\(\int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx\) [25]

Optimal result

Integrand size = 11, antiderivative size = 40 \[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\frac {2 i \sqrt {2} E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {\sinh (x)}}{\sqrt {i \sinh (x)}} \] Output:

2*I*(sin(1/4*Pi+1/2*I*x)^2)^(1/2)/sin(1/4*Pi+1/2*I*x)*EllipticE(cos(1/4*Pi 
+1/2*I*x),2^(1/2))*2^(1/2)*sinh(x)^(1/2)/(I*sinh(x))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.90 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.15 \[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\frac {2}{3} \left (-3+\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\tanh ^2\left (\frac {x}{2}\right )\right ) \sqrt {\text {sech}^2\left (\frac {x}{2}\right )}+4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\tanh ^2\left (\frac {x}{2}\right )\right ) \sqrt {\text {sech}^2\left (\frac {x}{2}\right )}\right ) \sqrt {\text {sech}(x) \sinh (2 x)} \tanh \left (\frac {x}{2}\right ) \] Input:

Integrate[Sqrt[Sech[x]*Sinh[2*x]],x]
 

Output:

(2*(-3 + Hypergeometric2F1[1/2, 3/4, 7/4, Tanh[x/2]^2]*Sqrt[Sech[x/2]^2] + 
 4*Hypergeometric2F1[3/4, 3/2, 7/4, Tanh[x/2]^2]*Sqrt[Sech[x/2]^2])*Sqrt[S 
ech[x]*Sinh[2*x]]*Tanh[x/2])/3
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.909, Rules used = {3042, 4898, 3042, 4900, 3042, 4709, 3042, 4797, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {-i \sin (2 i x) \sec (i x)}dx\)

\(\Big \downarrow \) 4898

\(\displaystyle \frac {\sqrt {\sinh (2 x) \text {sech}(x)} \int \sqrt {i \text {sech}(x) \sinh (2 x)}dx}{\sqrt {i \sinh (2 x) \text {sech}(x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sinh (2 x) \text {sech}(x)} \int \sqrt {\sec (i x) \sin (2 i x)}dx}{\sqrt {i \sinh (2 x) \text {sech}(x)}}\)

\(\Big \downarrow \) 4900

\(\displaystyle \frac {\sqrt {\sinh (2 x) \text {sech}(x)} \int \sqrt {\text {sech}(x)} \sqrt {i \sinh (2 x)}dx}{\sqrt {i \sinh (2 x)} \sqrt {\text {sech}(x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sinh (2 x) \text {sech}(x)} \int \sqrt {\sec (i x)} \sqrt {\sin (2 i x)}dx}{\sqrt {i \sinh (2 x)} \sqrt {\text {sech}(x)}}\)

\(\Big \downarrow \) 4709

\(\displaystyle \frac {\sqrt {\cosh (x)} \sqrt {\sinh (2 x) \text {sech}(x)} \int \frac {\sqrt {i \sinh (2 x)}}{\sqrt {\cosh (x)}}dx}{\sqrt {i \sinh (2 x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cosh (x)} \sqrt {\sinh (2 x) \text {sech}(x)} \int \frac {\sqrt {\sin (2 i x)}}{\sqrt {\cos (i x)}}dx}{\sqrt {i \sinh (2 x)}}\)

\(\Big \downarrow \) 4797

\(\displaystyle \frac {\sqrt {\sinh (2 x) \text {sech}(x)} \int \sqrt {i \sinh (x)}dx}{\sqrt {i \sinh (x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sinh (2 x) \text {sech}(x)} \int \sqrt {\sin (i x)}dx}{\sqrt {i \sinh (x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 i E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {\sinh (2 x) \text {sech}(x)}}{\sqrt {i \sinh (x)}}\)

Input:

Int[Sqrt[Sech[x]*Sinh[2*x]],x]
 

Output:

((2*I)*EllipticE[Pi/4 - (I/2)*x, 2]*Sqrt[Sech[x]*Sinh[2*x]])/Sqrt[I*Sinh[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 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

rule 4797

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( 
p_), x_Symbol] :> Simp[(g*Sin[c + d*x])^p/((e*Cos[a + b*x])^p*Sin[a + b*x]^ 
p)   Int[(e*Cos[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p]
 

rule 4898

Int[(u_.)*((a_)*(v_))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = A 
ctivateTrig[v]}, Simp[a^IntPart[p]*((a*vv)^FracPart[p]/vv^FracPart[p])   In 
t[uu*vv^p, x], x]] /; FreeQ[{a, p}, x] &&  !IntegerQ[p] &&  !InertTrigFreeQ 
[v]
 

rule 4900

Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTri 
g[u], vv = ActivateTrig[v], ww = ActivateTrig[w]}, Simp[(vv^m*ww^n)^FracPar 
t[p]/(vv^(m*FracPart[p])*ww^(n*FracPart[p]))   Int[uu*vv^(m*p)*ww^(n*p), x] 
, x]] /; FreeQ[{m, n, p}, x] &&  !IntegerQ[p] && ( !InertTrigFreeQ[v] ||  ! 
InertTrigFreeQ[w])
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.88

method	result	size
default	\(\frac {2 \sqrt {-i \left (\sinh \left (x \right )+i\right )}\, \sqrt {-i \left (-\sinh \left (x \right )+i\right )}\, \sqrt {i \sinh \left (x \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {1-i \sinh \left (x \right )}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (x \right )}, \frac {\sqrt {2}}{2}\right )\right )}{\cosh \left (x \right ) \sqrt {\sinh \left (x \right )}}\)	\(75\)
risch	\(2 \sqrt {{\mathrm e}^{-x} \left (-1+{\mathrm e}^{2 x}\right )}+\frac {\left (-\frac {4 \left (-1+{\mathrm e}^{2 x}\right )}{\sqrt {{\mathrm e}^{x} \left (-1+{\mathrm e}^{2 x}\right )}}+\frac {2 \sqrt {1+{\mathrm e}^{x}}\, \sqrt {-2 \,{\mathrm e}^{x}+2}\, \sqrt {-{\mathrm e}^{x}}\, \left (-2 \operatorname {EllipticE}\left (\sqrt {1+{\mathrm e}^{x}}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticF}\left (\sqrt {1+{\mathrm e}^{x}}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {{\mathrm e}^{3 x}-{\mathrm e}^{x}}}\right ) \sqrt {{\mathrm e}^{-x} \left (-1+{\mathrm e}^{2 x}\right )}\, \sqrt {{\mathrm e}^{x} \left (-1+{\mathrm e}^{2 x}\right )}}{-1+{\mathrm e}^{2 x}}\)	\(130\)

Input:

int((sinh(2*x)/cosh(x))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2*(-I*(sinh(x)+I))^(1/2)*(-I*(-sinh(x)+I))^(1/2)*(I*sinh(x))^(1/2)*(2*Elli 
pticE((1-I*sinh(x))^(1/2),1/2*2^(1/2))-EllipticF((1-I*sinh(x))^(1/2),1/2*2 
^(1/2)))/cosh(x)/sinh(x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.58 \[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=-2 \, \sqrt {2} \sqrt {\sinh \left (x\right )} - 4 \, {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right )\right ) \] Input:

integrate((sinh(2*x)/cosh(x))^(1/2),x, algorithm="fricas")
 

Output:

-2*sqrt(2)*sqrt(sinh(x)) - 4*weierstrassZeta(4, 0, weierstrassPInverse(4, 
0, cosh(x) + sinh(x)))
 

Sympy [F]

\[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\int \sqrt {\frac {\sinh {\left (2 x \right )}}{\cosh {\left (x \right )}}}\, dx \] Input:

integrate((sinh(2*x)/cosh(x))**(1/2),x)
 

Output:

Integral(sqrt(sinh(2*x)/cosh(x)), x)
 

Maxima [F]

\[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\int { \sqrt {\frac {\sinh \left (2 \, x\right )}{\cosh \left (x\right )}} \,d x } \] Input:

integrate((sinh(2*x)/cosh(x))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(sinh(2*x)/cosh(x)), x)
 

Giac [F]

\[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\int { \sqrt {\frac {\sinh \left (2 \, x\right )}{\cosh \left (x\right )}} \,d x } \] Input:

integrate((sinh(2*x)/cosh(x))^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(sinh(2*x)/cosh(x)), x)
 

Mupad [F(-1)]

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

int((sinh(2*x)/cosh(x))^(1/2),x)
 

Output:

int((sinh(2*x)/cosh(x))^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\int \frac {\sqrt {\sinh \left (2 x \right )}\, \sqrt {\cosh \left (x \right )}}{\cosh \left (x \right )}d x \] Input:

int((sinh(2*x)/cosh(x))^(1/2),x)
 

Output:

int((sqrt(sinh(2*x))*sqrt(cosh(x)))/cosh(x),x)