Integrand size = 11, antiderivative size = 40 \[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\frac {2 i \sqrt {2} E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {\sinh (x)}}{\sqrt {i \sinh (x)}} \]
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)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.91 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.15 \[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\frac {2}{3} \left (-3+\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\tanh ^2\left (\frac {x}{2}\right )\right ) \sqrt {\text {sech}^2\left (\frac {x}{2}\right )}+4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\tanh ^2\left (\frac {x}{2}\right )\right ) \sqrt {\text {sech}^2\left (\frac {x}{2}\right )}\right ) \sqrt {\text {sech}(x) \sinh (2 x)} \tanh \left (\frac {x}{2}\right ) \]
(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
Time = 0.45 (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 |
\(\displaystyle \frac {2 i E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {\sinh (2 x) \text {sech}(x)}}{\sqrt {i \sinh (x)}}\) |
3.1.25.3.1 Defintions of rubi rules used
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]
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]
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]
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]
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])
Time = 0.66 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.88
method | result | size |
default | \(\frac {2 \sqrt {-i \left (\sinh \left (x \right )+i\right )}\, \sqrt {-i \left (-\sinh \left (x \right )+i\right )}\, \sqrt {i \sinh \left (x \right )}\, \left (2 E\left (\sqrt {1-i \sinh \left (x \right )}, \frac {\sqrt {2}}{2}\right )-F\left (\sqrt {1-i \sinh \left (x \right )}, \frac {\sqrt {2}}{2}\right )\right )}{\cosh \left (x \right ) \sqrt {\sinh \left (x \right )}}\) | \(75\) |
risch | \(2 \sqrt {{\mathrm e}^{-x} \left ({\mathrm e}^{2 x}-1\right )}+\frac {\left (-\frac {4 \left ({\mathrm e}^{2 x}-1\right )}{\sqrt {{\mathrm e}^{x} \left ({\mathrm e}^{2 x}-1\right )}}+\frac {2 \sqrt {1+{\mathrm e}^{x}}\, \sqrt {-2 \,{\mathrm e}^{x}+2}\, \sqrt {-{\mathrm e}^{x}}\, \left (-2 E\left (\sqrt {1+{\mathrm e}^{x}}, \frac {\sqrt {2}}{2}\right )+F\left (\sqrt {1+{\mathrm e}^{x}}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {{\mathrm e}^{3 x}-{\mathrm e}^{x}}}\right ) \sqrt {{\mathrm e}^{-x} \left ({\mathrm e}^{2 x}-1\right )}\, \sqrt {{\mathrm e}^{x} \left ({\mathrm e}^{2 x}-1\right )}}{{\mathrm e}^{2 x}-1}\) | \(130\) |
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)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.58 \[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=-2 \, \sqrt {2} \sqrt {\sinh \left (x\right )} - 4 \, {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right )\right ) \]
-2*sqrt(2)*sqrt(sinh(x)) - 4*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(x) + sinh(x)))
\[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\int \sqrt {\frac {\sinh {\left (2 x \right )}}{\cosh {\left (x \right )}}}\, dx \]
\[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\int { \sqrt {\frac {\sinh \left (2 \, x\right )}{\cosh \left (x\right )}} \,d x } \]
\[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\int { \sqrt {\frac {\sinh \left (2 \, x\right )}{\cosh \left (x\right )}} \,d x } \]
Timed out. \[ \int \sqrt {\text {sech}(x) \sinh (2 x)} \, dx=\int \sqrt {\frac {\mathrm {sinh}\left (2\,x\right )}{\mathrm {cosh}\left (x\right )}} \,d x \]