3.1.28 \(\int \frac {1}{\sqrt {\text {sech}^2(a+b x)}} \, dx\) [28]

3.1.28.1 Optimal result

Integrand size = 12, antiderivative size = 22 \[ \int \frac {1}{\sqrt {\text {sech}^2(a+b x)}} \, dx=\frac {\tanh (a+b x)}{b \sqrt {\text {sech}^2(a+b x)}} \]

output

tanh(b*x+a)/b/(sech(b*x+a)^2)^(1/2)
 

3.1.28.2 Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {\text {sech}^2(a+b x)}} \, dx=\frac {\tanh (a+b x)}{b \sqrt {\text {sech}^2(a+b x)}} \]

input

Integrate[1/Sqrt[Sech[a + b*x]^2],x]
 

output

Tanh[a + b*x]/(b*Sqrt[Sech[a + b*x]^2])
 

3.1.28.3 Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4610, 208}

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.

\(\displaystyle \int \frac {1}{\sqrt {\text {sech}^2(a+b x)}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4610

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

\(\Big \downarrow \) 208

\(\displaystyle \frac {\tanh (a+b x)}{b \sqrt {1-\tanh ^2(a+b x)}}\)

input

Int[1/Sqrt[Sech[a + b*x]^2],x]
 

output

Tanh[a + b*x]/(b*Sqrt[1 - Tanh[a + b*x]^2])
 

3.1.28.3.1 Defintions of rubi rules used

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 3042

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

rule 4610

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

3.1.28.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(96\) vs. \(2(20)=40\).

Time = 0.46 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.41

method	result	size
risch	\(\frac {{\mathrm e}^{2 b x +2 a}}{2 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {1}{2 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}\)	\(97\)

input

int(1/(sech(b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

output

1/2/b/(1+exp(2*b*x+2*a))/(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*exp 
(2*b*x+2*a)-1/2/b/(1+exp(2*b*x+2*a))/(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a 
))^(1/2)
 

3.1.28.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\sqrt {\text {sech}^2(a+b x)}} \, dx=\frac {\sinh \left (b x + a\right )}{b} \]

input

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

output

sinh(b*x + a)/b
 

3.1.28.6 Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\sqrt {\text {sech}^2(a+b x)}} \, dx=\begin {cases} \frac {\tanh {\left (a + b x \right )}}{b \sqrt {\operatorname {sech}^{2}{\left (a + b x \right )}}} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {\operatorname {sech}^{2}{\left (a \right )}}} & \text {otherwise} \end {cases} \]

input

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

output

Piecewise((tanh(a + b*x)/(b*sqrt(sech(a + b*x)**2)), Ne(b, 0)), (x/sqrt(se 
ch(a)**2), True))
 

3.1.28.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\sqrt {\text {sech}^2(a+b x)}} \, dx=\frac {e^{\left (b x + a\right )}}{2 \, b} - \frac {e^{\left (-b x - a\right )}}{2 \, b} \]

input

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

output

1/2*e^(b*x + a)/b - 1/2*e^(-b*x - a)/b
 

3.1.28.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\sqrt {\text {sech}^2(a+b x)}} \, dx=\frac {e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}}{2 \, b} \]

input

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

output

1/2*(e^(b*x + a) - e^(-b*x - a))/b
 

3.1.28.9 Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.41 \[ \int \frac {1}{\sqrt {\text {sech}^2(a+b x)}} \, dx=\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-1\right )\,\sqrt {\frac {4\,{\mathrm {e}}^{2\,a+2\,b\,x}}{{\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}^2}}}{4\,b} \]

input

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

output

(exp(- 2*a - 2*b*x)*(exp(4*a + 4*b*x) - 1)*((4*exp(2*a + 2*b*x))/(exp(2*a 
+ 2*b*x) + 1)^2)^(1/2))/(4*b)