\(\int \frac {1}{\sqrt {i \sinh (c+d x)}} \, dx\) [27]

Optimal result

Integrand size = 14, antiderivative size = 30 \[ \int \frac {1}{\sqrt {i \sinh (c+d x)}} \, dx=-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right )}{d} \]

[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))/d

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2720} \[ \int \frac {1}{\sqrt {i \sinh (c+d x)}} \, dx=-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right )}{d} \]

[In]

Int[1/Sqrt[I*Sinh[c + d*x]],x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\sqrt {i \sinh (c+d x)}} \, dx=\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (\frac {\pi }{2}-i (c+d x)\right ),2\right )}{d} \]

[In]

Integrate[1/Sqrt[I*Sinh[c + d*x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.27

method	result	size
default	\(\frac {i \sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (i-\sinh \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}, \frac {\sqrt {2}}{2}\right )}{\cosh \left (d x +c \right ) d}\)	\(68\)

[In]

int(1/(I*sinh(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {i \sinh (c+d x)}} \, dx=-\frac {2 i \, \sqrt {2} \sqrt {i} {\rm weierstrassPInverse}\left (4, 0, e^{\left (d x + c\right )}\right )}{d} \]

[In]

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

[Out]

-2*I*sqrt(2)*sqrt(I)*weierstrassPInverse(4, 0, e^(d*x + c))/d

Sympy [F]

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

[In]

integrate(1/(I*sinh(d*x+c))**(1/2),x)

[Out]

Integral(1/sqrt(I*sinh(c + d*x)), x)

Maxima [F]

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

[In]

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

[Out]

integrate(1/sqrt(I*sinh(d*x + c)), x)

Giac [F]

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

[In]

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

[Out]

integrate(1/sqrt(I*sinh(d*x + c)), x)

Mupad [F(-1)]

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

[In]

int(1/(sinh(c + d*x)*1i)^(1/2),x)

[Out]

int(1/(sinh(c + d*x)*1i)^(1/2), x)