Integrand size = 20, antiderivative size = 136 \[ \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}+\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{32 b}-\frac {5 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}+\frac {5 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}-\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b} \]
-5/32*arcsin(cos(b*x+a)-sin(b*x+a))/b+5/32*ln(cos(b*x+a)+sin(b*x+a)+sin(2* b*x+2*a)^(1/2))/b+5/24*sin(b*x+a)*sin(2*b*x+2*a)^(3/2)/b-1/6*cos(b*x+a)*si n(2*b*x+2*a)^(5/2)/b-5/16*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b
Time = 0.46 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.72 \[ \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {15 \left (-\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )-2 (14 \cos (a+b x)+3 \cos (3 (a+b x))-2 \cos (5 (a+b x))) \sqrt {\sin (2 (a+b x))}}{96 b} \]
(15*(-ArcSin[Cos[a + b*x] - Sin[a + b*x]] + Log[Cos[a + b*x] + Sin[a + b*x ] + Sqrt[Sin[2*(a + b*x)]]]) - 2*(14*Cos[a + b*x] + 3*Cos[3*(a + b*x)] - 2 *Cos[5*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/(96*b)
Time = 0.53 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4790, 3042, 4789, 3042, 4790, 3042, 4793}
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.
\(\displaystyle \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x) \sin (2 a+2 b x)^{5/2}dx\) |
\(\Big \downarrow \) 4790 |
\(\displaystyle \frac {5}{6} \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \int \cos (a+b x) \sin (2 a+2 b x)^{3/2}dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 4789 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 4790 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\) |
\(\Big \downarrow \) 4793 |
\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}\right )-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\) |
-1/6*(Cos[a + b*x]*Sin[2*a + 2*b*x]^(5/2))/b + (5*((3*((-1/2*ArcSin[Cos[a + b*x] - Sin[a + b*x]]/b + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(2*b))/2 - (Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)))/4 + (S in[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(4*b)))/6
3.1.73.3.1 Defintions of rubi rules used
Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[2*Sin[a + b*x]*((g*Sin[c + d*x])^p/(d*(2*p + 1))), x] + Simp[2*p*( g/(2*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] && !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]
Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[-2*Cos[a + b*x]*((g*Sin[c + d*x])^p/(d*(2*p + 1))), x] + Simp[2*p* (g/(2*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] && GtQ[p, 0] && IntegerQ[2*p]
Int[cos[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Sim p[-ArcSin[Cos[a + b*x] - Sin[a + b*x]]/d, x] + Simp[Log[Cos[a + b*x] + Sin[ a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 76.22 (sec) , antiderivative size = 183661410, normalized size of antiderivative = 1350451.54
Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (118) = 236\).
Time = 0.26 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.14 \[ \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{5} - 52 \, \cos \left (b x + a\right )^{3} + 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 30 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 30 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 15 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{384 \, b} \]
1/384*(8*sqrt(2)*(32*cos(b*x + a)^5 - 52*cos(b*x + a)^3 + 5*cos(b*x + a))* sqrt(cos(b*x + a)*sin(b*x + a)) + 30*arctan(-(sqrt(2)*sqrt(cos(b*x + a)*si n(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + cos(b*x + a)*sin(b*x + a))/(co s(b*x + a)^2 + 2*cos(b*x + a)*sin(b*x + a) - 1)) - 30*arctan(-(2*sqrt(2)*s qrt(cos(b*x + a)*sin(b*x + a)) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + a ) - sin(b*x + a))) - 15*log(-32*cos(b*x + a)^4 + 4*sqrt(2)*(4*cos(b*x + a) ^3 - (4*cos(b*x + a)^2 + 1)*sin(b*x + a) - 5*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*cos(b*x + a)^2 + 16*cos(b*x + a)*sin(b*x + a) + 1))/ b
Timed out. \[ \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]
\[ \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \sin \left (b x + a\right ) \,d x } \]
\[ \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \sin \left (b x + a\right ) \,d x } \]
Timed out. \[ \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \sin \left (a+b\,x\right )\,{\sin \left (2\,a+2\,b\,x\right )}^{5/2} \,d x \]