\(\int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx\) [19]

Optimal result

Integrand size = 12, antiderivative size = 90 \[ \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx=\frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}+\frac {6 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right )}{5 b^2 d \sqrt {b \text {csch}(c+d x)} \sqrt {i \sinh (c+d x)}} \] Output:

2/5*cosh(d*x+c)/b/d/(b*csch(d*x+c))^(3/2)-6/5*I*EllipticE(cos(1/2*I*c+1/4* 
Pi+1/2*I*d*x),2^(1/2))/b^2/d/(b*csch(d*x+c))^(1/2)/(I*sinh(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx=\frac {-\frac {6 i E\left (\left .\frac {1}{4} (-2 i c+\pi -2 i d x)\right |2\right )}{\sqrt {i \sinh (c+d x)}}+\sinh (2 (c+d x))}{5 b^2 d \sqrt {b \text {csch}(c+d x)}} \] Input:

Integrate[(b*Csch[c + d*x])^(-5/2),x]
 

Output:

(((-6*I)*EllipticE[((-2*I)*c + Pi - (2*I)*d*x)/4, 2])/Sqrt[I*Sinh[c + d*x] 
] + Sinh[2*(c + d*x)])/(5*b^2*d*Sqrt[b*Csch[c + d*x]])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4256, 3042, 4258, 3042, 3119}

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.

\(\displaystyle \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(i b \csc (i c+i d x))^{5/2}}dx\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}-\frac {3 \int \frac {1}{\sqrt {b \text {csch}(c+d x)}}dx}{5 b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}-\frac {3 \int \frac {1}{\sqrt {i b \csc (i c+i d x)}}dx}{5 b^2}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}-\frac {3 \int \sqrt {i \sinh (c+d x)}dx}{5 b^2 \sqrt {i \sinh (c+d x)} \sqrt {b \text {csch}(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}-\frac {3 \int \sqrt {\sin (i c+i d x)}dx}{5 b^2 \sqrt {i \sinh (c+d x)} \sqrt {b \text {csch}(c+d x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}+\frac {6 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{5 b^2 d \sqrt {i \sinh (c+d x)} \sqrt {b \text {csch}(c+d x)}}\)

Input:

Int[(b*Csch[c + d*x])^(-5/2),x]
 

Output:

(2*Cosh[c + d*x])/(5*b*d*(b*Csch[c + d*x])^(3/2)) + (((6*I)/5)*EllipticE[( 
I*c - Pi/2 + I*d*x)/2, 2])/(b^2*d*Sqrt[b*Csch[c + d*x]]*Sqrt[I*Sinh[c + d* 
x]])
 

Defintions of rubi rules used

rule 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* 
n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

Maple [F]

\[\int \frac {1}{\left (\operatorname {csch}\left (d x +c \right ) b \right )^{\frac {5}{2}}}d x\]

Input:

int(1/(csch(d*x+c)*b)^(5/2),x)
 

Output:

int(1/(csch(d*x+c)*b)^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (69) = 138\).

Time = 0.09 (sec) , antiderivative size = 378, normalized size of antiderivative = 4.20 \[ \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx=\frac {24 \, \sqrt {2} {\left (\cosh \left (d x + c\right )^{3} + 3 \, \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + \sinh \left (d x + c\right )^{3}\right )} \sqrt {b} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (\cosh \left (d x + c\right )^{6} + 6 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + \sinh \left (d x + c\right )^{6} + {\left (15 \, \cosh \left (d x + c\right )^{2} + 11\right )} \sinh \left (d x + c\right )^{4} + 11 \, \cosh \left (d x + c\right )^{4} + 4 \, {\left (5 \, \cosh \left (d x + c\right )^{3} + 11 \, \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + {\left (15 \, \cosh \left (d x + c\right )^{4} + 66 \, \cosh \left (d x + c\right )^{2} - 13\right )} \sinh \left (d x + c\right )^{2} - 13 \, \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, \cosh \left (d x + c\right )^{5} + 22 \, \cosh \left (d x + c\right )^{3} - 13 \, \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 1\right )} \sqrt {\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1}}}{20 \, {\left (b^{3} d \cosh \left (d x + c\right )^{3} + 3 \, b^{3} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, b^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + b^{3} d \sinh \left (d x + c\right )^{3}\right )}} \] Input:

integrate(1/(b*csch(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/20*(24*sqrt(2)*(cosh(d*x + c)^3 + 3*cosh(d*x + c)^2*sinh(d*x + c) + 3*co 
sh(d*x + c)*sinh(d*x + c)^2 + sinh(d*x + c)^3)*sqrt(b)*weierstrassZeta(4, 
0, weierstrassPInverse(4, 0, cosh(d*x + c) + sinh(d*x + c))) + sqrt(2)*(co 
sh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5 + sinh(d*x + c)^6 + (15*co 
sh(d*x + c)^2 + 11)*sinh(d*x + c)^4 + 11*cosh(d*x + c)^4 + 4*(5*cosh(d*x + 
 c)^3 + 11*cosh(d*x + c))*sinh(d*x + c)^3 + (15*cosh(d*x + c)^4 + 66*cosh( 
d*x + c)^2 - 13)*sinh(d*x + c)^2 - 13*cosh(d*x + c)^2 + 2*(3*cosh(d*x + c) 
^5 + 22*cosh(d*x + c)^3 - 13*cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt((b*cos 
h(d*x + c) + b*sinh(d*x + c))/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x 
+ c) + sinh(d*x + c)^2 - 1)))/(b^3*d*cosh(d*x + c)^3 + 3*b^3*d*cosh(d*x + 
c)^2*sinh(d*x + c) + 3*b^3*d*cosh(d*x + c)*sinh(d*x + c)^2 + b^3*d*sinh(d* 
x + c)^3)
 

Sympy [F]

\[ \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx=\int \frac {1}{\left (b \operatorname {csch}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(b*csch(d*x+c))**(5/2),x)
 

Output:

Integral((b*csch(c + d*x))**(-5/2), x)
 

Maxima [F]

\[ \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx=\int { \frac {1}{\left (b \operatorname {csch}\left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(b*csch(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*csch(d*x + c))^(-5/2), x)
 

Giac [F]

\[ \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx=\int { \frac {1}{\left (b \operatorname {csch}\left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(b*csch(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate((b*csch(d*x + c))^(-5/2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (\frac {b}{\mathrm {sinh}\left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int(1/(b/sinh(c + d*x))^(5/2),x)
 

Output:

int(1/(b/sinh(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\mathrm {csch}\left (d x +c \right )}}{\mathrm {csch}\left (d x +c \right )^{3}}d x \right )}{b^{3}} \] Input:

int(1/(b*csch(d*x+c))^(5/2),x)
 

Output:

(sqrt(b)*int(sqrt(csch(c + d*x))/csch(c + d*x)**3,x))/b**3