\(\int \cos (a+b x) \sin ^q(2 a+2 b x) \, dx\) [598]

Optimal result

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

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

Mathematica [C] (warning: unable to verify)

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

Time = 3.32 (sec) , antiderivative size = 567, normalized size of antiderivative = 6.83 \[ \int \cos (a+b x) \sin ^q(2 a+2 b x) \, dx=\frac {(3+q) \left (2 \operatorname {AppellF1}\left (\frac {1+q}{2},-q,2 (1+q),\frac {3+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-\operatorname {AppellF1}\left (\frac {1+q}{2},-q,1+2 q,\frac {3+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \sin ^{1+q}(2 (a+b x))}{2 b (1+q) \left (2 (3+q) \operatorname {AppellF1}\left (\frac {1+q}{2},-q,2 (1+q),\frac {3+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-(3+q) \operatorname {AppellF1}\left (\frac {1+q}{2},-q,1+2 q,\frac {3+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 \left (-2 q \operatorname {AppellF1}\left (\frac {3+q}{2},1-q,2 (1+q),\frac {5+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+q \operatorname {AppellF1}\left (\frac {3+q}{2},1-q,1+2 q,\frac {5+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+\operatorname {AppellF1}\left (\frac {3+q}{2},-q,2 (1+q),\frac {5+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 q \operatorname {AppellF1}\left (\frac {3+q}{2},-q,2 (1+q),\frac {5+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-4 \operatorname {AppellF1}\left (\frac {3+q}{2},-q,3+2 q,\frac {5+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )-4 q \operatorname {AppellF1}\left (\frac {3+q}{2},-q,3+2 q,\frac {5+q}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (a+b x)\right )\right )} \] Input:

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

Output:

((3 + q)*(2*AppellF1[(1 + q)/2, -q, 2*(1 + q), (3 + q)/2, Tan[(a + b*x)/2] 
^2, -Tan[(a + b*x)/2]^2] - AppellF1[(1 + q)/2, -q, 1 + 2*q, (3 + q)/2, Tan 
[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Sin[2*(a + b*x)]^(1 + q))/(2*b*(1 + 
 q)*(2*(3 + q)*AppellF1[(1 + q)/2, -q, 2*(1 + q), (3 + q)/2, Tan[(a + b*x) 
/2]^2, -Tan[(a + b*x)/2]^2] - (3 + q)*AppellF1[(1 + q)/2, -q, 1 + 2*q, (3 
+ q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*(-2*q*AppellF1[(3 + q 
)/2, 1 - q, 2*(1 + q), (5 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] 
 + q*AppellF1[(3 + q)/2, 1 - q, 1 + 2*q, (5 + q)/2, Tan[(a + b*x)/2]^2, -T 
an[(a + b*x)/2]^2] + AppellF1[(3 + q)/2, -q, 2*(1 + q), (5 + q)/2, Tan[(a 
+ b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*q*AppellF1[(3 + q)/2, -q, 2*(1 + q), 
 (5 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*AppellF1[(3 + q)/ 
2, -q, 3 + 2*q, (5 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*q* 
AppellF1[(3 + q)/2, -q, 3 + 2*q, (5 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + 
b*x)/2]^2])*Tan[(a + b*x)/2]^2))
 

Rubi [A] (verified)

Time = 0.33 (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.

\(\displaystyle \int \cos (a+b x) \sin ^q(2 a+2 b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (a+b x) \sin (2 a+2 b x)^qdx\)

\(\Big \downarrow \) 4797

\(\displaystyle \sin ^{-q}(a+b x) \sin ^q(2 a+2 b x) \cos ^{-q}(a+b x) \int \cos ^{q+1}(a+b x) \sin ^q(a+b x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sin ^{-q}(a+b x) \sin ^q(2 a+2 b x) \cos ^{-q}(a+b x) \int \cos (a+b x)^{q+1} \sin (a+b x)^qdx\)

\(\Big \downarrow \) 3056

\(\displaystyle -\frac {\cos (a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac {1-q}{2}} \sin ^q(2 a+2 b x) \operatorname {Hypergeometric2F1}\left (\frac {1-q}{2},\frac {q+2}{2},\frac {q+4}{2},\cos ^2(a+b x)\right )}{b (q+2)}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3056

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]
 

rule 4797

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(sin(2*a + 2*b*x)**q*cos(a + b*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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