Integrand size = 12, antiderivative size = 77 \[ \int \frac {1}{(c \csc (a+b x))^{5/2}} \, dx=-\frac {2 \cos (a+b x)}{5 b c (c \csc (a+b x))^{3/2}}+\frac {6 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right )}{5 b c^2 \sqrt {c \csc (a+b x)} \sqrt {\sin (a+b x)}} \]
-2/5*cos(b*x+a)/b/c/(c*csc(b*x+a))^(3/2)-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))/b/c^2/(c*csc(b*x+a))^(1/2)/sin(b*x+a)^(1/2)
Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(c \csc (a+b x))^{5/2}} \, dx=\frac {-\frac {12 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right )}{\sqrt {\sin (a+b x)}}-2 \sin (2 (a+b x))}{10 b c^2 \sqrt {c \csc (a+b x)}} \]
((-12*EllipticE[(-2*a + Pi - 2*b*x)/4, 2])/Sqrt[Sin[a + b*x]] - 2*Sin[2*(a + b*x)])/(10*b*c^2*Sqrt[c*Csc[a + b*x]])
Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4256, 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.
\(\displaystyle \int \frac {1}{(c \csc (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(c \csc (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {3 \int \frac {1}{\sqrt {c \csc (a+b x)}}dx}{5 c^2}-\frac {2 \cos (a+b x)}{5 b c (c \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {1}{\sqrt {c \csc (a+b x)}}dx}{5 c^2}-\frac {2 \cos (a+b x)}{5 b c (c \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {3 \int \sqrt {\sin (a+b x)}dx}{5 c^2 \sqrt {\sin (a+b x)} \sqrt {c \csc (a+b x)}}-\frac {2 \cos (a+b x)}{5 b c (c \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \sqrt {\sin (a+b x)}dx}{5 c^2 \sqrt {\sin (a+b x)} \sqrt {c \csc (a+b x)}}-\frac {2 \cos (a+b x)}{5 b c (c \csc (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {6 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{5 b c^2 \sqrt {\sin (a+b x)} \sqrt {c \csc (a+b x)}}-\frac {2 \cos (a+b x)}{5 b c (c \csc (a+b x))^{3/2}}\) |
(-2*Cos[a + b*x])/(5*b*c*(c*Csc[a + b*x])^(3/2)) + (6*EllipticE[(a - Pi/2 + b*x)/2, 2])/(5*b*c^2*Sqrt[c*Csc[a + b*x]]*Sqrt[Sin[a + b*x]])
3.1.23.3.1 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[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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]
Result contains complex when optimal does not.
Time = 3.56 (sec) , antiderivative size = 445, normalized size of antiderivative = 5.78
method | result | size |
default | \(\frac {\csc \left (x b +a \right ) \left (3 \sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (-i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (x b +a \right )-6 \sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (-i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}\, \operatorname {EllipticE}\left (\sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (x b +a \right )+\sqrt {2}\, \cos \left (x b +a \right )^{3}+3 \sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (-i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}, \frac {\sqrt {2}}{2}\right )-6 \sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (-i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}\, \operatorname {EllipticE}\left (\sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}, \frac {\sqrt {2}}{2}\right )-4 \cos \left (x b +a \right ) \sqrt {2}+3 \sqrt {2}\right ) \sqrt {2}}{5 b \sqrt {c \csc \left (x b +a \right )}\, c^{2}}\) | \(445\) |
1/5/b*csc(b*x+a)*(3*(-I*(I-cot(b*x+a)+csc(b*x+a)))^(1/2)*(I*(-I-cot(b*x+a) +csc(b*x+a)))^(1/2)*(I*(csc(b*x+a)-cot(b*x+a)))^(1/2)*EllipticF((-I*(I-cot (b*x+a)+csc(b*x+a)))^(1/2),1/2*2^(1/2))*cos(b*x+a)-6*(-I*(I-cot(b*x+a)+csc (b*x+a)))^(1/2)*(I*(-I-cot(b*x+a)+csc(b*x+a)))^(1/2)*(I*(csc(b*x+a)-cot(b* x+a)))^(1/2)*EllipticE((-I*(I-cot(b*x+a)+csc(b*x+a)))^(1/2),1/2*2^(1/2))*c os(b*x+a)+2^(1/2)*cos(b*x+a)^3+3*(-I*(I-cot(b*x+a)+csc(b*x+a)))^(1/2)*(I*( -I-cot(b*x+a)+csc(b*x+a)))^(1/2)*(I*(csc(b*x+a)-cot(b*x+a)))^(1/2)*Ellipti cF((-I*(I-cot(b*x+a)+csc(b*x+a)))^(1/2),1/2*2^(1/2))-6*(-I*(I-cot(b*x+a)+c sc(b*x+a)))^(1/2)*(I*(-I-cot(b*x+a)+csc(b*x+a)))^(1/2)*(I*(csc(b*x+a)-cot( b*x+a)))^(1/2)*EllipticE((-I*(I-cot(b*x+a)+csc(b*x+a)))^(1/2),1/2*2^(1/2)) -4*cos(b*x+a)*2^(1/2)+3*2^(1/2))/(c*csc(b*x+a))^(1/2)/c^2*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25 \[ \int \frac {1}{(c \csc (a+b x))^{5/2}} \, dx=\frac {2 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {\frac {c}{\sin \left (b x + a\right )}} + 3 \, \sqrt {2 i \, c} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 \, \sqrt {-2 i \, c} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}{5 \, b c^{3}} \]
1/5*(2*(cos(b*x + a)^3 - cos(b*x + a))*sqrt(c/sin(b*x + a)) + 3*sqrt(2*I*c )*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*sqrt(-2*I*c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, c os(b*x + a) - I*sin(b*x + a))))/(b*c^3)
\[ \int \frac {1}{(c \csc (a+b x))^{5/2}} \, dx=\int \frac {1}{\left (c \csc {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{(c \csc (a+b x))^{5/2}} \, dx=\int { \frac {1}{\left (c \csc \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{(c \csc (a+b x))^{5/2}} \, dx=\int { \frac {1}{\left (c \csc \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(c \csc (a+b x))^{5/2}} \, dx=\int \frac {1}{{\left (\frac {c}{\sin \left (a+b\,x\right )}\right )}^{5/2}} \,d x \]