Integrand size = 18, antiderivative size = 82 \[ \int \sin (a+b x) \sin ^q(2 a+2 b x) \, dx=\frac {\cos ^2(a+b x)^{\frac {1-q}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-q}{2},\frac {2+q}{2},\frac {4+q}{2},\sin ^2(a+b x)\right ) \sin (a+b x) \sin ^q(2 a+2 b x) \tan (a+b x)}{b (2+q)} \] Output:
(cos(b*x+a)^2)^(1/2-1/2*q)*hypergeom([1+1/2*q, 1/2-1/2*q],[2+1/2*q],sin(b* x+a)^2)*sin(b*x+a)*sin(2*b*x+2*a)^q*tan(b*x+a)/b/(2+q)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.91 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.39 \[ \int \sin (a+b x) \sin ^q(2 a+2 b x) \, dx=\frac {8 (4+q) \operatorname {AppellF1}\left (1+\frac {q}{2},-q,2+2 q,2+\frac {q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) \cos ^4\left (\frac {1}{2} (a+b x)\right ) \sin ^2\left (\frac {1}{2} (a+b x)\right ) \sin ^q(2 (a+b x))}{b (2+q) \left (2 \left (q \operatorname {AppellF1}\left (2+\frac {q}{2},1-q,2+2 q,3+\frac {q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 (1+q) \operatorname {AppellF1}\left (2+\frac {q}{2},-q,3+2 q,3+\frac {q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) (-1+\cos (a+b x))+(4+q) \operatorname {AppellF1}\left (1+\frac {q}{2},-q,2+2 q,2+\frac {q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) (1+\cos (a+b x))\right )} \] Input:
Integrate[Sin[a + b*x]*Sin[2*a + 2*b*x]^q,x]
Output:
(8*(4 + q)*AppellF1[1 + q/2, -q, 2 + 2*q, 2 + q/2, Tan[(a + b*x)/2]^2, -Ta n[(a + b*x)/2]^2]*Cos[(a + b*x)/2]^4*Sin[(a + b*x)/2]^2*Sin[2*(a + b*x)]^q )/(b*(2 + q)*(2*(q*AppellF1[2 + q/2, 1 - q, 2 + 2*q, 3 + q/2, Tan[(a + b*x )/2]^2, -Tan[(a + b*x)/2]^2] + 2*(1 + q)*AppellF1[2 + q/2, -q, 3 + 2*q, 3 + q/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*(-1 + Cos[a + b*x]) + (4 + q)*AppellF1[1 + q/2, -q, 2 + 2*q, 2 + q/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*(1 + Cos[a + b*x])))
Time = 0.30 (sec) , antiderivative size = 82, 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, 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.
\(\displaystyle \int \sin (a+b x) \sin ^q(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x) \sin (2 a+2 b x)^qdx\) |
\(\Big \downarrow \) 4798 |
\(\displaystyle \sin ^{-q}(a+b x) \sin ^q(2 a+2 b x) \cos ^{-q}(a+b x) \int \cos ^q(a+b x) \sin ^{q+1}(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^{-q}(a+b x) \sin ^q(2 a+2 b x) \cos ^{-q}(a+b x) \int \cos (a+b x)^q \sin (a+b x)^{q+1}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {\sin (a+b x) \tan (a+b x) \sin ^q(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-q}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-q}{2},\frac {q+2}{2},\frac {q+4}{2},\sin ^2(a+b x)\right )}{b (q+2)}\) |
Input:
Int[Sin[a + b*x]*Sin[2*a + 2*b*x]^q,x]
Output:
((Cos[a + b*x]^2)^((1 - q)/2)*Hypergeometric2F1[(1 - q)/2, (2 + q)/2, (4 + q)/2, Sin[a + b*x]^2]*Sin[a + b*x]*Sin[2*a + 2*b*x]^q*Tan[a + b*x])/(b*(2 + q))
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]
\[\int \sin \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{q}d x\]
Input:
int(sin(b*x+a)*sin(2*b*x+2*a)^q,x)
Output:
int(sin(b*x+a)*sin(2*b*x+2*a)^q,x)
\[ \int \sin (a+b x) \sin ^q(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{q} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)^q,x, algorithm="fricas")
Output:
integral(sin(2*b*x + 2*a)^q*sin(b*x + a), x)
\[ \int \sin (a+b x) \sin ^q(2 a+2 b x) \, dx=\int \sin {\left (a + b x \right )} \sin ^{q}{\left (2 a + 2 b x \right )}\, dx \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)**q,x)
Output:
Integral(sin(a + b*x)*sin(2*a + 2*b*x)**q, x)
\[ \int \sin (a+b x) \sin ^q(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{q} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)^q,x, algorithm="maxima")
Output:
integrate(sin(2*b*x + 2*a)^q*sin(b*x + a), x)
\[ \int \sin (a+b x) \sin ^q(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{q} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(sin(b*x+a)*sin(2*b*x+2*a)^q,x, algorithm="giac")
Output:
integrate(sin(2*b*x + 2*a)^q*sin(b*x + a), x)
Timed out. \[ \int \sin (a+b x) \sin ^q(2 a+2 b x) \, dx=\int \sin \left (a+b\,x\right )\,{\sin \left (2\,a+2\,b\,x\right )}^q \,d x \] Input:
int(sin(a + b*x)*sin(2*a + 2*b*x)^q,x)
Output:
int(sin(a + b*x)*sin(2*a + 2*b*x)^q, x)
\[ \int \sin (a+b x) \sin ^q(2 a+2 b x) \, dx=\int \sin \left (2 b x +2 a \right )^{q} \sin \left (b x +a \right )d x \] Input:
int(sin(b*x+a)*sin(2*b*x+2*a)^q,x)
Output:
int(sin(2*a + 2*b*x)**q*sin(a + b*x),x)