Integrand size = 12, antiderivative size = 105 \[ \int \frac {1}{(c \csc (a+b x))^{7/2}} \, dx=-\frac {2 \cos (a+b x)}{7 b c (c \csc (a+b x))^{5/2}}-\frac {10 \cos (a+b x)}{21 b c^3 \sqrt {c \csc (a+b x)}}+\frac {10 \sqrt {c \csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{21 b c^4} \]
-2/7*cos(b*x+a)/b/c/(c*csc(b*x+a))^(5/2)-10/21*cos(b*x+a)/b/c^3/(c*csc(b*x +a))^(1/2)-10/21*(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(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*(c*csc(b*x+a))^(1/2)*sin (b*x+a)^(1/2)/b/c^4
Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(c \csc (a+b x))^{7/2}} \, dx=-\frac {\sqrt {c \csc (a+b x)} \left (40 \operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right ) \sqrt {\sin (a+b x)}+26 \sin (2 (a+b x))-3 \sin (4 (a+b x))\right )}{84 b c^4} \]
-1/84*(Sqrt[c*Csc[a + b*x]]*(40*EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[S in[a + b*x]] + 26*Sin[2*(a + b*x)] - 3*Sin[4*(a + b*x)]))/(b*c^4)
Time = 0.43 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4256, 3042, 4256, 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.
\(\displaystyle \int \frac {1}{(c \csc (a+b x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(c \csc (a+b x))^{7/2}}dx\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {5 \int \frac {1}{(c \csc (a+b x))^{3/2}}dx}{7 c^2}-\frac {2 \cos (a+b x)}{7 b c (c \csc (a+b x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \int \frac {1}{(c \csc (a+b x))^{3/2}}dx}{7 c^2}-\frac {2 \cos (a+b x)}{7 b c (c \csc (a+b x))^{5/2}}\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {5 \left (\frac {\int \sqrt {c \csc (a+b x)}dx}{3 c^2}-\frac {2 \cos (a+b x)}{3 b c \sqrt {c \csc (a+b x)}}\right )}{7 c^2}-\frac {2 \cos (a+b x)}{7 b c (c \csc (a+b x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \left (\frac {\int \sqrt {c \csc (a+b x)}dx}{3 c^2}-\frac {2 \cos (a+b x)}{3 b c \sqrt {c \csc (a+b x)}}\right )}{7 c^2}-\frac {2 \cos (a+b x)}{7 b c (c \csc (a+b x))^{5/2}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {5 \left (\frac {\sqrt {\sin (a+b x)} \sqrt {c \csc (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}}dx}{3 c^2}-\frac {2 \cos (a+b x)}{3 b c \sqrt {c \csc (a+b x)}}\right )}{7 c^2}-\frac {2 \cos (a+b x)}{7 b c (c \csc (a+b x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \left (\frac {\sqrt {\sin (a+b x)} \sqrt {c \csc (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}}dx}{3 c^2}-\frac {2 \cos (a+b x)}{3 b c \sqrt {c \csc (a+b x)}}\right )}{7 c^2}-\frac {2 \cos (a+b x)}{7 b c (c \csc (a+b x))^{5/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {5 \left (\frac {2 \sqrt {\sin (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right ) \sqrt {c \csc (a+b x)}}{3 b c^2}-\frac {2 \cos (a+b x)}{3 b c \sqrt {c \csc (a+b x)}}\right )}{7 c^2}-\frac {2 \cos (a+b x)}{7 b c (c \csc (a+b x))^{5/2}}\) |
(-2*Cos[a + b*x])/(7*b*c*(c*Csc[a + b*x])^(5/2)) + (5*((-2*Cos[a + b*x])/( 3*b*c*Sqrt[c*Csc[a + b*x]]) + (2*Sqrt[c*Csc[a + b*x]]*EllipticF[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/(3*b*c^2)))/(7*c^2)
3.1.24.3.1 Defintions of rubi rules used
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[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 = 3.86 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.70
method | result | size |
default | \(\frac {\sin \left (x b +a \right )^{3} \left (5 i \sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (-i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (x b +a \right )+3 \sqrt {2}\, \cos \left (x b +a \right )^{3} \sin \left (x b +a \right )+5 i \sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (-i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}\, \sqrt {i \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (i-\cot \left (x b +a \right )+\csc \left (x b +a \right )\right )}, \frac {\sqrt {2}}{2}\right )-8 \sqrt {2}\, \cos \left (x b +a \right ) \sin \left (x b +a \right )\right ) \sqrt {2}}{21 b \,c^{3} \sqrt {c \csc \left (x b +a \right )}\, \left (\cos \left (x b +a \right )-1\right )^{2} \left (1+\cos \left (x b +a \right )\right )^{2}}\) | \(283\) |
1/21/b*sin(b*x+a)^3*(5*I*(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))*(-I*(I-cot(b*x+a)+csc(b*x+a)))^(1/2)*cos(b*x+a)+3*2^(1/2)*cos(b*x+ a)^3*sin(b*x+a)+5*I*(I*(-I-cot(b*x+a)+csc(b*x+a)))^(1/2)*(I*(csc(b*x+a)-co t(b*x+a)))^(1/2)*EllipticF((-I*(I-cot(b*x+a)+csc(b*x+a)))^(1/2),1/2*2^(1/2 ))*(-I*(I-cot(b*x+a)+csc(b*x+a)))^(1/2)-8*2^(1/2)*cos(b*x+a)*sin(b*x+a))/c ^3/(c*csc(b*x+a))^(1/2)/(cos(b*x+a)-1)^2/(1+cos(b*x+a))^2*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(c \csc (a+b x))^{7/2}} \, dx=\frac {2 \, {\left (3 \, \cos \left (b x + a\right )^{3} - 8 \, \cos \left (b x + a\right )\right )} \sqrt {\frac {c}{\sin \left (b x + a\right )}} \sin \left (b x + a\right ) - 5 i \, \sqrt {2 i \, c} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 5 i \, \sqrt {-2 i \, c} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )}{21 \, b c^{4}} \]
1/21*(2*(3*cos(b*x + a)^3 - 8*cos(b*x + a))*sqrt(c/sin(b*x + a))*sin(b*x + a) - 5*I*sqrt(2*I*c)*weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a)) + 5*I*sqrt(-2*I*c)*weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a)))/(b*c^4)
\[ \int \frac {1}{(c \csc (a+b x))^{7/2}} \, dx=\int \frac {1}{\left (c \csc {\left (a + b x \right )}\right )^{\frac {7}{2}}}\, dx \]
\[ \int \frac {1}{(c \csc (a+b x))^{7/2}} \, dx=\int { \frac {1}{\left (c \csc \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {1}{(c \csc (a+b x))^{7/2}} \, dx=\int { \frac {1}{\left (c \csc \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(c \csc (a+b x))^{7/2}} \, dx=\int \frac {1}{{\left (\frac {c}{\sin \left (a+b\,x\right )}\right )}^{7/2}} \,d x \]