\(\int \sqrt {b \text {csch}(c+d x)} \, dx\) [16]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 12, antiderivative size = 56 \[ \int \sqrt {b \text {csch}(c+d x)} \, dx=-\frac {2 i \sqrt {b \text {csch}(c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right ) \sqrt {i \sinh (c+d x)}}{d} \]

[Out]

2*I*(sin(1/2*I*c+1/4*Pi+1/2*I*d*x)^2)^(1/2)/sin(1/2*I*c+1/4*Pi+1/2*I*d*x)*EllipticF(cos(1/2*I*c+1/4*Pi+1/2*I*d
*x),2^(1/2))*(b*csch(d*x+c))^(1/2)*(I*sinh(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, 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} \[ \int \sqrt {b \text {csch}(c+d x)} \, dx=-\frac {2 i \sqrt {i \sinh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right ) \sqrt {b \text {csch}(c+d x)}}{d} \]

[In]

Int[Sqrt[b*Csch[c + d*x]],x]

[Out]

((-2*I)*Sqrt[b*Csch[c + d*x]]*EllipticF[(I*c - Pi/2 + I*d*x)/2, 2]*Sqrt[I*Sinh[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 {b \text {csch}(c+d x)} \sqrt {i \sinh (c+d x)}\right ) \int \frac {1}{\sqrt {i \sinh (c+d x)}} \, dx \\ & = -\frac {2 i \sqrt {b \text {csch}(c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right ) \sqrt {i \sinh (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int \sqrt {b \text {csch}(c+d x)} \, dx=\frac {2 i \sqrt {b \text {csch}(c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (\frac {\pi }{2}-i (c+d x)\right ),2\right ) \sqrt {i \sinh (c+d x)}}{d} \]

[In]

Integrate[Sqrt[b*Csch[c + d*x]],x]

[Out]

((2*I)*Sqrt[b*Csch[c + d*x]]*EllipticF[(Pi/2 - I*(c + d*x))/2, 2]*Sqrt[I*Sinh[c + d*x]])/d

Maple [F]

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

[In]

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

[Out]

int((b*csch(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 = 0.48 \[ \int \sqrt {b \text {csch}(c+d x)} \, dx=\frac {2 \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )}{d} \]

[In]

integrate((b*csch(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]

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

[In]

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

[Out]

Integral(sqrt(b*csch(c + d*x)), x)

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(b*csch(d*x + c)), x)

Giac [F]

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

[In]

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

[Out]

integrate(sqrt(b*csch(d*x + c)), x)

Mupad [F(-1)]

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

[In]

int((b/sinh(c + d*x))^(1/2),x)

[Out]

int((b/sinh(c + d*x))^(1/2), x)