Integrand size = 18, antiderivative size = 83 \[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=-\frac {\cos (a+b x) \cot (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(a+b x)\right ) \sin ^2(a+b x)^{\frac {1-m}{2}} \sin ^m(2 a+2 b x)}{b (2+m)} \]
-cos(b*x+a)*cot(b*x+a)*hypergeom([1+1/2*m, -1/2*m+1/2],[2+1/2*m],cos(b*x+a )^2)*(sin(b*x+a)^2)^(-1/2*m+1/2)*sin(2*b*x+2*a)^m/b/(2+m)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 5.28 (sec) , antiderivative size = 567, normalized size of antiderivative = 6.83 \[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {(3+m) \left (2 \operatorname {AppellF1}\left (\frac {1+m}{2},-m,2 (1+m),\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-\operatorname {AppellF1}\left (\frac {1+m}{2},-m,1+2 m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \sin ^{1+m}(2 (a+b x))}{2 b (1+m) \left (2 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-m,2 (1+m),\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-(3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-m,1+2 m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 \left (-2 m \operatorname {AppellF1}\left (\frac {3+m}{2},1-m,2 (1+m),\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+m \operatorname {AppellF1}\left (\frac {3+m}{2},1-m,1+2 m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+\operatorname {AppellF1}\left (\frac {3+m}{2},-m,2 (1+m),\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 m \operatorname {AppellF1}\left (\frac {3+m}{2},-m,2 (1+m),\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-4 \operatorname {AppellF1}\left (\frac {3+m}{2},-m,3+2 m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-4 m \operatorname {AppellF1}\left (\frac {3+m}{2},-m,3+2 m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (a+b x)\right )\right )} \]
((3 + m)*(2*AppellF1[(1 + m)/2, -m, 2*(1 + m), (3 + m)/2, Tan[(a + b*x)/2] ^2, -Tan[(a + b*x)/2]^2] - AppellF1[(1 + m)/2, -m, 1 + 2*m, (3 + m)/2, Tan [(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Sin[2*(a + b*x)]^(1 + m))/(2*b*(1 + m)*(2*(3 + m)*AppellF1[(1 + m)/2, -m, 2*(1 + m), (3 + m)/2, Tan[(a + b*x) /2]^2, -Tan[(a + b*x)/2]^2] - (3 + m)*AppellF1[(1 + m)/2, -m, 1 + 2*m, (3 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*(-2*m*AppellF1[(3 + m )/2, 1 - m, 2*(1 + m), (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + m*AppellF1[(3 + m)/2, 1 - m, 1 + 2*m, (5 + m)/2, Tan[(a + b*x)/2]^2, -T an[(a + b*x)/2]^2] + AppellF1[(3 + m)/2, -m, 2*(1 + m), (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*m*AppellF1[(3 + m)/2, -m, 2*(1 + m), (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*AppellF1[(3 + m)/ 2, -m, 3 + 2*m, (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*m* AppellF1[(3 + m)/2, -m, 3 + 2*m, (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Tan[(a + b*x)/2]^2))
Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4797, 3042, 3056}
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.
\(\displaystyle \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (a+b x) \sin (2 a+2 b x)^mdx\) |
\(\Big \downarrow \) 4797 |
\(\displaystyle \sin ^{-m}(a+b x) \sin ^m(2 a+2 b x) \cos ^{-m}(a+b x) \int \cos ^{m+1}(a+b x) \sin ^m(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^{-m}(a+b x) \sin ^m(2 a+2 b x) \cos ^{-m}(a+b x) \int \cos (a+b x)^{m+1} \sin (a+b x)^mdx\) |
\(\Big \downarrow \) 3056 |
\(\displaystyle -\frac {\cos (a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac {1-m}{2}} \sin ^m(2 a+2 b x) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(a+b x)\right )}{b (m+2)}\) |
-((Cos[a + b*x]*Cot[a + b*x]*Hypergeometric2F1[(1 - m)/2, (2 + m)/2, (4 + m)/2, Cos[a + b*x]^2]*(Sin[a + b*x]^2)^((1 - m)/2)*Sin[2*a + 2*b*x]^m)/(b* (2 + m)))
3.2.89.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-b^(2*IntPart[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*F racPart[(n - 1)/2])*((a*Cos[e + f*x])^(m + 1)/(a*f*(m + 1)*(Sin[e + f*x]^2) ^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, C os[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && SimplerQ[n, m]
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( p_), x_Symbol] :> Simp[(g*Sin[c + d*x])^p/((e*Cos[a + b*x])^p*Sin[a + b*x]^ p) Int[(e*Cos[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p]
\[\int \cos \left (x b +a \right ) \sin \left (2 x b +2 a \right )^{m}d x\]
\[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \cos \left (b x + a\right ) \,d x } \]
Timed out. \[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=\text {Timed out} \]
\[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \cos \left (b x + a\right ) \,d x } \]
\[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \cos \left (b x + a\right ) \,d x } \]
Timed out. \[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=\int \cos \left (a+b\,x\right )\,{\sin \left (2\,a+2\,b\,x\right )}^m \,d x \]