\(\int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx\) [263]

Optimal result

Integrand size = 19, antiderivative size = 67 \[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {4 \sqrt {\csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{3 b} \]

[Out]

2/3*cos(b*x+a)/b/csc(b*x+a)^(1/2)-4/3*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)*EllipticF(
cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*csc(b*x+a)^(1/2)*sin(b*x+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2708, 3856, 2720} \[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {4 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{3 b} \]

[In]

Int[Cos[a + b*x]^2*Sqrt[Csc[a + b*x]],x]

[Out]

(2*Cos[a + b*x])/(3*b*Sqrt[Csc[a + b*x]]) + (4*Sqrt[Csc[a + b*x]]*EllipticF[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a
+ b*x]])/(3*b)

Rule 2708

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a)*(a*Csc[
e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n + 1)/(b*f*(m + n))), x] + Dist[(n + 1)/(b^2*(m + n)), Int[(a*Csc[e + f*
x])^m*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n, 0] && Integers
Q[2*m, 2*n]

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}& = \frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {2}{3} \int \sqrt {\csc (a+b x)} \, dx \\ & = \frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {1}{3} \left (2 \sqrt {\csc (a+b x)} \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx \\ & = \frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {4 \sqrt {\csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.79 \[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {\sqrt {\csc (a+b x)} \left (-4 \operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right ) \sqrt {\sin (a+b x)}+\sin (2 (a+b x))\right )}{3 b} \]

[In]

Integrate[Cos[a + b*x]^2*Sqrt[Csc[a + b*x]],x]

[Out]

(Sqrt[Csc[a + b*x]]*(-4*EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[Sin[a + b*x]] + Sin[2*(a + b*x)]))/(3*b)

Maple [A] (verified)

Time = 1.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31

method	result	size
default	\(\frac {\frac {2 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, F\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 \left (\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}}{\cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b}\)	\(88\)

[In]

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

[Out]

(2/3*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*EllipticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/
2))+2/3*cos(b*x+a)^2*sin(b*x+a))/cos(b*x+a)/sin(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 = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {2 \, {\left (\cos \left (b x + a\right ) \sqrt {\sin \left (b x + a\right )} - i \, \sqrt {2 i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, \sqrt {-2 i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}}{3 \, b} \]

[In]

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

[Out]

2/3*(cos(b*x + a)*sqrt(sin(b*x + a)) - I*sqrt(2*I)*weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a)) +
I*sqrt(-2*I)*weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a)))/b

Sympy [F]

\[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\int \cos ^{2}{\left (a + b x \right )} \sqrt {\csc {\left (a + b x \right )}}\, dx \]

[In]

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

[Out]

Integral(cos(a + b*x)**2*sqrt(csc(a + b*x)), x)

Maxima [F]

\[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\csc \left (b x + a\right )} \,d x } \]

[In]

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

[Out]

integrate(cos(b*x + a)^2*sqrt(csc(b*x + a)), x)

Giac [F]

\[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\csc \left (b x + a\right )} \,d x } \]

[In]

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

[Out]

integrate(cos(b*x + a)^2*sqrt(csc(b*x + a)), x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\int {\cos \left (a+b\,x\right )}^2\,\sqrt {\frac {1}{\sin \left (a+b\,x\right )}} \,d x \]

[In]

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

[Out]

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