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

Optimal. Leaf size=40 \[ \frac{2 i \sqrt{2} \sqrt{\sinh (x)} E\left (\left .\frac{\pi }{4}-\frac{i x}{2}\right |2\right )}{\sqrt{i \sinh (x)}} \]

[Out]

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

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Rubi [A]  time = 0.0812508, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {4398, 4400, 4221, 4309, 2639} \[ \frac{2 i E\left (\left .\frac{\pi }{4}-\frac{i x}{2}\right |2\right ) \sqrt{\sinh (2 x) \text{sech}(x)}}{\sqrt{i \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4398

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

Rule 4400

Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v], ww = Ac
tivateTrig[w]}, Dist[(vv^m*ww^n)^FracPart[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])

Rule 4221

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

Rule 4309

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[(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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \sqrt{\text{sech}(x) \sinh (2 x)} \, dx &=\frac{\sqrt{\text{sech}(x) \sinh (2 x)} \int \sqrt{i \text{sech}(x) \sinh (2 x)} \, dx}{\sqrt{i \text{sech}(x) \sinh (2 x)}}\\ &=\frac{\sqrt{\text{sech}(x) \sinh (2 x)} \int \sqrt{\text{sech}(x)} \sqrt{i \sinh (2 x)} \, dx}{\sqrt{\text{sech}(x)} \sqrt{i \sinh (2 x)}}\\ &=\frac{\left (\sqrt{\cosh (x)} \sqrt{\text{sech}(x) \sinh (2 x)}\right ) \int \frac{\sqrt{i \sinh (2 x)}}{\sqrt{\cosh (x)}} \, dx}{\sqrt{i \sinh (2 x)}}\\ &=\frac{\sqrt{\text{sech}(x) \sinh (2 x)} \int \sqrt{i \sinh (x)} \, dx}{\sqrt{i \sinh (x)}}\\ &=\frac{2 i E\left (\left .\frac{\pi }{4}-\frac{i x}{2}\right |2\right ) \sqrt{\text{sech}(x) \sinh (2 x)}}{\sqrt{i \sinh (x)}}\\ \end{align*}

Mathematica [C]  time = 1.78887, size = 54, normalized size = 1.35 \[ -\frac{2}{3} \sqrt{2} \sqrt{\sinh (x)} \tanh \left (\frac{x}{2}\right ) \left (\sqrt{\text{sech}^2\left (\frac{x}{2}\right )} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},\tanh ^2\left (\frac{x}{2}\right )\right )-3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.1, size = 75, normalized size = 1.9 \begin{align*} 2\,{\frac{\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\,{\it EllipticE} \left ( \sqrt{1-i\sinh \left ( x \right ) },1/2\,\sqrt{2} \right ) -{\it EllipticF} \left ( \sqrt{1-i\sinh \left ( x \right ) },1/2\,\sqrt{2} \right ) \right ) }{\cosh \left ( x \right ) \sqrt{\sinh \left ( x \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-I*(sinh(x)+I))^(1/2)*(-I*(-sinh(x)+I))^(1/2)*(I*sinh(x))^(1/2)*(2*EllipticE((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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\frac{\sinh \left (2 \, x\right )}{\cosh \left (x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sinh(2*x)/cosh(x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{\frac{\sinh \left (2 \, x\right )}{\cosh \left (x\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(sinh(2*x)/cosh(x)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\frac{\sinh{\left (2 x \right )}}{\cosh{\left (x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sinh(2*x)/cosh(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\frac{\sinh \left (2 \, x\right )}{\cosh \left (x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sinh(2*x)/cosh(x)), x)