Integrand size = 22, antiderivative size = 81 \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {16 \cos (a+b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^3(a+b x)}{7 b \sqrt {\sin (2 a+2 b x)}}+\frac {32 \sin (a+b x)}{21 b \sqrt {\sin (2 a+2 b x)}} \]
-16/21*cos(b*x+a)/b/sin(2*b*x+2*a)^(3/2)-1/7*csc(b*x+a)^3/b/sin(2*b*x+2*a) ^(1/2)+32/21*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)
Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68 \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {(5-12 \cos (2 (a+b x))+4 \cos (4 (a+b x))) \csc ^4(a+b x) \sec (a+b x) \sqrt {\sin (2 (a+b x))}}{42 b} \]
((5 - 12*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)])*Csc[a + b*x]^4*Sec[a + b*x ]*Sqrt[Sin[2*(a + b*x)]])/(42*b)
Time = 0.47 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4788, 3042, 4796, 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.
| \(\displaystyle \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (a+b x)^3 \sin (2 a+2 b x)^{3/2}}dx\) |
| \(\Big \downarrow \) 4788 |
| \(\displaystyle \frac {8}{7} \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx-\frac {\csc ^3(a+b x)}{7 b \sqrt {\sin (2 a+2 b x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8}{7} \int \frac {1}{\sin (a+b x) \sin (2 a+2 b x)^{3/2}}dx-\frac {\csc ^3(a+b x)}{7 b \sqrt {\sin (2 a+2 b x)}}\) |
| \(\Big \downarrow \) 4796 |
| \(\displaystyle \frac {16}{7} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx-\frac {\csc ^3(a+b x)}{7 b \sqrt {\sin (2 a+2 b x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {16}{7} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx-\frac {\csc ^3(a+b x)}{7 b \sqrt {\sin (2 a+2 b x)}}\) |
| \(\Big \downarrow \) 4791 |
| \(\displaystyle \frac {16}{7} \left (\frac {2}{3} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\csc ^3(a+b x)}{7 b \sqrt {\sin (2 a+2 b x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {16}{7} \left (\frac {2}{3} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\csc ^3(a+b x)}{7 b \sqrt {\sin (2 a+2 b x)}}\) |
| \(\Big \downarrow \) 4780 |
| \(\displaystyle \frac {16}{7} \left (\frac {2 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\csc ^3(a+b x)}{7 b \sqrt {\sin (2 a+2 b x)}}\) |
(16*(-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]])))/7 - Csc[a + b*x]^3/(7*b*Sqrt[Sin[2*a + 2*b*x]])
3.2.21.3.1 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[((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)/(2*b*g*( m + p + 1))), x] + Simp[(m + 2*p + 2)/(e^2*(m + p + 1)) Int[(e*Sin[a + b* x])^(m + 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, x] & & EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p] && LtQ[m, -1] && NeQ[m + 2*p + 2, 0] && NeQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
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[((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 = 60.11 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.74
| method | result | size |
| default | \(-\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right ) \left (-3 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{8}+16 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}+2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+3\right )}{336 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, b}\) | \(222\) |
-1/336*(-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+16*(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- 2*tan(1/2*a+1/2*x*b)^6+2*tan(1/2*a+1/2*x*b)^2+3)/(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) /b
Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.28 \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {32 \, \cos \left (b x + a\right )^{5} - 64 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 56 \, \cos \left (b x + a\right )^{2} + 21\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )}{42 \, {\left (b \cos \left (b x + a\right )^{5} - 2 \, b \cos \left (b x + a\right )^{3} + b \cos \left (b x + a\right )\right )}} \]
1/42*(32*cos(b*x + a)^5 - 64*cos(b*x + a)^3 + sqrt(2)*(32*cos(b*x + a)^4 - 56*cos(b*x + a)^2 + 21)*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*cos(b*x + a) )/(b*cos(b*x + a)^5 - 2*b*cos(b*x + a)^3 + b*cos(b*x + a))
Timed out. \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]
Time = 24.17 (sec) , antiderivative size = 302, normalized size of antiderivative = 3.73 \[ \int \frac {\csc ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {10\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{21\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,12{}\mathrm {i}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3}-\frac {8\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {10{}\mathrm {i}}{21\,b}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,32{}\mathrm {i}}{21\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )} \]
(exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i) /2)^(1/2)*12i)/(7*b*(exp(a*2i + b*x*2i)*1i - 1i)^3) - (10*exp(a*1i + b*x*1 i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(21*b* (exp(a*2i + b*x*2i)*1i - 1i)^2) - (8*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x *2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(7*b*(exp(a*2i + b*x*2i)*1i - 1i)^4) - (exp(a*1i + b*x*1i)*(10i/(21*b) - (exp(a*2i + b*x*2i)*32i)/(21 *b))*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/((ex p(a*2i + b*x*2i) + 1)*(exp(a*2i + b*x*2i)*1i - 1i))