3.1.34 \(\int \sqrt {a \text {sech}^2(x)} \, dx\) [34]

3.1.34.1 Optimal result

Integrand size = 10, antiderivative size = 25 \[ \int \sqrt {a \text {sech}^2(x)} \, dx=\sqrt {a} \arctan \left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a \text {sech}^2(x)}}\right ) \]

output

arctan(a^(1/2)*tanh(x)/(a*sech(x)^2)^(1/2))*a^(1/2)
 

3.1.34.2 Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \sqrt {a \text {sech}^2(x)} \, dx=\arctan (\sinh (x)) \cosh (x) \sqrt {a \text {sech}^2(x)} \]

input

Integrate[Sqrt[a*Sech[x]^2],x]
 

output

ArcTan[Sinh[x]]*Cosh[x]*Sqrt[a*Sech[x]^2]
 

3.1.34.3 Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4610, 224, 216}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a \sec (i x)^2}dx\)

\(\Big \downarrow \) 4610

\(\displaystyle a \int \frac {1}{\sqrt {a-a \tanh ^2(x)}}d\tanh (x)\)

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

\(\displaystyle \sqrt {a} \arctan \left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-a \tanh ^2(x)}}\right )\)

input

Int[Sqrt[a*Sech[x]^2],x]
 

output

Sqrt[a]*ArcTan[(Sqrt[a]*Tanh[x])/Sqrt[a - a*Tanh[x]^2]]
 

3.1.34.3.1 Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224

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

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]
 

3.1.34.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.88

method	result	size
risch	\(i \sqrt {\frac {{\mathrm e}^{2 x} a}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \ln \left ({\mathrm e}^{x}+i\right )-i \sqrt {\frac {{\mathrm e}^{2 x} a}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \ln \left ({\mathrm e}^{x}-i\right )\)	\(72\)

input

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

output

I*(exp(2*x)*a/(1+exp(2*x))^2)^(1/2)*exp(-x)*(1+exp(2*x))*ln(exp(x)+I)-I*(e 
xp(2*x)*a/(1+exp(2*x))^2)^(1/2)*exp(-x)*(1+exp(2*x))*ln(exp(x)-I)
 

3.1.34.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 5.80 \[ \int \sqrt {a \text {sech}^2(x)} \, dx=\left [\sqrt {-a} \log \left (\frac {2 \, a \cosh \left (x\right ) e^{x} \sinh \left (x\right ) + a e^{x} \sinh \left (x\right )^{2} + 2 \, {\left (\cosh \left (x\right ) e^{\left (2 \, x\right )} + {\left (e^{\left (2 \, x\right )} + 1\right )} \sinh \left (x\right ) + \cosh \left (x\right )\right )} \sqrt {-a} \sqrt {\frac {a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}} e^{x} + {\left (a \cosh \left (x\right )^{2} - a\right )} e^{x}}{2 \, \cosh \left (x\right ) e^{x} \sinh \left (x\right ) + e^{x} \sinh \left (x\right )^{2} + {\left (\cosh \left (x\right )^{2} + 1\right )} e^{x}}\right ), 2 \, \sqrt {\frac {a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}} {\left (e^{\left (2 \, x\right )} + 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )\right ] \]

input

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

output

[sqrt(-a)*log((2*a*cosh(x)*e^x*sinh(x) + a*e^x*sinh(x)^2 + 2*(cosh(x)*e^(2 
*x) + (e^(2*x) + 1)*sinh(x) + cosh(x))*sqrt(-a)*sqrt(a/(e^(4*x) + 2*e^(2*x 
) + 1))*e^x + (a*cosh(x)^2 - a)*e^x)/(2*cosh(x)*e^x*sinh(x) + e^x*sinh(x)^ 
2 + (cosh(x)^2 + 1)*e^x)), 2*sqrt(a/(e^(4*x) + 2*e^(2*x) + 1))*(e^(2*x) + 
1)*arctan(cosh(x) + sinh(x))]
 

3.1.34.6 Sympy [F]

\[ \int \sqrt {a \text {sech}^2(x)} \, dx=\int \sqrt {a \operatorname {sech}^{2}{\left (x \right )}}\, dx \]

input

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

output

Integral(sqrt(a*sech(x)**2), x)
 

3.1.34.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.32 \[ \int \sqrt {a \text {sech}^2(x)} \, dx=2 \, \sqrt {a} \arctan \left (e^{x}\right ) \]

input

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

output

2*sqrt(a)*arctan(e^x)
 

3.1.34.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.32 \[ \int \sqrt {a \text {sech}^2(x)} \, dx=2 \, \sqrt {a} \arctan \left (e^{x}\right ) \]

input

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

output

2*sqrt(a)*arctan(e^x)
 

3.1.34.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {a \text {sech}^2(x)} \, dx=\int \sqrt {\frac {a}{{\mathrm {cosh}\left (x\right )}^2}} \,d x \]

input

int((a/cosh(x)^2)^(1/2),x)
 

output

int((a/cosh(x)^2)^(1/2), x)