Integrand size = 19, antiderivative size = 92 \[ \int \frac {\sec ^4(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\frac {\sec (a+b x)}{2 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {\sec ^3(a+b x)}{3 b \csc ^{\frac {3}{2}}(a+b x)}-\frac {\sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{2 b} \]
1/2*sec(b*x+a)/b/csc(b*x+a)^(3/2)+1/3*sec(b*x+a)^3/b/csc(b*x+a)^(3/2)+1/2* (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)*EllipticE(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 = 0.35 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^4(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\frac {\cos (a+b x) \sqrt {\csc (a+b x)} \left (-3+\sec ^2(a+b x)+2 \sec ^4(a+b x)+3 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right ) \sec (a+b x) \sqrt {\sin (a+b x)}\right )}{6 b} \]
(Cos[a + b*x]*Sqrt[Csc[a + b*x]]*(-3 + Sec[a + b*x]^2 + 2*Sec[a + b*x]^4 + 3*EllipticE[(-2*a + Pi - 2*b*x)/4, 2]*Sec[a + b*x]*Sqrt[Sin[a + b*x]]))/( 6*b)
Time = 0.46 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, 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, 3119}
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.
\(\displaystyle \int \frac {\sec ^4(a+b x)}{\sqrt {\csc (a+b x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (a+b x)^4}{\sqrt {\csc (a+b x)}}dx\) |
\(\Big \downarrow \) 3106 |
\(\displaystyle \frac {1}{2} \int \frac {\sec ^2(a+b x)}{\sqrt {\csc (a+b x)}}dx+\frac {\sec ^3(a+b x)}{3 b \csc ^{\frac {3}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {\sec (a+b x)^2}{\sqrt {\csc (a+b x)}}dx+\frac {\sec ^3(a+b x)}{3 b \csc ^{\frac {3}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3106 |
\(\displaystyle \frac {1}{2} \left (\frac {\sec (a+b x)}{b \csc ^{\frac {3}{2}}(a+b x)}-\frac {1}{2} \int \frac {1}{\sqrt {\csc (a+b x)}}dx\right )+\frac {\sec ^3(a+b x)}{3 b \csc ^{\frac {3}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {\sec (a+b x)}{b \csc ^{\frac {3}{2}}(a+b x)}-\frac {1}{2} \int \frac {1}{\sqrt {\csc (a+b x)}}dx\right )+\frac {\sec ^3(a+b x)}{3 b \csc ^{\frac {3}{2}}(a+b x)}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{2} \left (\frac {\sec (a+b x)}{b \csc ^{\frac {3}{2}}(a+b x)}-\frac {1}{2} \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \sqrt {\sin (a+b x)}dx\right )+\frac {\sec ^3(a+b x)}{3 b \csc ^{\frac {3}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {\sec (a+b x)}{b \csc ^{\frac {3}{2}}(a+b x)}-\frac {1}{2} \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \sqrt {\sin (a+b x)}dx\right )+\frac {\sec ^3(a+b x)}{3 b \csc ^{\frac {3}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\sec ^3(a+b x)}{3 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {1}{2} \left (\frac {\sec (a+b x)}{b \csc ^{\frac {3}{2}}(a+b x)}-\frac {\sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{b}\right )\) |
Sec[a + b*x]^3/(3*b*Csc[a + b*x]^(3/2)) + (Sec[a + b*x]/(b*Csc[a + b*x]^(3 /2)) - (Sqrt[Csc[a + b*x]]*EllipticE[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b *x]])/b)/2
3.3.78.3.1 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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 = 1.77 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.74
method | result | size |
default | \(\frac {6 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, E\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\cos ^{2}\left (b x +a \right )\right )-3 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, F\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\cos ^{2}\left (b x +a \right )\right )-6 \left (\cos ^{4}\left (b x +a \right )\right )+2 \left (\cos ^{2}\left (b x +a \right )\right )+4}{12 \cos \left (b x +a \right )^{3} \sqrt {\sin \left (b x +a \right )}\, b}\) | \(160\) |
1/12/cos(b*x+a)^3/sin(b*x+a)^(1/2)*(6*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+ 2)^(1/2)*(-sin(b*x+a))^(1/2)*EllipticE((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))*c os(b*x+a)^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))*cos(b*x+a)^2-6*cos(b*x+a) ^4+2*cos(b*x+a)^2+4)/b
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24 \[ \int \frac {\sec ^4(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=-\frac {3 \, \sqrt {2 i} \cos \left (b x + a\right )^{3} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 \, \sqrt {-2 i} \cos \left (b x + a\right )^{3} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + \frac {2 \, {\left (3 \, \cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2} - 2\right )}}{\sqrt {\sin \left (b x + a\right )}}}{12 \, b \cos \left (b x + a\right )^{3}} \]
-1/12*(3*sqrt(2*I)*cos(b*x + a)^3*weierstrassZeta(4, 0, weierstrassPInvers e(4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*sqrt(-2*I)*cos(b*x + a)^3*weie rstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a)) ) + 2*(3*cos(b*x + a)^4 - cos(b*x + a)^2 - 2)/sqrt(sin(b*x + a)))/(b*cos(b *x + a)^3)
\[ \int \frac {\sec ^4(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\int \frac {\sec ^{4}{\left (a + b x \right )}}{\sqrt {\csc {\left (a + b x \right )}}}\, dx \]
\[ \int \frac {\sec ^4(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\int { \frac {\sec \left (b x + a\right )^{4}}{\sqrt {\csc \left (b x + a\right )}} \,d x } \]
\[ \int \frac {\sec ^4(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\int { \frac {\sec \left (b x + a\right )^{4}}{\sqrt {\csc \left (b x + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\sec ^4(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\int \frac {1}{{\cos \left (a+b\,x\right )}^4\,\sqrt {\frac {1}{\sin \left (a+b\,x\right )}}} \,d x \]