Integrand size = 19, antiderivative size = 67 \[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {4 \sqrt {\csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{3 b} \] Output:
2/3*cos(b*x+a)/b/csc(b*x+a)^(1/2)+4/3*csc(b*x+a)^(1/2)*InverseJacobiAM(1/2 *a-1/4*Pi+1/2*b*x,2^(1/2))*sin(b*x+a)^(1/2)/b
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.79 \[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {\sqrt {\csc (a+b x)} \left (-4 \operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right ) \sqrt {\sin (a+b x)}+\sin (2 (a+b x))\right )}{3 b} \] Input:
Integrate[Cos[a + b*x]^2*Sqrt[Csc[a + b*x]],x]
Output:
(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)
Time = 0.38 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3108, 3042, 4258, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\csc (a+b x)}}{\sec (a+b x)^2}dx\) |
\(\Big \downarrow \) 3108 |
\(\displaystyle \frac {2}{3} \int \sqrt {\csc (a+b x)}dx+\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \int \sqrt {\csc (a+b x)}dx+\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {2}{3} \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}}dx+\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}}dx+\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {4 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{3 b}\) |
Input:
Int[Cos[a + b*x]^2*Sqrt[Csc[a + b*x]],x]
Output:
(2*Cos[a + b*x])/(3*b*Sqrt[Csc[a + b*x]]) + (4*Sqrt[Csc[a + b*x]]*Elliptic F[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/(3*b)
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] + Simp[(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] && IntegersQ[2*m, 2*n]
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]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(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]
Time = 0.73 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31
method | result | size |
default | \(\frac {\frac {2 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )}{3}}{\cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b}\) | \(88\) |
Input:
int(cos(b*x+a)^2*csc(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
(2/3*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Elli pticF((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
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {2 \, {\left (\cos \left (b x + a\right ) \sqrt {\sin \left (b x + a\right )} - i \, \sqrt {2 i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, \sqrt {-2 i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}}{3 \, b} \] Input:
integrate(cos(b*x+a)^2*csc(b*x+a)^(1/2),x, algorithm="fricas")
Output:
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
\[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\int \cos ^{2}{\left (a + b x \right )} \sqrt {\csc {\left (a + b x \right )}}\, dx \] Input:
integrate(cos(b*x+a)**2*csc(b*x+a)**(1/2),x)
Output:
Integral(cos(a + b*x)**2*sqrt(csc(a + b*x)), x)
\[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\csc \left (b x + a\right )} \,d x } \] Input:
integrate(cos(b*x+a)^2*csc(b*x+a)^(1/2),x, algorithm="maxima")
Output:
integrate(cos(b*x + a)^2*sqrt(csc(b*x + a)), x)
\[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\csc \left (b x + a\right )} \,d x } \] Input:
integrate(cos(b*x+a)^2*csc(b*x+a)^(1/2),x, algorithm="giac")
Output:
integrate(cos(b*x + a)^2*sqrt(csc(b*x + a)), x)
Timed out. \[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\int {\cos \left (a+b\,x\right )}^2\,\sqrt {\frac {1}{\sin \left (a+b\,x\right )}} \,d x \] Input:
int(cos(a + b*x)^2*(1/sin(a + b*x))^(1/2),x)
Output:
int(cos(a + b*x)^2*(1/sin(a + b*x))^(1/2), x)
\[ \int \cos ^2(a+b x) \sqrt {\csc (a+b x)} \, dx=\int \sqrt {\csc \left (b x +a \right )}\, \cos \left (b x +a \right )^{2}d x \] Input:
int(cos(b*x+a)^2*csc(b*x+a)^(1/2),x)
Output:
int(sqrt(csc(a + b*x))*cos(a + b*x)**2,x)