\(\int \sin (a+b x) \sin ^q(c+d x) \, dx\) [143]

Optimal result

Integrand size = 15, antiderivative size = 230 \[ \int \sin (a+b x) \sin ^q(c+d x) \, dx=-\frac {e^{i (a-c q)+i (b-d q) x+i q (c+d x)} \left (1-e^{2 i (c+d x)}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q,\frac {b-d q}{2 d},\frac {1}{2} \left (2+\frac {b}{d}-q\right ),e^{2 i (c+d x)}\right ) \sin ^q(c+d x)}{2 (b-d q)}-\frac {e^{-i (a+c q)-i (b+d q) x+i q (c+d x)} \left (1-e^{2 i (c+d x)}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q,-\frac {b+d q}{2 d},-\frac {b-d (2-q)}{2 d},e^{2 i (c+d x)}\right ) \sin ^q(c+d x)}{2 (b+d q)} \] Output:

-1/2*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)-1/2*exp(-I*(c*q+a)-I*(d*q+b)*x+I*q*(d*x+c))*hypergeom([-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)
 

Mathematica [A] (warning: unable to verify)

Time = 1.50 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.79 \[ \int \sin (a+b x) \sin ^q(c+d x) \, dx=\frac {i 2^{-1-q} e^{-i (a-c+(b-d) x)} \left (-i e^{-i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )\right )^{1+q} \left ((b-d q) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (2-\frac {b}{d}+q\right ),-\frac {b+d (-2+q)}{2 d},e^{2 i (c+d x)}\right )+e^{2 i (a+b x)} (b+d q) \operatorname {Hypergeometric2F1}\left (1,\frac {b+d (2+q)}{2 d},\frac {1}{2} \left (2+\frac {b}{d}-q\right ),e^{2 i (c+d x)}\right )\right )}{(b-d q) (b+d q)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.25, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.

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

\(\Big \downarrow \) 5064

\(\displaystyle 2^{-q-1} \int \left (i e^{-i a-i b x} \left (i e^{-i (c+d x)}-i e^{i (c+d x)}\right )^q-i e^{i a+i b x} \left (i e^{-i (c+d x)}-i e^{i (c+d x)}\right )^q\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

2^(-1 - q)*(-((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))) - (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)))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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