3.19 \(\int \frac {1}{\sqrt {b \text {sech}(c+d x)}} \, dx\)

Optimal. Leaf size=42 \[ -\frac {2 i E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}} \]

[Out]

-2*I*(cosh(1/2*d*x+1/2*c)^2)^(1/2)/cosh(1/2*d*x+1/2*c)*EllipticE(I*sinh(1/2*d*x+1/2*c),2^(1/2))/d/cosh(d*x+c)^
(1/2)/(b*sech(d*x+c))^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3771, 2639} \[ -\frac {2 i E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Sech[c + d*x]],x]

[Out]

((-2*I)*EllipticE[(I/2)*(c + d*x), 2])/(d*Sqrt[Cosh[c + d*x]]*Sqrt[b*Sech[c + d*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 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {b \text {sech}(c+d x)}} \, dx &=\frac {\int \sqrt {\cosh (c+d x)} \, dx}{\sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}\\ &=-\frac {2 i E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 42, normalized size = 1.00 \[ -\frac {2 i E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Sech[c + d*x]],x]

[Out]

((-2*I)*EllipticE[(I/2)*(c + d*x), 2])/(d*Sqrt[Cosh[c + d*x]]*Sqrt[b*Sech[c + d*x]])

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fricas [F]  time = 0.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \operatorname {sech}\left (d x + c\right )}}{b \operatorname {sech}\left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sech(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sech(d*x + c))/(b*sech(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \operatorname {sech}\left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sech(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(b*sech(d*x + c)), x)

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maple [B]  time = 0.37, size = 244, normalized size = 5.81 \[ \frac {\sqrt {2}}{d \sqrt {\frac {b \,{\mathrm e}^{d x +c}}{1+{\mathrm e}^{2 d x +2 c}}}}+\frac {\left (-\frac {2 \left (b \,{\mathrm e}^{2 d x +2 c}+b \right )}{b \sqrt {{\mathrm e}^{d x +c} \left (b \,{\mathrm e}^{2 d x +2 c}+b \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{d x +c}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{d x +c}-i\right )}\, \sqrt {i {\mathrm e}^{d x +c}}\, \left (-2 i \EllipticE \left (\sqrt {-i \left ({\mathrm e}^{d x +c}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \EllipticF \left (\sqrt {-i \left ({\mathrm e}^{d x +c}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {{\mathrm e}^{3 d x +3 c} b +b \,{\mathrm e}^{d x +c}}}\right ) \sqrt {2}\, \sqrt {b \,{\mathrm e}^{d x +c} \left (1+{\mathrm e}^{2 d x +2 c}\right )}}{d \sqrt {\frac {b \,{\mathrm e}^{d x +c}}{1+{\mathrm e}^{2 d x +2 c}}}\, \left (1+{\mathrm e}^{2 d x +2 c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sech(d*x+c))^(1/2),x)

[Out]

1/d*2^(1/2)/(b*exp(d*x+c)/(exp(d*x+c)^2+1))^(1/2)+1/d*(-2*(b*exp(d*x+c)^2+b)/b/(exp(d*x+c)*(b*exp(d*x+c)^2+b))
^(1/2)+I*(-I*(exp(d*x+c)+I))^(1/2)*2^(1/2)*(I*(exp(d*x+c)-I))^(1/2)*(I*exp(d*x+c))^(1/2)/(exp(d*x+c)^3*b+b*exp
(d*x+c))^(1/2)*(-2*I*EllipticE((-I*(exp(d*x+c)+I))^(1/2),1/2*2^(1/2))+I*EllipticF((-I*(exp(d*x+c)+I))^(1/2),1/
2*2^(1/2))))*2^(1/2)/(b*exp(d*x+c)/(exp(d*x+c)^2+1))^(1/2)*(b*exp(d*x+c)*(exp(d*x+c)^2+1))^(1/2)/(exp(d*x+c)^2
+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \operatorname {sech}\left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sech(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(b*sech(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/cosh(c + d*x))^(1/2),x)

[Out]

int(1/(b/cosh(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \operatorname {sech}{\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sech(d*x+c))**(1/2),x)

[Out]

Integral(1/sqrt(b*sech(c + d*x)), x)

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