\(\int \sin ^3(a+b x) \sin ^q(c+d x) \, dx\) [141]

Optimal result

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

1/8*exp(I*(-c*q+3*a)+I*(-d*q+3*b)*x+I*q*(d*x+c))*hypergeom([-q, 3/2*b/d-1/ 
2*q],[1+3/2*b/d-1/2*q],exp(2*I*(d*x+c)))*sin(d*x+c)^q/((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))*hypergeom([-q 
, 1/2*(-d*q+b)/d],[1+1/2*b/d-1/2*q],exp(2*I*(d*x+c)))*sin(d*x+c)^q/((1-exp 
(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))*hyp 
ergeom([-q, -1/2*(d*q+b)/d],[-1/2*(b-d*(2-q))/d],exp(2*I*(d*x+c)))*sin(d*x 
+c)^q/((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))*hypergeom([-q, -1/2*(d*q+3*b)/d],[1-3/2*b/d-1/2*q],exp(2*I*(d 
*x+c)))*sin(d*x+c)^q/((1-exp(2*I*(d*x+c)))^q)/(d*q+3*b)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 1.90 (sec) , antiderivative size = 580, normalized size of antiderivative = 1.23, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {5064, 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.

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

\(\Big \downarrow \) 5064

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

2^(-3 - q)*((E^(I*(3*a - c*q) + I*(3*b - d*q)*x + I*q*(c + d*x))*(I/E^(I*( 
c + d*x)) - I*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))*(I/E^(I* 
(c + d*x)) - I*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))*(I/E^(I*(c 
 + d*x)) - I*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))*(I/E^ 
(I*(c + d*x)) - I*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 5064

Int[Sin[(a_.) + (b_.)*(x_)]^(p_.)*Sin[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] 
:> Simp[1/2^(p + q)   Int[ExpandIntegrand[(I/E^(I*(c + d*x)) - I*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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