
]*Log[Sqrt[2]*Sqrt[a]*Cosh[c + d*x] + Sqrt[-a + 2*b + a*Cosh[2*(c + d*x)]] 
]))/(4*a^(3/2)*d*(a + b*Csch[c + d*x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3042, 4616, 296, 291, 219}

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.

\(\displaystyle \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4616

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\int \frac {1}{1-\frac {a \coth ^2(c+d x)}{b \coth ^2(c+d x)+a-b}}d\frac {\coth (c+d x)}{\sqrt {b \coth ^2(c+d x)+a-b}}}{a}+\frac {b \coth (c+d x)}{a (a-b) \sqrt {a+b \coth ^2(c+d x)-b}}}{d}\)

\(\Big \downarrow \) 219

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

Input:

Int[(a + b*Csch[c + d*x]^2)^(-3/2),x]
 

Output:

(ArcTanh[(Sqrt[a]*Coth[c + d*x])/Sqrt[a - b + b*Coth[c + d*x]^2]]/a^(3/2) 
+ (b*Coth[c + d*x])/(a*(a - b)*Sqrt[a - b + b*Coth[c + d*x]^2]))/d
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 296

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[p, -1]
 

rule 3042

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

rule 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Subst[Int[(a + b + b*ff^2*x^2)^p 
/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] 
&& NeQ[a + b, 0] && NeQ[p, -1]
 

Maple [F]

\[\int \frac {1}{\left (a +b \operatorname {csch}\left (d x +c \right )^{2}\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1166 vs. \(2 (74) = 148\).

Time = 0.19 (sec) , antiderivative size = 3009, normalized size of antiderivative = 36.70 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {csch}^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

Integral((a + b*csch(c + d*x)**2)**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {csch}\left (d x + c\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

integrate((b*csch(d*x + c)^2 + a)^(-3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio 
n over extensionNot implemented, e.g. for multivariate mod/approx polynomi 
alsError:
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\mathrm {sinh}\left (c+d\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\mathrm {csch}\left (d x +c \right )^{2} b +a}}{\mathrm {csch}\left (d x +c \right )^{4} b^{2}+2 \mathrm {csch}\left (d x +c \right )^{2} a b +a^{2}}d x \] Input:

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

Output:

int(sqrt(csch(c + d*x)**2*b + a)/(csch(c + d*x)**4*b**2 + 2*csch(c + d*x)* 
*2*a*b + a**2),x)