Integrand size = 20, antiderivative size = 53 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}} \]
Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {\sqrt {\sin (2 (a+b x))} \left (-\frac {1}{4} \csc (a+b x)+\frac {1}{12} \sec (a+b x) \tan (a+b x)\right )}{b} \]
Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.
\(\displaystyle \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx\) |
\(\Big \downarrow \) 4792 |
\(\displaystyle \frac {2}{3} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx+\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx+\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 4779 |
\(\displaystyle \frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}\) |
3.1.78.3.1 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[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]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 27.10 (sec) , antiderivative size = 597, normalized size of antiderivative = 11.26
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 (6 \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}-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 )+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}+6 \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 )-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 )+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 )+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 )^{4}+2 \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}-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}-2 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\right )}{8 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right )}\) | \(597\) |
1/8/b*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(6*(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*t an(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)^2-3*(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)+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+6*(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))-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)+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))+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)^4+2*(ta n(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)^4-2*(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)^2-2*(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)/(tan(1/2*a+ 1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)/(1+tan(1/2*a+1/2*x*b)^2)
Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.30 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {4 \, \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) + \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 1\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{12 \, b \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )} \]
-1/12*(4*cos(b*x + a)^2*sin(b*x + a) + sqrt(2)*(4*cos(b*x + a)^2 - 1)*sqrt (cos(b*x + a)*sin(b*x + a)))/(b*cos(b*x + a)^2*sin(b*x + a))
Timed out. \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\sin \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 7875 vs. \(2 (45) = 90\).
Time = 52.85 (sec) , antiderivative size = 7875, normalized size of antiderivative = 148.58 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Too large to display} \]
-1/24*sqrt(2)*sqrt(-tan(1/2*b*x)^4*tan(1/2*a)^3 - tan(1/2*b*x)^3*tan(1/2*a )^4 + tan(1/2*b*x)^4*tan(1/2*a) + 6*tan(1/2*b*x)^3*tan(1/2*a)^2 + 6*tan(1/ 2*b*x)^2*tan(1/2*a)^3 + tan(1/2*b*x)*tan(1/2*a)^4 - tan(1/2*b*x)^3 - 6*tan (1/2*b*x)^2*tan(1/2*a) - 6*tan(1/2*b*x)*tan(1/2*a)^2 - tan(1/2*a)^3 + tan( 1/2*b*x) + tan(1/2*a))*((((((2*(sqrt(2)*tan(1/2*a)^56 + 23*sqrt(2)*tan(1/2 *a)^54 + 251*sqrt(2)*tan(1/2*a)^52 + 1725*sqrt(2)*tan(1/2*a)^50 + 8350*sqr t(2)*tan(1/2*a)^48 + 30130*sqrt(2)*tan(1/2*a)^46 + 83490*sqrt(2)*tan(1/2*a )^44 + 179630*sqrt(2)*tan(1/2*a)^42 + 297275*sqrt(2)*tan(1/2*a)^40 + 36052 5*sqrt(2)*tan(1/2*a)^38 + 264385*sqrt(2)*tan(1/2*a)^36 - 37145*sqrt(2)*tan (1/2*a)^34 - 445740*sqrt(2)*tan(1/2*a)^32 - 742900*sqrt(2)*tan(1/2*a)^30 - 742900*sqrt(2)*tan(1/2*a)^28 - 445740*sqrt(2)*tan(1/2*a)^26 - 37145*sqrt( 2)*tan(1/2*a)^24 + 264385*sqrt(2)*tan(1/2*a)^22 + 360525*sqrt(2)*tan(1/2*a )^20 + 297275*sqrt(2)*tan(1/2*a)^18 + 179630*sqrt(2)*tan(1/2*a)^16 + 83490 *sqrt(2)*tan(1/2*a)^14 + 30130*sqrt(2)*tan(1/2*a)^12 + 8350*sqrt(2)*tan(1/ 2*a)^10 + 1725*sqrt(2)*tan(1/2*a)^8 + 251*sqrt(2)*tan(1/2*a)^6 + 23*sqrt(2 )*tan(1/2*a)^4 + sqrt(2)*tan(1/2*a)^2)*tan(1/2*b*x)/(tan(1/2*a)^51 + 23*ta n(1/2*a)^49 + 252*tan(1/2*a)^47 + 1748*tan(1/2*a)^45 + 8602*tan(1/2*a)^43 + 31878*tan(1/2*a)^41 + 92092*tan(1/2*a)^39 + 211508*tan(1/2*a)^37 + 38936 7*tan(1/2*a)^35 + 572033*tan(1/2*a)^33 + 653752*tan(1/2*a)^31 + 534888*tan (1/2*a)^29 + 208012*tan(1/2*a)^27 - 208012*tan(1/2*a)^25 - 534888*tan(1...
Time = 24.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.04 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {2\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{3\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2} \]