Integrand size = 20, antiderivative size = 105 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {2 \cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {12 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {32 \sin (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}} \]
-2/7*cos(b*x+a)/b/sin(2*b*x+2*a)^(7/2)+12/35*sin(b*x+a)/b/sin(2*b*x+2*a)^( 5/2)-16/35*cos(b*x+a)/b/sin(2*b*x+2*a)^(3/2)+32/35*sin(b*x+a)/b/sin(2*b*x+ 2*a)^(1/2)
Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.64 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\frac {(5-10 \cos (2 (a+b x))-4 \cos (4 (a+b x))+4 \cos (6 (a+b x))) \csc ^4(a+b x) \sec ^3(a+b x) \sqrt {\sin (2 (a+b x))}}{280 b} \]
((5 - 10*Cos[2*(a + b*x)] - 4*Cos[4*(a + b*x)] + 4*Cos[6*(a + b*x)])*Csc[a + b*x]^4*Sec[a + b*x]^3*Sqrt[Sin[2*(a + b*x)]])/(280*b)
Time = 0.58 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4796, 3042, 4791, 3042, 4792, 3042, 4791, 3042, 4780}
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.
| \(\displaystyle \int \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (a+b x) \sin (2 a+2 b x)^{7/2}}dx\) |
| \(\Big \downarrow \) 4796 |
| \(\displaystyle 2 \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{9/2}}dx\) |
| \(\Big \downarrow \) 4791 |
| \(\displaystyle 2 \left (\frac {6}{7} \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \left (\frac {6}{7} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{7/2}}dx-\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\right )\) |
| \(\Big \downarrow \) 4792 |
| \(\displaystyle 2 \left (\frac {6}{7} \left (\frac {4}{5} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx+\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \left (\frac {6}{7} \left (\frac {4}{5} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx+\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\right )\) |
| \(\Big \downarrow \) 4791 |
| \(\displaystyle 2 \left (\frac {6}{7} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )+\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \left (\frac {6}{7} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )+\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\right )\) |
| \(\Big \downarrow \) 4780 |
| \(\displaystyle 2 \left (\frac {6}{7} \left (\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4}{5} \left (\frac {2 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )\right )-\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}\right )\) |
2*((6*((4*(-1/3*Cos[a + b*x]/(b*Sin[2*a + 2*b*x]^(3/2)) + (2*Sin[a + b*x]) /(3*b*Sqrt[Sin[2*a + 2*b*x]])))/5 + Sin[a + b*x]/(5*b*Sin[2*a + 2*b*x]^(5/ 2))))/7 - Cos[a + b*x]/(7*b*Sin[2*a + 2*b*x]^(7/2)))
3.2.4.3.1 Defintions of rubi rules used
Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( p_), x_Symbol] :> Simp[(e*Sin[a + b*x])^m*((g*Sin[c + d*x])^(p + 1)/(b*g*m) ), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b , 2] && !IntegerQ[p] && EqQ[m + 2*p + 2, 0]
Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[Cos[a + b*x]*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Simp [(2*p + 3)/(2*g*(p + 1)) Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x], x ] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !Int egerQ[p] && LtQ[p, -1] && IntegerQ[2*p]
Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-Sin[a + b*x])*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + S imp[(2*p + 3)/(2*g*(p + 1)) Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p + 1), x] , x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && ! IntegerQ[p] && LtQ[p, -1] && IntegerQ[2*p]
Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[2*g Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; FreeQ[{ a, b, c, d, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && IntegerQ[2*p]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 89.89 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.11
| method | result | size |
| default | \(\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right ) \left (3 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{8}+40 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-26 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}+26 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-3\right )}{1344 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}}\) | \(222\) |
1/1344/b*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1 /2*x*b)^2-1)/tan(1/2*a+1/2*x*b)^3*(3*tan(1/2*a+1/2*x*b)^8+40*(tan(1/2*a+1/ 2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2 )*EllipticF((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^3 -26*tan(1/2*a+1/2*x*b)^6+26*tan(1/2*a+1/2*x*b)^2-3)/(tan(1/2*a+1/2*x*b)*(t an(1/2*a+1/2*x*b)^2-1))^(1/2)/(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1 /2)
Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.12 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\frac {128 \, \cos \left (b x + a\right )^{7} - 256 \, \cos \left (b x + a\right )^{5} + 128 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 224 \, \cos \left (b x + a\right )^{4} + 84 \, \cos \left (b x + a\right )^{2} + 7\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{280 \, {\left (b \cos \left (b x + a\right )^{7} - 2 \, b \cos \left (b x + a\right )^{5} + b \cos \left (b x + a\right )^{3}\right )}} \]
1/280*(128*cos(b*x + a)^7 - 256*cos(b*x + a)^5 + 128*cos(b*x + a)^3 + sqrt (2)*(128*cos(b*x + a)^6 - 224*cos(b*x + a)^4 + 84*cos(b*x + a)^2 + 7)*sqrt (cos(b*x + a)*sin(b*x + a)))/(b*cos(b*x + a)^7 - 2*b*cos(b*x + a)^5 + b*co s(b*x + a)^3)
Timed out. \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
\[ \int \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}}} \,d x } \]
Time = 24.44 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.33 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {2\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4}+\frac {{\mathrm {e}}^{a\,3{}\mathrm {i}+b\,x\,3{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,32{}\mathrm {i}}{35\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {2}{7\,b}-\frac {16\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}}{35\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {32{}\mathrm {i}}{35\,b}+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,88{}\mathrm {i}}{35\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3} \]
(exp(a*3i + b*x*3i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i) /2)^(1/2)*32i)/(35*b*(exp(a*2i + b*x*2i) + 1)*(exp(a*2i + b*x*2i)*1i - 1i) ) - (2*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2 i)*1i)/2)^(1/2))/(7*b*(exp(a*2i + b*x*2i)*1i - 1i)^4) - (exp(a*1i + b*x*1i )*(2/(7*b) - (16*exp(a*2i + b*x*2i))/(35*b))*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/((exp(a*2i + b*x*2i) + 1)^2*(exp(a*2i + b*x*2i)*1i - 1i)^2) + (exp(a*1i + b*x*1i)*(32i/(35*b) + (exp(a*2i + b*x* 2i)*88i)/(35*b))*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2) ^(1/2))/((exp(a*2i + b*x*2i) + 1)^3*(exp(a*2i + b*x*2i)*1i - 1i)^3)