Integrand size = 19, antiderivative size = 92 \[ \int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx=\frac {5 \sec (a+b x)}{6 b \sqrt {\csc (a+b x)}}+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {5 \sqrt {\csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{6 b} \]
5/6*sec(b*x+a)/b/csc(b*x+a)^(1/2)+1/3*sec(b*x+a)^3/b/csc(b*x+a)^(1/2)-5/6* (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(co s(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
Time = 0.53 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx=\frac {\sqrt {\csc (a+b x)} \left (-5 \operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right ) \sqrt {\sin (a+b x)}+\left (5+2 \sec ^2(a+b x)\right ) \tan (a+b x)\right )}{6 b} \]
(Sqrt[Csc[a + b*x]]*(-5*EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[Sin[a + b *x]] + (5 + 2*Sec[a + b*x]^2)*Tan[a + b*x]))/(6*b)
Time = 0.44 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3106, 3042, 3106, 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.
\(\displaystyle \int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\csc (a+b x)} \sec (a+b x)^4dx\) |
\(\Big \downarrow \) 3106 |
\(\displaystyle \frac {5}{6} \int \sqrt {\csc (a+b x)} \sec ^2(a+b x)dx+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \int \sqrt {\csc (a+b x)} \sec (a+b x)^2dx+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 3106 |
\(\displaystyle \frac {5}{6} \left (\frac {1}{2} \int \sqrt {\csc (a+b x)}dx+\frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}\right )+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {1}{2} \int \sqrt {\csc (a+b x)}dx+\frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}\right )+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {5}{6} \left (\frac {1}{2} \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}}dx+\frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}\right )+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {1}{2} \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}}dx+\frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}\right )+\frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\sec ^3(a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {5}{6} \left (\frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}+\frac {\sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{b}\right )\) |
Sec[a + b*x]^3/(3*b*Sqrt[Csc[a + b*x]]) + (5*(Sec[a + b*x]/(b*Sqrt[Csc[a + b*x]]) + (Sqrt[Csc[a + b*x]]*EllipticF[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/b))/6
3.3.77.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a*b*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(n - 1))), x] + Simp[b^2*((m + n - 2)/(n - 1)) Int[(a*Csc[e + f*x]) ^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && 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 = 1.32 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.83
method | result | size |
default | \(-\frac {\sqrt {\left (\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}\, \left (5 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, F\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\cos ^{2}\left (b x +a \right )\right )+10 \left (\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )+4 \sin \left (b x +a \right )\right )}{12 \left (\sin \left (b x +a \right )+1\right ) \left (\sin \left (b x +a \right )-1\right ) \sqrt {-\sin \left (b x +a \right ) \left (\sin \left (b x +a \right )-1\right ) \left (\sin \left (b x +a \right )+1\right )}\, \cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b}\) | \(168\) |
-1/12*(cos(b*x+a)^2*sin(b*x+a))^(1/2)*(5*(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) )*cos(b*x+a)^2+10*cos(b*x+a)^2*sin(b*x+a)+4*sin(b*x+a))/(sin(b*x+a)+1)/(si n(b*x+a)-1)/(-sin(b*x+a)*(sin(b*x+a)-1)*(sin(b*x+a)+1))^(1/2)/cos(b*x+a)/s in(b*x+a)^(1/2)/b
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx=\frac {-5 i \, \sqrt {2 i} \cos \left (b x + a\right )^{3} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 5 i \, \sqrt {-2 i} \cos \left (b x + a\right )^{3} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 2 \, {\left (5 \, \cos \left (b x + a\right )^{2} + 2\right )} \sqrt {\sin \left (b x + a\right )}}{12 \, b \cos \left (b x + a\right )^{3}} \]
1/12*(-5*I*sqrt(2*I)*cos(b*x + a)^3*weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a)) + 5*I*sqrt(-2*I)*cos(b*x + a)^3*weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a)) + 2*(5*cos(b*x + a)^2 + 2)*sqrt(sin(b*x + a)))/(b*cos(b*x + a)^3)
\[ \int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx=\int \sqrt {\csc {\left (a + b x \right )}} \sec ^{4}{\left (a + b x \right )}\, dx \]
\[ \int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx=\int { \sqrt {\csc \left (b x + a\right )} \sec \left (b x + a\right )^{4} \,d x } \]
\[ \int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx=\int { \sqrt {\csc \left (b x + a\right )} \sec \left (b x + a\right )^{4} \,d x } \]
Timed out. \[ \int \sqrt {\csc (a+b x)} \sec ^4(a+b x) \, dx=\int \frac {\sqrt {\frac {1}{\sin \left (a+b\,x\right )}}}{{\cos \left (a+b\,x\right )}^4} \,d x \]