\(\int \frac {1}{(b \text {sech}(c+d x))^{5/2}} \, dx\) [21]

Optimal result

Integrand size = 12, antiderivative size = 76 \[ \int \frac {1}{(b \text {sech}(c+d x))^{5/2}} \, dx=-\frac {6 i E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}+\frac {2 \sinh (c+d x)}{5 b d (b \text {sech}(c+d x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(b \text {sech}(c+d x))^{5/2}} \, dx=\frac {\sqrt {b \text {sech}(c+d x)} \left (-12 i \sqrt {\cosh (c+d x)} E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )+\sinh (c+d x)+\sinh (3 (c+d x))\right )}{10 b^3 d} \] Input:

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

Output:

(Sqrt[b*Sech[c + d*x]]*((-12*I)*Sqrt[Cosh[c + d*x]]*EllipticE[(I/2)*(c + d 
*x), 2] + Sinh[c + d*x] + Sinh[3*(c + d*x)]))/(10*b^3*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 76, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sinh (c+d x)}{5 b d (b \text {sech}(c+d x))^{3/2}}+\frac {3 \int \frac {1}{\sqrt {b \csc \left (i c+i d x+\frac {\pi }{2}\right )}}dx}{5 b^2}\)

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (66) = 132\).

Time = 0.08 (sec) , antiderivative size = 379, normalized size of antiderivative = 4.99 \[ \int \frac {1}{(b \text {sech}(c+d x))^{5/2}} \, dx=-\frac {24 \, \sqrt {2} {\left (\cosh \left (d x + c\right )^{3} + 3 \, \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + \sinh \left (d x + c\right )^{3}\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) - \sqrt {2} {\left (\cosh \left (d x + c\right )^{6} + 6 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + \sinh \left (d x + c\right )^{6} + {\left (15 \, \cosh \left (d x + c\right )^{2} - 11\right )} \sinh \left (d x + c\right )^{4} - 11 \, \cosh \left (d x + c\right )^{4} + 4 \, {\left (5 \, \cosh \left (d x + c\right )^{3} - 11 \, \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + {\left (15 \, \cosh \left (d x + c\right )^{4} - 66 \, \cosh \left (d x + c\right )^{2} - 13\right )} \sinh \left (d x + c\right )^{2} - 13 \, \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, \cosh \left (d x + c\right )^{5} - 22 \, \cosh \left (d x + c\right )^{3} - 13 \, \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 1\right )} \sqrt {\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1}}}{20 \, {\left (b^{3} d \cosh \left (d x + c\right )^{3} + 3 \, b^{3} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, b^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + b^{3} d \sinh \left (d x + c\right )^{3}\right )}} \] Input:

integrate(1/(b*sech(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*c 
osh(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)* 
(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5 + sinh(d*x + c)^6 + (15 
*cosh(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*co 
sh(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* 
cosh(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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