\(\int \sin ^2(a+b x) \sin ^q(c+d x) \, dx\) [142]

Optimal result

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

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

Mathematica [F]

\[ \int \sin ^2(a+b x) \sin ^q(c+d x) \, dx=\int \sin ^2(a+b x) \sin ^q(c+d x) \, dx \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.26, 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.

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

\(\Big \downarrow \) 5064

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

\(\Big \downarrow \) 2009

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

Input:

Int[Sin[a + b*x]^2*Sin[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))*( 
I/E^(I*(c + d*x)) - I*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))*(I/E^(I*(c + d*x)) - I*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)*(I/E^(I*(c + d*x)) - I*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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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