Integrand size = 10, antiderivative size = 90 \[ \int \csc ^{\frac {7}{2}}(a+b x) \, dx=-\frac {6 \cos (a+b x) \sqrt {\csc (a+b x)}}{5 b}-\frac {2 \cos (a+b x) \csc ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {6 \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{5 b} \]
-2/5*cos(b*x+a)*csc(b*x+a)^(5/2)/b-6/5*cos(b*x+a)*csc(b*x+a)^(1/2)/b+6/5*( 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(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
Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.70 \[ \int \csc ^{\frac {7}{2}}(a+b x) \, dx=\frac {\csc ^{\frac {5}{2}}(a+b x) \left (-7 \cos (a+b x)+3 \cos (3 (a+b x))+12 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right ) \sin ^{\frac {5}{2}}(a+b x)\right )}{10 b} \]
(Csc[a + b*x]^(5/2)*(-7*Cos[a + b*x] + 3*Cos[3*(a + b*x)] + 12*EllipticE[( -2*a + Pi - 2*b*x)/4, 2]*Sin[a + b*x]^(5/2)))/(10*b)
Time = 0.39 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 4255, 3042, 4255, 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.
\(\displaystyle \int \csc ^{\frac {7}{2}}(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x)^{7/2}dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {3}{5} \int \csc ^{\frac {3}{2}}(a+b x)dx-\frac {2 \cos (a+b x) \csc ^{\frac {5}{2}}(a+b x)}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} \int \csc (a+b x)^{3/2}dx-\frac {2 \cos (a+b x) \csc ^{\frac {5}{2}}(a+b x)}{5 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {3}{5} \left (-\int \frac {1}{\sqrt {\csc (a+b x)}}dx-\frac {2 \cos (a+b x) \sqrt {\csc (a+b x)}}{b}\right )-\frac {2 \cos (a+b x) \csc ^{\frac {5}{2}}(a+b x)}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} \left (-\int \frac {1}{\sqrt {\csc (a+b x)}}dx-\frac {2 \cos (a+b x) \sqrt {\csc (a+b x)}}{b}\right )-\frac {2 \cos (a+b x) \csc ^{\frac {5}{2}}(a+b x)}{5 b}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {3}{5} \left (-\sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \sqrt {\sin (a+b x)}dx-\frac {2 \cos (a+b x) \sqrt {\csc (a+b x)}}{b}\right )-\frac {2 \cos (a+b x) \csc ^{\frac {5}{2}}(a+b x)}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} \left (-\sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \sqrt {\sin (a+b x)}dx-\frac {2 \cos (a+b x) \sqrt {\csc (a+b x)}}{b}\right )-\frac {2 \cos (a+b x) \csc ^{\frac {5}{2}}(a+b x)}{5 b}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {3}{5} \left (-\frac {2 \cos (a+b x) \sqrt {\csc (a+b x)}}{b}-\frac {2 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{b}\right )-\frac {2 \cos (a+b x) \csc ^{\frac {5}{2}}(a+b x)}{5 b}\) |
(-2*Cos[a + b*x]*Csc[a + b*x]^(5/2))/(5*b) + (3*((-2*Cos[a + b*x]*Sqrt[Csc [a + b*x]])/b - (2*Sqrt[Csc[a + b*x]]*EllipticE[(a - Pi/2 + b*x)/2, 2]*Sqr t[Sin[a + b*x]])/b))/5
3.1.9.3.1 Defintions of rubi rules used
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)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
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 = 0.66 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.78
method | result | size |
default | \(\frac {6 \sqrt {\sin \left (x b +a \right )+1}\, \sqrt {-2 \sin \left (x b +a \right )+2}\, \sqrt {-\sin \left (x b +a \right )}\, \sin \left (x b +a \right )^{2} \operatorname {EllipticE}\left (\sqrt {\sin \left (x b +a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\sin \left (x b +a \right )+1}\, \sqrt {-2 \sin \left (x b +a \right )+2}\, \sqrt {-\sin \left (x b +a \right )}\, \sin \left (x b +a \right )^{2} \operatorname {EllipticF}\left (\sqrt {\sin \left (x b +a \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \sin \left (x b +a \right )^{4}-4 \sin \left (x b +a \right )^{2}-2}{5 \sin \left (x b +a \right )^{\frac {5}{2}} \cos \left (x b +a \right ) b}\) | \(160\) |
1/5/sin(b*x+a)^(5/2)*(6*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin (b*x+a))^(1/2)*sin(b*x+a)^2*EllipticE((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))-3* (sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*sin(b*x+a )^2*EllipticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))+6*sin(b*x+a)^4-4*sin(b*x+a )^2-2)/cos(b*x+a)/b
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.33 \[ \int \csc ^{\frac {7}{2}}(a+b x) \, dx=-\frac {3 \, \sqrt {2 i} {\left (\cos \left (b x + a\right )^{2} - 1\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 \, \sqrt {-2 i} {\left (\cos \left (b x + a\right )^{2} - 1\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + \frac {2 \, {\left (3 \, \cos \left (b x + a\right )^{3} - 4 \, \cos \left (b x + a\right )\right )}}{\sqrt {\sin \left (b x + a\right )}}}{5 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]
-1/5*(3*sqrt(2*I)*(cos(b*x + a)^2 - 1)*weierstrassZeta(4, 0, weierstrassPI nverse(4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*sqrt(-2*I)*(cos(b*x + a)^ 2 - 1)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x + a) - I*si n(b*x + a))) + 2*(3*cos(b*x + a)^3 - 4*cos(b*x + a))/sqrt(sin(b*x + a)))/( b*cos(b*x + a)^2 - b)
\[ \int \csc ^{\frac {7}{2}}(a+b x) \, dx=\int \csc ^{\frac {7}{2}}{\left (a + b x \right )}\, dx \]
\[ \int \csc ^{\frac {7}{2}}(a+b x) \, dx=\int { \csc \left (b x + a\right )^{\frac {7}{2}} \,d x } \]
\[ \int \csc ^{\frac {7}{2}}(a+b x) \, dx=\int { \csc \left (b x + a\right )^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int \csc ^{\frac {7}{2}}(a+b x) \, dx=\int {\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^{7/2} \,d x \]