Integrand size = 20, antiderivative size = 79 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}} \]
-2/5*cos(b*x+a)/b/sin(2*b*x+2*a)^(5/2)+8/15*sin(b*x+a)/b/sin(2*b*x+2*a)^(3 /2)-16/15*cos(b*x+a)/b/sin(2*b*x+2*a)^(1/2)
Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {\sin (2 (a+b x))} \left (27 \csc (a+b x)+3 \csc ^3(a+b x)-5 \sec (a+b x) \tan (a+b x)\right )}{60 b} \]
-1/60*(Sqrt[Sin[2*(a + b*x)]]*(27*Csc[a + b*x] + 3*Csc[a + b*x]^3 - 5*Sec[ a + b*x]*Tan[a + b*x]))/b
Time = 0.45 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4796, 3042, 4791, 3042, 4792, 3042, 4779}
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.
\(\displaystyle \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x) \sin (2 a+2 b x)^{5/2}}dx\) |
\(\Big \downarrow \) 4796 |
\(\displaystyle 2 \int \frac {\cos (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{7/2}}dx\) |
\(\Big \downarrow \) 4791 |
\(\displaystyle 2 \left (\frac {4}{5} \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \left (\frac {4}{5} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )\) |
\(\Big \downarrow \) 4792 |
\(\displaystyle 2 \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx+\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx+\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )\) |
\(\Big \downarrow \) 4779 |
\(\displaystyle 2 \left (\frac {4}{5} \left (\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )\) |
2*((4*(Sin[a + b*x]/(3*b*Sin[2*a + 2*b*x]^(3/2)) - (2*Cos[a + b*x])/(3*b*S qrt[Sin[2*a + 2*b*x]])))/5 - Cos[a + b*x]/(5*b*Sin[2*a + 2*b*x]^(5/2)))
3.2.3.3.1 Defintions of rubi rules used
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( p_), x_Symbol] :> Simp[(-(e*Cos[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 = 16.42 (sec) , antiderivative size = 481, normalized size of antiderivative = 6.09
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 (24 \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 )+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 {EllipticE}\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 )^{2}-12 \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 )+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 )^{2}+\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 )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}+12 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-\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 )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-12 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-\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 )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+\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 )}\right )}{80 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}}\) | \(481\) |
-1/80/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)^3*(24*(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)+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)*EllipticE((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b )^2-12*(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)+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)^2+( 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)^6+12 *(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^4-(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)^4-12*(t an(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^2-(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+(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)^3-tan(1/2 *a+1/2*x*b))^(1/2)
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.30 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 40 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{60 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \]
-1/60*(sqrt(2)*(32*cos(b*x + a)^4 - 40*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*(cos(b*x + a)^4 - cos(b*x + a)^2)*sin(b*x + a))/((b* cos(b*x + a)^4 - b*cos(b*x + a)^2)*sin(b*x + a))
Timed out. \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
\[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]
Time = 23.76 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.72 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {8\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,2{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,2{}\mathrm {i}-{\mathrm {e}}^{a\,8{}\mathrm {i}+b\,x\,8{}\mathrm {i}}\,2{}\mathrm {i}-2{}\mathrm {i}\right )}{15\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2} \]