\(\int \cot (c+d x) \sin ^2(a+b x) \, dx\) [260]

Optimal result

Integrand size = 15, antiderivative size = 293 \[ \int \cot (c+d x) \sin ^2(a+b x) \, dx=\frac {e^{-2 i (a-c)-2 i (b-d) x} \operatorname {Hypergeometric2F1}\left (1,1-\frac {b}{d},2-\frac {b}{d},e^{2 i (c+d x)}\right )}{4 (b-d) \left (-1+e^{2 i c}\right )}-\frac {e^{-2 i (a-c)-2 i b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{d},1-\frac {b}{d},e^{2 i (c+d x)}\right )}{4 b \left (-1+e^{2 i c}\right )}+\frac {e^{2 i (a+c)+2 i b x} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{d},\frac {b+d}{d},e^{2 i (c+d x)}\right )}{4 b \left (-1+e^{2 i c}\right )}-\frac {e^{2 i (a+c)+2 i (b+d) x} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{d},2+\frac {b}{d},e^{2 i (c+d x)}\right )}{4 (b+d) \left (-1+e^{2 i c}\right )}+\frac {\log (\sin (c+d x))}{2 d}-\frac {\cot (c) \sin (2 a+2 b x)}{4 b} \] Output:

1/4*exp(-2*I*(a-c)-2*I*(b-d)*x)*hypergeom([1, 1-b/d],[2-b/d],exp(2*I*(d*x+ 
c)))/(b-d)/(-1+exp(2*I*c))-1/4*exp(-2*I*(a-c)-2*I*b*x)*hypergeom([1, -b/d] 
,[1-b/d],exp(2*I*(d*x+c)))/b/(-1+exp(2*I*c))+1/4*exp(2*I*(a+c)+2*I*b*x)*hy 
pergeom([1, b/d],[(b+d)/d],exp(2*I*(d*x+c)))/b/(-1+exp(2*I*c))-1/4*exp(2*I 
*(a+c)+2*I*(b+d)*x)*hypergeom([1, (b+d)/d],[2+b/d],exp(2*I*(d*x+c)))/(b+d) 
/(-1+exp(2*I*c))+1/2*ln(sin(d*x+c))/d-1/4*cot(c)*sin(2*b*x+2*a)/b
 

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.92 \[ \int \cot (c+d x) \sin ^2(a+b x) \, dx=\frac {1}{4} \left (\frac {e^{-2 i (a-c)} \left (\frac {e^{-2 i (b-d) x} \operatorname {Hypergeometric2F1}\left (1,1-\frac {b}{d},2-\frac {b}{d},e^{2 i (c+d x)}\right )}{b-d}-\frac {e^{-2 i b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {b}{d},1-\frac {b}{d},e^{2 i (c+d x)}\right )}{b}\right )}{-1+e^{2 i c}}+\frac {e^{2 i (a+c)} \left ((b+d) e^{2 i b x} \operatorname {Hypergeometric2F1}\left (1,\frac {b}{d},\frac {b+d}{d},e^{2 i (c+d x)}\right )-b e^{2 i (b+d) x} \operatorname {Hypergeometric2F1}\left (1,\frac {b+d}{d},2+\frac {b}{d},e^{2 i (c+d x)}\right )\right )}{b (b+d) \left (-1+e^{2 i c}\right )}+\frac {2 \log (\sin (c+d x))}{d}-\frac {\cos (2 b x) \cot (c) \sin (2 a)}{b}-\frac {\cos (2 a) \cot (c) \sin (2 b x)}{b}\right ) \] Input:

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

Output:

((Hypergeometric2F1[1, 1 - b/d, 2 - b/d, E^((2*I)*(c + d*x))]/((b - d)*E^( 
(2*I)*(b - d)*x)) - Hypergeometric2F1[1, -(b/d), 1 - b/d, E^((2*I)*(c + d* 
x))]/(b*E^((2*I)*b*x)))/(E^((2*I)*(a - c))*(-1 + E^((2*I)*c))) + (E^((2*I) 
*(a + c))*((b + d)*E^((2*I)*b*x)*Hypergeometric2F1[1, b/d, (b + d)/d, E^(( 
2*I)*(c + d*x))] - b*E^((2*I)*(b + d)*x)*Hypergeometric2F1[1, (b + d)/d, 2 
 + b/d, E^((2*I)*(c + d*x))]))/(b*(b + d)*(-1 + E^((2*I)*c))) + (2*Log[Sin 
[c + d*x]])/d - (Cos[2*b*x]*Cot[c]*Sin[2*a])/b - (Cos[2*a]*Cot[c]*Sin[2*b* 
x])/b)/4
 

Rubi [F]

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.

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

\(\Big \downarrow \) 7299

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

Input:

Int[Cot[c + d*x]*Sin[a + b*x]^2,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(sin(a + b*x)**2*cot(c + d*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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