\(\int \cos ^q(c+d x) \sin ^3(a+b x) \, dx\) [226]

Optimal result

Integrand size = 17, antiderivative size = 473 \[ \int \cos ^q(c+d x) \sin ^3(a+b x) \, dx=\frac {e^{i (3 a-c q)+i (3 b-d q) x+i q (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-q} \cos ^q(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (\frac {3 b}{d}-q\right ),-q,\frac {1}{2} \left (2+\frac {3 b}{d}-q\right ),-e^{2 i (c+d x)}\right )}{8 (3 b-d q)}-\frac {3 e^{i (a-c q)+i (b-d q) x+i q (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-q} \cos ^q(c+d x) \operatorname {Hypergeometric2F1}\left (-q,\frac {b-d q}{2 d},\frac {1}{2} \left (2+\frac {b}{d}-q\right ),-e^{2 i (c+d x)}\right )}{8 (b-d q)}-\frac {3 e^{-i (a+c q)-i (b+d q) x+i q (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-q} \cos ^q(c+d x) \operatorname {Hypergeometric2F1}\left (-q,-\frac {b+d q}{2 d},-\frac {b-d (2-q)}{2 d},-e^{2 i (c+d x)}\right )}{8 (b+d q)}+\frac {e^{-i (3 a+c q)-i (3 b+d q) x+i q (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-q} \cos ^q(c+d x) \operatorname {Hypergeometric2F1}\left (-q,-\frac {3 b+d q}{2 d},\frac {1}{2} \left (2-\frac {3 b}{d}-q\right ),-e^{2 i (c+d x)}\right )}{8 (3 b+d q)} \] Output:

1/8*exp(I*(-c*q+3*a)+I*(-d*q+3*b)*x+I*q*(d*x+c))*cos(d*x+c)^q*hypergeom([- 
q, 3/2*b/d-1/2*q],[1+3/2*b/d-1/2*q],-exp(2*I*(d*x+c)))/((1+exp(2*I*(d*x+c) 
))^q)/(-d*q+3*b)-3/8*exp(I*(-c*q+a)+I*(-d*q+b)*x+I*q*(d*x+c))*cos(d*x+c)^q 
*hypergeom([-q, 1/2*(-d*q+b)/d],[1+1/2*b/d-1/2*q],-exp(2*I*(d*x+c)))/((1+e 
xp(2*I*(d*x+c)))^q)/(-d*q+b)-3/8*exp(-I*(c*q+a)-I*(d*q+b)*x+I*q*(d*x+c))*c 
os(d*x+c)^q*hypergeom([-q, -1/2*(d*q+b)/d],[-1/2*(b-d*(2-q))/d],-exp(2*I*( 
d*x+c)))/((1+exp(2*I*(d*x+c)))^q)/(d*q+b)+1/8*exp(-I*(c*q+3*a)-I*(d*q+3*b) 
*x+I*q*(d*x+c))*cos(d*x+c)^q*hypergeom([-q, -1/2*(d*q+3*b)/d],[1-3/2*b/d-1 
/2*q],-exp(2*I*(d*x+c)))/((1+exp(2*I*(d*x+c)))^q)/(d*q+3*b)
 

Mathematica [A] (warning: unable to verify)

Time = 1.57 (sec) , antiderivative size = 329, normalized size of antiderivative = 0.70 \[ \int \cos ^q(c+d x) \sin ^3(a+b x) \, dx=2^{-3-q} e^{i (-3 a+c+d (1+q) x)} \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^{1+q} \left (\frac {e^{-i (3 b+d q) x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (2-\frac {3 b}{d}+q\right ),1-\frac {3 b}{2 d}-\frac {q}{2},-e^{2 i (c+d x)}\right )}{3 b+d q}-\frac {3 e^{2 i a-i (b+d q) x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (2-\frac {b}{d}+q\right ),-\frac {b+d (-2+q)}{2 d},-e^{2 i (c+d x)}\right )}{b+d q}+e^{i (4 a+b x-d q x)} \left (\frac {e^{2 i (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (2+\frac {3 b}{d}+q\right ),1+\frac {3 b}{2 d}-\frac {q}{2},-e^{2 i (c+d x)}\right )}{3 b-d q}-\frac {3 \operatorname {Hypergeometric2F1}\left (1,\frac {b+d (2+q)}{2 d},\frac {1}{2} \left (2+\frac {b}{d}-q\right ),-e^{2 i (c+d x)}\right )}{b-d q}\right )\right ) \] Input:

Integrate[Cos[c + d*x]^q*Sin[a + b*x]^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 548, normalized size of antiderivative = 1.16, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {5066, 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.

\(\displaystyle \int \sin ^3(a+b x) \cos ^q(c+d x) \, dx\)

\(\Big \downarrow \) 5066

\(\displaystyle 2^{-q-3} \int \left (3 i e^{-i a-i b x} \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q-3 i e^{i a+i b x} \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q-i e^{-3 i a-3 i b x} \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q+i e^{3 i a+3 i b x} \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2^{-q-3} \left (\frac {\left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q \left (1+e^{2 i c+2 i d x}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (\frac {3 b}{d}-q\right ),-q,\frac {1}{2} \left (\frac {3 b}{d}-q+2\right ),-e^{2 i (c+d x)}\right ) \exp (i (3 a-c q)+i x (3 b-d q)+i q (c+d x))}{3 b-d q}-\frac {3 \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q \left (1+e^{2 i c+2 i d x}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q,\frac {b-d q}{2 d},\frac {1}{2} \left (\frac {b}{d}-q+2\right ),-e^{2 i (c+d x)}\right ) \exp (i (a-c q)+i x (b-d q)+i q (c+d x))}{b-d q}-\frac {3 \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q \left (1+e^{2 i c+2 i d x}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q,-\frac {b+d q}{2 d},1-\frac {b+d q}{2 d},-e^{2 i (c+d x)}\right ) \exp (-i (a+c q)-i x (b+d q)+i q (c+d x))}{b+d q}+\frac {\left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q \left (1+e^{2 i c+2 i d x}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q,-\frac {3 b+d q}{2 d},\frac {1}{2} \left (-\frac {3 b}{d}-q+2\right ),-e^{2 i (c+d x)}\right ) \exp (-i (3 a+c q)-i x (3 b+d q)+i q (c+d x))}{3 b+d q}\right )\)

Input:

Int[Cos[c + d*x]^q*Sin[a + b*x]^3,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 5066

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

Maple [F]

\[\int \cos \left (d x +c \right )^{q} \sin \left (b x +a \right )^{3}d x\]

Input:

int(cos(d*x+c)^q*sin(b*x+a)^3,x)
 

Output:

int(cos(d*x+c)^q*sin(b*x+a)^3,x)
 

Fricas [F]

\[ \int \cos ^q(c+d x) \sin ^3(a+b x) \, dx=\int { \cos \left (d x + c\right )^{q} \sin \left (b x + a\right )^{3} \,d x } \] Input:

integrate(cos(d*x+c)^q*sin(b*x+a)^3,x, algorithm="fricas")
 

Output:

integral(-(cos(b*x + a)^2 - 1)*cos(d*x + c)^q*sin(b*x + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^q(c+d x) \sin ^3(a+b x) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**q*sin(b*x+a)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^q(c+d x) \sin ^3(a+b x) \, dx=\int { \cos \left (d x + c\right )^{q} \sin \left (b x + a\right )^{3} \,d x } \] Input:

integrate(cos(d*x+c)^q*sin(b*x+a)^3,x, algorithm="maxima")
 

Output:

integrate(cos(d*x + c)^q*sin(b*x + a)^3, x)
 

Giac [F]

\[ \int \cos ^q(c+d x) \sin ^3(a+b x) \, dx=\int { \cos \left (d x + c\right )^{q} \sin \left (b x + a\right )^{3} \,d x } \] Input:

integrate(cos(d*x+c)^q*sin(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate(cos(d*x + c)^q*sin(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^q(c+d x) \sin ^3(a+b x) \, dx=\int {\cos \left (c+d\,x\right )}^q\,{\sin \left (a+b\,x\right )}^3 \,d x \] Input:

int(cos(c + d*x)^q*sin(a + b*x)^3,x)
 

Output:

int(cos(c + d*x)^q*sin(a + b*x)^3, x)
 

Reduce [F]

\[ \int \cos ^q(c+d x) \sin ^3(a+b x) \, dx=\int \cos \left (d x +c \right )^{q} \sin \left (b x +a \right )^{3}d x \] Input:

int(cos(d*x+c)^q*sin(b*x+a)^3,x)
 

Output:

int(cos(c + d*x)**q*sin(a + b*x)**3,x)