Integrand size = 15, antiderivative size = 230 \[ \int \cos ^q(c+d x) \sin (a+b x) \, dx=-\frac {e^{i (a-c q)+i (b-d q) x+i q (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-q} \cos ^q(c+d x) \operatorname {Hypergeometric2F1}\left (-q,\frac {b-d q}{2 d},\frac {1}{2} \left (2+\frac {b}{d}-q\right ),-e^{2 i (c+d x)}\right )}{2 (b-d q)}-\frac {e^{-i (a+c q)-i (b+d q) x+i q (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-q} \cos ^q(c+d x) \operatorname {Hypergeometric2F1}\left (-q,-\frac {b+d q}{2 d},-\frac {b-d (2-q)}{2 d},-e^{2 i (c+d x)}\right )}{2 (b+d q)} \] Output:
-1/2*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+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))*cos(d*x+c)^q*hype rgeom([-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)
Time = 1.45 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.79 \[ \int \cos ^q(c+d x) \sin (a+b x) \, dx=-\frac {2^{-1-q} e^{-i (a-c+(b-d) x)} \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^{1+q} \left ((b-d q) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (2-\frac {b}{d}+q\right ),-\frac {b+d (-2+q)}{2 d},-e^{2 i (c+d x)}\right )+e^{2 i (a+b x)} (b+d q) \operatorname {Hypergeometric2F1}\left (1,\frac {b+d (2+q)}{2 d},\frac {1}{2} \left (2+\frac {b}{d}-q\right ),-e^{2 i (c+d x)}\right )\right )}{(b-d q) (b+d q)} \] Input:
Integrate[Cos[c + d*x]^q*Sin[a + b*x],x]
Output:
-((2^(-1 - q)*((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)))
Time = 0.73 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.
\(\displaystyle \int \sin (a+b x) \cos ^q(c+d x) \, dx\) |
\(\Big \downarrow \) 5066 |
\(\displaystyle 2^{-q-1} \int \left (i e^{-i a-i b x} \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q-i e^{i a+i b x} \left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2^{-q-1} \left (-\frac {\left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q \left (1+e^{2 i c+2 i d x}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q,\frac {b-d q}{2 d},\frac {1}{2} \left (\frac {b}{d}-q+2\right ),-e^{2 i (c+d x)}\right ) \exp (i (a-c q)+i x (b-d q)+i q (c+d x))}{b-d q}-\frac {\left (e^{-i (c+d x)}+e^{i (c+d x)}\right )^q \left (1+e^{2 i c+2 i d x}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q,-\frac {b+d q}{2 d},1-\frac {b+d q}{2 d},-e^{2 i (c+d x)}\right ) \exp (-i (a+c q)-i x (b+d q)+i q (c+d x))}{b+d q}\right )\) |
Input:
Int[Cos[c + d*x]^q*Sin[a + b*x],x]
Output:
2^(-1 - q)*(-((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))) - (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)))
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]
\[\int \cos \left (d x +c \right )^{q} \sin \left (b x +a \right )d x\]
Input:
int(cos(d*x+c)^q*sin(b*x+a),x)
Output:
int(cos(d*x+c)^q*sin(b*x+a),x)
\[ \int \cos ^q(c+d x) \sin (a+b x) \, dx=\int { \cos \left (d x + c\right )^{q} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(cos(d*x+c)^q*sin(b*x+a),x, algorithm="fricas")
Output:
integral(cos(d*x + c)^q*sin(b*x + a), x)
Timed out. \[ \int \cos ^q(c+d x) \sin (a+b x) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**q*sin(b*x+a),x)
Output:
Timed out
\[ \int \cos ^q(c+d x) \sin (a+b x) \, dx=\int { \cos \left (d x + c\right )^{q} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(cos(d*x+c)^q*sin(b*x+a),x, algorithm="maxima")
Output:
integrate(cos(d*x + c)^q*sin(b*x + a), x)
\[ \int \cos ^q(c+d x) \sin (a+b x) \, dx=\int { \cos \left (d x + c\right )^{q} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(cos(d*x+c)^q*sin(b*x+a),x, algorithm="giac")
Output:
integrate(cos(d*x + c)^q*sin(b*x + a), x)
Timed out. \[ \int \cos ^q(c+d x) \sin (a+b x) \, dx=\int {\cos \left (c+d\,x\right )}^q\,\sin \left (a+b\,x\right ) \,d x \] Input:
int(cos(c + d*x)^q*sin(a + b*x),x)
Output:
int(cos(c + d*x)^q*sin(a + b*x), x)
\[ \int \cos ^q(c+d x) \sin (a+b x) \, dx=\int \cos \left (d x +c \right )^{q} \sin \left (b x +a \right )d x \] Input:
int(cos(d*x+c)^q*sin(b*x+a),x)
Output:
int(cos(c + d*x)**q*sin(a + b*x),x)