\(\int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx\) [125]

Optimal result

Integrand size = 18, antiderivative size = 82 \[ \int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {2+m}{2},\frac {4+m}{2},\sin ^2(a+b x)\right ) \sin (a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (2+m)} \]

[Out]

(cos(b*x+a)^2)^(1/2-1/2*m)*hypergeom([1+1/2*m, 1/2-1/2*m],[2+1/2*m],sin(b*x+a)^2)*sin(b*x+a)*sin(2*b*x+2*a)^m*
tan(b*x+a)/b/(2+m)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4395, 2657} \[ \int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {\sin (a+b x) \tan (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m+2}{2},\frac {m+4}{2},\sin ^2(a+b x)\right )}{b (m+2)} \]

[In]

Int[Sin[a + b*x]*Sin[2*a + 2*b*x]^m,x]

[Out]

((Cos[a + b*x]^2)^((1 - m)/2)*Hypergeometric2F1[(1 - m)/2, (2 + m)/2, (4 + m)/2, Sin[a + b*x]^2]*Sin[a + b*x]*
Sin[2*a + 2*b*x]^m*Tan[a + b*x])/(b*(2 + m))

Rule 2657

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*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(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]

Rule 4395

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \left (\cos ^{-m}(a+b x) \sin ^{-m}(a+b x) \sin ^m(2 a+2 b x)\right ) \int \cos ^m(a+b x) \sin ^{1+m}(a+b x) \, dx \\ & = \frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {2+m}{2},\frac {4+m}{2},\sin ^2(a+b x)\right ) \sin (a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (2+m)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 1.43 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.39 \[ \int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {8 (4+m) \operatorname {AppellF1}\left (1+\frac {m}{2},-m,2+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) \cos ^4\left (\frac {1}{2} (a+b x)\right ) \sin ^2\left (\frac {1}{2} (a+b x)\right ) \sin ^m(2 (a+b x))}{b (2+m) \left (2 \left (m \operatorname {AppellF1}\left (2+\frac {m}{2},1-m,2+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 (1+m) \operatorname {AppellF1}\left (2+\frac {m}{2},-m,3+2 m,3+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) (-1+\cos (a+b x))+(4+m) \operatorname {AppellF1}\left (1+\frac {m}{2},-m,2+2 m,2+\frac {m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) (1+\cos (a+b x))\right )} \]

[In]

Integrate[Sin[a + b*x]*Sin[2*a + 2*b*x]^m,x]

[Out]

(8*(4 + m)*AppellF1[1 + m/2, -m, 2 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Cos[(a + b*x)/2]^4
*Sin[(a + b*x)/2]^2*Sin[2*(a + b*x)]^m)/(b*(2 + m)*(2*(m*AppellF1[2 + m/2, 1 - m, 2 + 2*m, 3 + m/2, Tan[(a + b
*x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*(1 + m)*AppellF1[2 + m/2, -m, 3 + 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, 2 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[
(a + b*x)/2]^2]*(1 + Cos[a + b*x])))

Maple [F]

\[\int \sin \left (x b +a \right ) \sin \left (2 x b +2 a \right )^{m}d x\]

[In]

int(sin(b*x+a)*sin(2*b*x+2*a)^m,x)

[Out]

int(sin(b*x+a)*sin(2*b*x+2*a)^m,x)

Fricas [F]

\[ \int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right ) \,d x } \]

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^m,x, algorithm="fricas")

[Out]

integral(sin(2*b*x + 2*a)^m*sin(b*x + a), x)

Sympy [F(-1)]

Timed out. \[ \int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)**m,x)

[Out]

Timed out

Maxima [F]

\[ \int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right ) \,d x } \]

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^m,x, algorithm="maxima")

[Out]

integrate(sin(2*b*x + 2*a)^m*sin(b*x + a), x)

Giac [F]

\[ \int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right ) \,d x } \]

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^m,x, algorithm="giac")

[Out]

integrate(sin(2*b*x + 2*a)^m*sin(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx=\int \sin \left (a+b\,x\right )\,{\sin \left (2\,a+2\,b\,x\right )}^m \,d x \]

[In]

int(sin(a + b*x)*sin(2*a + 2*b*x)^m,x)

[Out]

int(sin(a + b*x)*sin(2*a + 2*b*x)^m, x)