\(\int \cos ^2(a+b x) \cos ^q(c+d x) \, dx\) [359]

Optimal result

Integrand size = 17, antiderivative size = 314 \[ \int \cos ^2(a+b x) \cos ^q(c+d x) \, dx=\frac {i e^{-i (2 a+c q)-i (2 b+d q) x+i q (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-q} \cos ^q(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (-\frac {2 b}{d}-q\right ),-q,\frac {1}{2} \left (2-\frac {2 b}{d}-q\right ),-e^{2 i (c+d x)}\right )}{4 (2 b+d q)}-\frac {i e^{i (2 a-c q)+i (2 b-d q) x+i q (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-q} \cos ^q(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (\frac {2 b}{d}-q\right ),-q,\frac {1}{2} \left (2+\frac {2 b}{d}-q\right ),-e^{2 i (c+d x)}\right )}{4 (2 b-d q)}-\frac {\cos ^{1+q}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+q}{2},\frac {3+q}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (1+q) \sqrt {\sin ^2(c+d x)}} \] Output:

1/4*I*exp(-I*(c*q+2*a)-I*(d*q+2*b)*x+I*q*(d*x+c))*cos(d*x+c)^q*hypergeom([ 
-q, -b/d-1/2*q],[1-b/d-1/2*q],-exp(2*I*(d*x+c)))/((1+exp(2*I*(d*x+c)))^q)/ 
(d*q+2*b)-1/4*I*exp(I*(-c*q+2*a)+I*(-d*q+2*b)*x+I*q*(d*x+c))*cos(d*x+c)^q* 
hypergeom([-q, b/d-1/2*q],[1+b/d-1/2*q],-exp(2*I*(d*x+c)))/((1+exp(2*I*(d* 
x+c)))^q)/(-d*q+2*b)-1/2*cos(d*x+c)^(1+q)*hypergeom([1/2, 1/2+1/2*q],[3/2+ 
1/2*q],cos(d*x+c)^2)*sin(d*x+c)/d/(1+q)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.73 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.77 \[ \int \cos ^2(a+b x) \cos ^q(c+d x) \, dx=\frac {i 2^{-2-q} e^{-2 i (a+b x)+i (c+d x)} \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^{1+q} \left (d q (-2 b+d q) \operatorname {Hypergeometric2F1}\left (1,1-\frac {b}{d}+\frac {q}{2},1-\frac {b}{d}-\frac {q}{2},-e^{2 i (c+d x)}\right )+e^{2 i (a+b x)} (2 b+d q) \left (d e^{2 i (a+b x)} q \operatorname {Hypergeometric2F1}\left (1,1+\frac {b}{d}+\frac {q}{2},1+\frac {b}{d}-\frac {q}{2},-e^{2 i (c+d x)}\right )+2 (-2 b+d q) \operatorname {Hypergeometric2F1}\left (1,\frac {2+q}{2},1-\frac {q}{2},-e^{2 i (c+d x)}\right )\right )\right )}{-4 b^2 d q+d^3 q^3} \] Input:

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

Output:

(I*2^(-2 - q)*E^((-2*I)*(a + b*x) + I*(c + d*x))*((1 + E^((2*I)*(c + d*x)) 
)/E^(I*(c + d*x)))^(1 + q)*(d*q*(-2*b + d*q)*Hypergeometric2F1[1, 1 - b/d 
+ q/2, 1 - b/d - q/2, -E^((2*I)*(c + d*x))] + E^((2*I)*(a + b*x))*(2*b + d 
*q)*(d*E^((2*I)*(a + b*x))*q*Hypergeometric2F1[1, 1 + b/d + q/2, 1 + b/d - 
 q/2, -E^((2*I)*(c + d*x))] + 2*(-2*b + d*q)*Hypergeometric2F1[1, (2 + q)/ 
2, 1 - q/2, -E^((2*I)*(c + d*x))])))/(-4*b^2*d*q + d^3*q^3)
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {5065, 2009}

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.

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

\(\Big \downarrow \) 5065

\(\displaystyle 2^{-q-2} \int \left (e^{-2 i a-2 i b x} \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q+e^{2 i a+2 i b x} \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q+2 \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2^{-q-2} \left (\frac {i \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q \left (1+e^{2 i c+2 i d x}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (-\frac {2 b}{d}-q\right ),-q,\frac {1}{2} \left (-\frac {2 b}{d}-q+2\right ),-e^{2 i (c+d x)}\right ) \exp (-i (2 a+c q)-i x (2 b+d q)+i q (c+d x))}{2 b+d q}-\frac {i \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q \left (1+e^{2 i c+2 i d x}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (\frac {2 b}{d}-q\right ),-q,\frac {1}{2} \left (\frac {2 b}{d}-q+2\right ),-e^{2 i (c+d x)}\right ) \exp (i (2 a-c q)+i x (2 b-d q)+i q (c+d x))}{2 b-d q}+\frac {2 i \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q \left (1+e^{2 i (c+d x)}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q,-\frac {q}{2},1-\frac {q}{2},-e^{2 i (c+d x)}\right )}{d q}\right )\)

Input:

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

Output:

2^(-2 - q)*((I*E^((-I)*(2*a + c*q) - I*(2*b + d*q)*x + I*q*(c + d*x))*(E^( 
(-I)*(c + d*x)) + E^(I*(c + d*x)))^q*Hypergeometric2F1[((-2*b)/d - q)/2, - 
q, (2 - (2*b)/d - q)/2, -E^((2*I)*(c + d*x))])/((1 + E^((2*I)*c + (2*I)*d* 
x))^q*(2*b + d*q)) - (I*E^(I*(2*a - c*q) + I*(2*b - d*q)*x + I*q*(c + d*x) 
)*(E^((-I)*(c + d*x)) + E^(I*(c + d*x)))^q*Hypergeometric2F1[((2*b)/d - q) 
/2, -q, (2 + (2*b)/d - q)/2, -E^((2*I)*(c + d*x))])/((1 + E^((2*I)*c + (2* 
I)*d*x))^q*(2*b - d*q)) + ((2*I)*(E^((-I)*(c + d*x)) + E^(I*(c + d*x)))^q* 
Hypergeometric2F1[-q, -1/2*q, 1 - q/2, -E^((2*I)*(c + d*x))])/(d*(1 + E^(( 
2*I)*(c + d*x)))^q*q))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5065

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*Cos[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] 
:> Simp[1/2^(p + q)   Int[ExpandIntegrand[(E^((-I)*(c + d*x)) + E^(I*(c + d 
*x)))^q, (E^((-I)*(a + b*x)) + E^(I*(a + b*x)))^p, x], x], x] /; FreeQ[{a, 
b, c, d, q}, x] && IGtQ[p, 0] &&  !IntegerQ[q]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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