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*(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)

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Rubi [A]  time = 0.08, antiderivative size = 40, normalized size of antiderivative = 1.00, 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 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]

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 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])

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*}

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Mathematica [C]  time = 1.93, size = 54, normalized size = 1.35 \[ -\frac {2}{3} \sqrt {2} \sqrt {\sinh (x)} \tanh \left (\frac {x}{2}\right ) \left (\sqrt {\text {sech}^2\left (\frac {x}{2}\right )} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\tanh ^2\left (\frac {x}{2}\right )\right )-3\right ) \]

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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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\frac {\sinh \left (2 \, x\right )}{\cosh \relax (x)}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {\sinh \left (2 \, x\right )}{\cosh \relax (x)}}\,{d x} \]

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)

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maple [A]  time = 0.16, size = 75, normalized size = 1.88 \[ \frac {2 \sqrt {-i \left (\sinh \relax (x )+i\right )}\, \sqrt {-i \left (-\sinh \relax (x )+i\right )}\, \sqrt {i \sinh \relax (x )}\, \left (2 \EllipticE \left (\sqrt {-i \sinh \relax (x )+1}, \frac {\sqrt {2}}{2}\right )-\EllipticF \left (\sqrt {-i \sinh \relax (x )+1}, \frac {\sqrt {2}}{2}\right )\right )}{\cosh \relax (x ) \sqrt {\sinh \relax (x )}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {\sinh \left (2 \, x\right )}{\cosh \relax (x)}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \sqrt {\frac {\mathrm {sinh}\left (2\,x\right )}{\mathrm {cosh}\relax (x)}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((sinh(2*x)/cosh(x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {\sinh {\left (2 x \right )}}{\cosh {\relax (x )}}}\, dx \]

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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