\(\int \frac {1}{\sqrt {\sin (b x)}} \, dx\) [13]

Optimal result

Integrand size = 8, antiderivative size = 19 \[ \int \frac {1}{\sqrt {\sin (b x)}} \, dx=-\frac {2 \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {b x}{2},2\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2720} \[ \int \frac {1}{\sqrt {\sin (b x)}} \, dx=-\frac {2 \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {b x}{2},2\right )}{b} \]

[In]

Int[1/Sqrt[Sin[b*x]],x]

[Out]

(-2*EllipticF[Pi/4 - (b*x)/2, 2])/b

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 \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {b x}{2},2\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {1}{\sqrt {\sin (b x)}} \, dx=-\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (\frac {\pi }{2}-b x\right ),2\right )}{b} \]

[In]

Integrate[1/Sqrt[Sin[b*x]],x]

[Out]

(-2*EllipticF[(Pi/2 - b*x)/2, 2])/b

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 3.00

method	result	size
default	\(\frac {\sqrt {\sin \left (b x \right )+1}\, \sqrt {-2 \sin \left (b x \right )+2}\, \sqrt {-\sin \left (b x \right )}\, F\left (\sqrt {\sin \left (b x \right )+1}, \frac {\sqrt {2}}{2}\right )}{\cos \left (b x \right ) \sqrt {\sin \left (b x \right )}\, b}\)	\(57\)

[In]

int(1/sin(b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(sin(b*x)+1)^(1/2)*(-2*sin(b*x)+2)^(1/2)*(-sin(b*x))^(1/2)*EllipticF((sin(b*x)+1)^(1/2),1/2*2^(1/2))/cos(b*x)/
sin(b*x)^(1/2)/b

Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.47 \[ \int \frac {1}{\sqrt {\sin (b x)}} \, dx=\frac {\sqrt {2} \sqrt {-i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x\right ) + i \, \sin \left (b x\right )\right ) + \sqrt {2} \sqrt {i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x\right ) - i \, \sin \left (b x\right )\right )}{b} \]

[In]

integrate(1/sin(b*x)^(1/2),x, algorithm="fricas")

[Out]

(sqrt(2)*sqrt(-I)*weierstrassPInverse(4, 0, cos(b*x) + I*sin(b*x)) + sqrt(2)*sqrt(I)*weierstrassPInverse(4, 0,
cos(b*x) - I*sin(b*x)))/b

Sympy [F]

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

[In]

integrate(1/sin(b*x)**(1/2),x)

[Out]

Integral(1/sqrt(sin(b*x)), x)

Maxima [F]

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

[In]

integrate(1/sin(b*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(sin(b*x)), x)

Giac [F]

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

[In]

integrate(1/sin(b*x)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(sin(b*x)), x)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\sqrt {\sin (b x)}} \, dx=-\frac {2\,\mathrm {F}\left (\frac {\pi }{4}-\frac {b\,x}{2}\middle |2\right )}{b} \]

[In]

int(1/sin(b*x)^(1/2),x)

[Out]

-(2*ellipticF(pi/4 - (b*x)/2, 2))/b