\(\int \sqrt {b \text {sech}(c+d x)} \, dx\) [18]

Optimal result

Integrand size = 12, antiderivative size = 42 \[ \int \sqrt {b \text {sech}(c+d x)} \, dx=-\frac {2 i \sqrt {\cosh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} i (c+d x),2\right ) \sqrt {b \text {sech}(c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3856, 2720} \[ \int \sqrt {b \text {sech}(c+d x)} \, dx=-\frac {2 i \sqrt {\cosh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} i (c+d x),2\right ) \sqrt {b \text {sech}(c+d x)}}{d} \]

[In]

Int[Sqrt[b*Sech[c + d*x]],x]

[Out]

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

Rule 2720

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

Rule 3856

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \sqrt {b \text {sech}(c+d x)} \, dx=-\frac {2 i \sqrt {\cosh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} i (c+d x),2\right ) \sqrt {b \text {sech}(c+d x)}}{d} \]

[In]

Integrate[Sqrt[b*Sech[c + d*x]],x]

[Out]

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

Maple [F]

\[\int \sqrt {b \,\operatorname {sech}\left (d x +c \right )}d x\]

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.64 \[ \int \sqrt {b \text {sech}(c+d x)} \, dx=\frac {2 \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )}{d} \]

[In]

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

[Out]

2*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cosh(d*x + c) + sinh(d*x + c))/d

Sympy [F]

\[ \int \sqrt {b \text {sech}(c+d x)} \, dx=\int \sqrt {b \operatorname {sech}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

Integral(sqrt(b*sech(c + d*x)), x)

Maxima [F]

\[ \int \sqrt {b \text {sech}(c+d x)} \, dx=\int { \sqrt {b \operatorname {sech}\left (d x + c\right )} \,d x } \]

[In]

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

[Out]

integrate(sqrt(b*sech(d*x + c)), x)

Giac [F]

\[ \int \sqrt {b \text {sech}(c+d x)} \, dx=\int { \sqrt {b \operatorname {sech}\left (d x + c\right )} \,d x } \]

[In]

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

[Out]

integrate(sqrt(b*sech(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {b \text {sech}(c+d x)} \, dx=\int \sqrt {\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}} \,d x \]

[In]

int((b/cosh(c + d*x))^(1/2),x)

[Out]

int((b/cosh(c + d*x))^(1/2), x)