\(\int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx\) [189]

Optimal result

Integrand size = 18, antiderivative size = 83 \[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=-\frac {\cos (a+b x) \cot (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(a+b x)\right ) \sin ^2(a+b x)^{\frac {1-m}{2}} \sin ^m(2 a+2 b x)}{b (2+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, 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 = {4394, 2656} \[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=-\frac {\cos (a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac {1-m}{2}} \sin ^m(2 a+2 b x) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(a+b x)\right )}{b (m+2)} \]

[In]

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

[Out]

-((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)))

Rule 2656

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^(2*IntPar
t[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*FracPart[(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, Cos[e + f*x]^2], x] /; FreeQ[{a
, b, e, f, m, n}, x] && SimplerQ[n, m]

Rule 4394

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

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 ^{1+m}(a+b x) \sin ^m(a+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)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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 \[ \int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {(3+m) \left (2 \operatorname {AppellF1}\left (\frac {1+m}{2},-m,2 (1+m),\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-\operatorname {AppellF1}\left (\frac {1+m}{2},-m,1+2 m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \sin ^{1+m}(2 (a+b x))}{2 b (1+m) \left (2 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-m,2 (1+m),\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-(3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-m,1+2 m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 \left (-2 m \operatorname {AppellF1}\left (\frac {3+m}{2},1-m,2 (1+m),\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+m \operatorname {AppellF1}\left (\frac {3+m}{2},1-m,1+2 m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+\operatorname {AppellF1}\left (\frac {3+m}{2},-m,2 (1+m),\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 m \operatorname {AppellF1}\left (\frac {3+m}{2},-m,2 (1+m),\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-4 \operatorname {AppellF1}\left (\frac {3+m}{2},-m,3+2 m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-4 m \operatorname {AppellF1}\left (\frac {3+m}{2},-m,3+2 m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (a+b x)\right )\right )} \]

[In]

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

[Out]

((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, -Tan[(a + b*x)/2]^2] + AppellF1[(3 + m)/2, -m, 2*(1 + m), (5 + m)/2, T
an[(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))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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