\(\int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx\) [11]

Optimal result

Integrand size = 10, antiderivative size = 54 \[ \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx=-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{b \sqrt {\sinh (a+b x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2721, 2720} \[ \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx=-\frac {2 i \sqrt {i \sinh (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right ),2\right )}{b \sqrt {\sinh (a+b x)}} \]

[In]

Int[1/Sqrt[Sinh[a + b*x]],x]

[Out]

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

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 2721

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {i \sinh (a+b x)} \int \frac {1}{\sqrt {i \sinh (a+b x)}} \, dx}{\sqrt {\sinh (a+b x)}} \\ & = -\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{b \sqrt {\sinh (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx=-\frac {2 \operatorname {EllipticF}\left (\frac {1}{4} (-2 i a+\pi -2 i b x),2\right ) \sqrt {\sinh (a+b x)}}{b \sqrt {i \sinh (a+b x)}} \]

[In]

Integrate[1/Sqrt[Sinh[a + b*x]],x]

[Out]

(-2*EllipticF[((-2*I)*a + Pi - (2*I)*b*x)/4, 2]*Sqrt[Sinh[a + b*x]])/(b*Sqrt[I*Sinh[a + b*x]])

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.61

method	result	size
default	\(\frac {i \sqrt {-i \left (\sinh \left (b x +a \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (-\sinh \left (b x +a \right )+i\right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (\sinh \left (b x +a \right )+i\right )}, \frac {\sqrt {2}}{2}\right )}{\cosh \left (b x +a \right ) \sqrt {\sinh \left (b x +a \right )}\, b}\)	\(87\)

[In]

int(1/sinh(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx=\frac {2 \, \sqrt {2} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}{b} \]

[In]

integrate(1/sinh(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

2*sqrt(2)*weierstrassPInverse(4, 0, cosh(b*x + a) + sinh(b*x + a))/b

Sympy [F]

\[ \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx=\int \frac {1}{\sqrt {\sinh {\left (a + b x \right )}}}\, dx \]

[In]

integrate(1/sinh(b*x+a)**(1/2),x)

[Out]

Integral(1/sqrt(sinh(a + b*x)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx=\int { \frac {1}{\sqrt {\sinh \left (b x + a\right )}} \,d x } \]

[In]

integrate(1/sinh(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(sinh(b*x + a)), x)

Giac [F]

\[ \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx=\int { \frac {1}{\sqrt {\sinh \left (b x + a\right )}} \,d x } \]

[In]

integrate(1/sinh(b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(sinh(b*x + a)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx=\int \frac {1}{\sqrt {\mathrm {sinh}\left (a+b\,x\right )}} \,d x \]

[In]

int(1/sinh(a + b*x)^(1/2),x)

[Out]

int(1/sinh(a + b*x)^(1/2), x)