Integrand size = 20, antiderivative size = 84 \[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {4+m}{2},\frac {6+m}{2},\sin ^2(a+b x)\right ) \sin ^3(a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (4+m)} \]
(cos(b*x+a)^2)^(-1/2*m+1/2)*hypergeom([2+1/2*m, -1/2*m+1/2],[3+1/2*m],sin( b*x+a)^2)*sin(b*x+a)^3*sin(2*b*x+2*a)^m*tan(b*x+a)/b/(4+m)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 6.70 (sec) , antiderivative size = 602, normalized size of antiderivative = 7.17 \[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {32 (4+m) \left (\operatorname {AppellF1}\left (1+\frac {m}{2},-m,3+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-\operatorname {AppellF1}\left (1+\frac {m}{2},-m,4+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \cos ^6\left (\frac {1}{2} (a+b x)\right ) \sin ^4\left (\frac {1}{2} (a+b x)\right ) \sin ^m(2 (a+b x))}{b (2+m) \left (-2 (4+m) \operatorname {AppellF1}\left (1+\frac {m}{2},-m,4+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) \cos ^2\left (\frac {1}{2} (a+b x)\right )+2 \left (m \operatorname {AppellF1}\left (2+\frac {m}{2},1-m,3+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-m \operatorname {AppellF1}\left (2+\frac {m}{2},1-m,4+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+3 \operatorname {AppellF1}\left (2+\frac {m}{2},-m,4+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 m \operatorname {AppellF1}\left (2+\frac {m}{2},-m,4+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-4 \operatorname {AppellF1}\left (2+\frac {m}{2},-m,5+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-2 m \operatorname {AppellF1}\left (2+\frac {m}{2},-m,5+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) (-1+\cos (a+b x))+(4+m) \operatorname {AppellF1}\left (1+\frac {m}{2},-m,3+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) (1+\cos (a+b x))\right )} \]
(32*(4 + m)*(AppellF1[1 + m/2, -m, 3 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, - Tan[(a + b*x)/2]^2] - AppellF1[1 + m/2, -m, 4 + 2*m, 2 + m/2, Tan[(a + b*x )/2]^2, -Tan[(a + b*x)/2]^2])*Cos[(a + b*x)/2]^6*Sin[(a + b*x)/2]^4*Sin[2* (a + b*x)]^m)/(b*(2 + m)*(-2*(4 + m)*AppellF1[1 + m/2, -m, 4 + 2*m, 2 + m/ 2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Cos[(a + b*x)/2]^2 + 2*(m*Appe llF1[2 + m/2, 1 - m, 3 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/ 2]^2] - m*AppellF1[2 + m/2, 1 - m, 4 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, - Tan[(a + b*x)/2]^2] + 3*AppellF1[2 + m/2, -m, 4 + 2*m, 3 + m/2, Tan[(a + b *x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*m*AppellF1[2 + m/2, -m, 4 + 2*m, 3 + m/ 2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*AppellF1[2 + m/2, -m, 5 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 2*m*AppellF1[2 + m/2, -m, 5 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*(-1 + Cos[a + b*x]) + (4 + m)*AppellF1[1 + m/2, -m, 3 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*(1 + Cos[a + b*x])))
Time = 0.33 (sec) , antiderivative size = 84, 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, 4798, 3042, 3057}
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.
\(\displaystyle \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^3 \sin (2 a+2 b x)^mdx\) |
\(\Big \downarrow \) 4798 |
\(\displaystyle \sin ^{-m}(a+b x) \sin ^m(2 a+2 b x) \cos ^{-m}(a+b x) \int \cos ^m(a+b x) \sin ^{m+3}(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^{-m}(a+b x) \sin ^m(2 a+2 b x) \cos ^{-m}(a+b x) \int \cos (a+b x)^m \sin (a+b x)^{m+3}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {\sin ^3(a+b x) \tan (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m+4}{2},\frac {m+6}{2},\sin ^2(a+b x)\right )}{b (m+4)}\) |
((Cos[a + b*x]^2)^((1 - m)/2)*Hypergeometric2F1[(1 - m)/2, (4 + m)/2, (6 + m)/2, Sin[a + b*x]^2]*Sin[a + b*x]^3*Sin[2*a + 2*b*x]^m*Tan[a + b*x])/(b* (4 + m))
3.2.23.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( p_), x_Symbol] :> Simp[(g*Sin[c + d*x])^p/(Cos[a + b*x]^p*(f*Sin[a + b*x])^ p) Int[Cos[a + b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, g, n, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p]
\[\int \sin \left (x b +a \right )^{3} \sin \left (2 x b +2 a \right )^{m}d x\]
\[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\text {Timed out} \]
\[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right )^{3} \,d x } \]
\[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx=\int {\sin \left (a+b\,x\right )}^3\,{\sin \left (2\,a+2\,b\,x\right )}^m \,d x \]