\(\int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx\) [1060]

Optimal result

Integrand size = 27, antiderivative size = 89 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=-2 a b x+\frac {\left (a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 d}+\frac {3 b^2 \cos (c+d x)}{2 d}-\frac {a b \cot (c+d x)}{d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d} \] Output:

-2*a*b*x+1/2*(a^2-2*b^2)*arctanh(cos(d*x+c))/d+3/2*b^2*cos(d*x+c)/d-a*b*co 
t(d*x+c)/d-1/2*cot(d*x+c)*csc(d*x+c)*(a+b*sin(d*x+c))^2/d
 

Mathematica [A] (verified)

Time = 2.19 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.74 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-16 a b c-16 a b d x+8 b^2 \cos (c+d x)-8 a b \cot \left (\frac {1}{2} (c+d x)\right )-a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )+4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+8 a b \tan \left (\frac {1}{2} (c+d x)\right )}{8 d} \] Input:

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

Output:

(-16*a*b*c - 16*a*b*d*x + 8*b^2*Cos[c + d*x] - 8*a*b*Cot[(c + d*x)/2] - a^ 
2*Csc[(c + d*x)/2]^2 + 4*a^2*Log[Cos[(c + d*x)/2]] - 8*b^2*Log[Cos[(c + d* 
x)/2]] - 4*a^2*Log[Sin[(c + d*x)/2]] + 8*b^2*Log[Sin[(c + d*x)/2]] + a^2*S 
ec[(c + d*x)/2]^2 + 8*a*b*Tan[(c + d*x)/2])/(8*d)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3368, 3042, 3527, 3042, 3510, 3042, 3502, 3042, 3214, 3042, 4257}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^2 (a+b \sin (c+d x))^2}{\sin (c+d x)^3}dx\)

\(\Big \downarrow \) 3368

\(\displaystyle \int \left (1-\sin ^2(c+d x)\right ) \csc ^3(c+d x) (a+b \sin (c+d x))^2dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (1-\sin (c+d x)^2\right ) (a+b \sin (c+d x))^2}{\sin (c+d x)^3}dx\)

\(\Big \downarrow \) 3527

\(\displaystyle \frac {1}{2} \int \csc ^2(c+d x) (a+b \sin (c+d x)) \left (-3 b \sin ^2(c+d x)-a \sin (c+d x)+2 b\right )dx-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {(a+b \sin (c+d x)) \left (-3 b \sin (c+d x)^2-a \sin (c+d x)+2 b\right )}{\sin (c+d x)^2}dx-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{2} \left (-\int \csc (c+d x) \left (a^2+4 b \sin (c+d x) a-2 b^2+3 b^2 \sin ^2(c+d x)\right )dx-\frac {2 a b \cot (c+d x)}{d}\right )-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (-\int \frac {a^2+4 b \sin (c+d x) a-2 b^2+3 b^2 \sin (c+d x)^2}{\sin (c+d x)}dx-\frac {2 a b \cot (c+d x)}{d}\right )-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{2} \left (-\int \csc (c+d x) \left (a^2+4 b \sin (c+d x) a-2 b^2\right )dx-\frac {2 a b \cot (c+d x)}{d}+\frac {3 b^2 \cos (c+d x)}{d}\right )-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (-\int \frac {a^2+4 b \sin (c+d x) a-2 b^2}{\sin (c+d x)}dx-\frac {2 a b \cot (c+d x)}{d}+\frac {3 b^2 \cos (c+d x)}{d}\right )-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {1}{2} \left (-\left (a^2-2 b^2\right ) \int \csc (c+d x)dx-\frac {2 a b \cot (c+d x)}{d}-4 a b x+\frac {3 b^2 \cos (c+d x)}{d}\right )-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (-\left (a^2-2 b^2\right ) \int \csc (c+d x)dx-\frac {2 a b \cot (c+d x)}{d}-4 a b x+\frac {3 b^2 \cos (c+d x)}{d}\right )-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{2} \left (\frac {\left (a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{d}-\frac {2 a b \cot (c+d x)}{d}-4 a b x+\frac {3 b^2 \cos (c+d x)}{d}\right )-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}\)

Input:

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

Output:

(-4*a*b*x + ((a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/d + (3*b^2*Cos[c + d*x]) 
/d - (2*a*b*Cot[c + d*x])/d)/2 - (Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + 
 d*x])^2)/(2*d)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3368

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a 
 + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n 
}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3510

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

rule 3527

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + 
 f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 
2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* 
d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b 
*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( 
A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a b \left (-\cot \left (d x +c \right )-d x -c \right )+b^{2} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(102\)
default	\(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a b \left (-\cot \left (d x +c \right )-d x -c \right )+b^{2} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(102\)
risch	\(-2 a b x +\frac {{\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}-\frac {i a \left (i a \,{\mathrm e}^{3 i \left (d x +c \right )}+i a \,{\mathrm e}^{i \left (d x +c \right )}+4 b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{d}\)	\(182\)

Input:

int(cot(d*x+c)^2*csc(d*x+c)*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^2*(-1/2/sin(d*x+c)^2*cos(d*x+c)^3-1/2*cos(d*x+c)-1/2*ln(csc(d*x+c)- 
cot(d*x+c)))+2*a*b*(-cot(d*x+c)-d*x-c)+b^2*(cos(d*x+c)+ln(csc(d*x+c)-cot(d 
*x+c))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (83) = 166\).

Time = 0.09 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.89 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {8 \, a b d x \cos \left (d x + c\right )^{2} - 4 \, b^{2} \cos \left (d x + c\right )^{3} - 8 \, a b d x - 8 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:

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

Output:

-1/4*(8*a*b*d*x*cos(d*x + c)^2 - 4*b^2*cos(d*x + c)^3 - 8*a*b*d*x - 8*a*b* 
cos(d*x + c)*sin(d*x + c) - 2*(a^2 - 2*b^2)*cos(d*x + c) - ((a^2 - 2*b^2)* 
cos(d*x + c)^2 - a^2 + 2*b^2)*log(1/2*cos(d*x + c) + 1/2) + ((a^2 - 2*b^2) 
*cos(d*x + c)^2 - a^2 + 2*b^2)*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + 
c)^2 - d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.16 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {8 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a b - a^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, b^{2} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \] Input:

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

Output:

-1/4*(8*(d*x + c + 1/tan(d*x + c))*a*b - a^2*(2*cos(d*x + c)/(cos(d*x + c) 
^2 - 1) + log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) - 2*b^2*(2*cos(d* 
x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)))/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.66 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, {\left (d x + c\right )} a b + 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, {\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {16 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:

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

Output:

1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 - 16*(d*x + c)*a*b + 8*a*b*tan(1/2*d*x + 1 
/2*c) - 4*(a^2 - 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + 16*b^2/(tan(1/2*d 
*x + 1/2*c)^2 + 1) + (6*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 
1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2)/d
 

Mupad [B] (verification not implemented)

Time = 34.68 (sec) , antiderivative size = 397, normalized size of antiderivative = 4.46 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\cos \left (c+d\,x\right )\,\left (\frac {a^2}{2}-\frac {b^2}{4}\right )-\frac {b^2}{2}+\frac {a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}-\frac {b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {b^2\,\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {b^2\,\cos \left (3\,c+3\,d\,x\right )}{4}-2\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )-\frac {a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{4}+\frac {b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{2}+a\,b\,\sin \left (2\,c+2\,d\,x\right )+2\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )\,\cos \left (2\,c+2\,d\,x\right )}{d\,\left ({\cos \left (c+d\,x\right )}^2-1\right )} \] Input:

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

Output:

(cos(c + d*x)*(a^2/2 - b^2/4) - b^2/2 + (a^2*log(sin(c/2 + (d*x)/2)/cos(c/ 
2 + (d*x)/2)))/4 - (b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (b 
^2*cos(2*c + 2*d*x))/2 + (b^2*cos(3*c + 3*d*x))/4 - 2*a*b*atan((a^2*sin(c/ 
2 + (d*x)/2) - 2*b^2*sin(c/2 + (d*x)/2) + 4*a*b*cos(c/2 + (d*x)/2))/(2*b^2 
*cos(c/2 + (d*x)/2) - a^2*cos(c/2 + (d*x)/2) + 4*a*b*sin(c/2 + (d*x)/2))) 
- (a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/4 + (b 
^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/2 + a*b*si 
n(2*c + 2*d*x) + 2*a*b*atan((a^2*sin(c/2 + (d*x)/2) - 2*b^2*sin(c/2 + (d*x 
)/2) + 4*a*b*cos(c/2 + (d*x)/2))/(2*b^2*cos(c/2 + (d*x)/2) - a^2*cos(c/2 + 
 (d*x)/2) + 4*a*b*sin(c/2 + (d*x)/2)))*cos(2*c + 2*d*x))/(d*(cos(c + d*x)^ 
2 - 1))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.63 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-16 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -4 \cos \left (d x +c \right ) a^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2}+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{2}+\sin \left (d x +c \right )^{2} a^{2}-16 \sin \left (d x +c \right )^{2} a b d x -8 \sin \left (d x +c \right )^{2} b^{2}}{8 \sin \left (d x +c \right )^{2} d} \] Input:

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

Output:

(8*cos(c + d*x)*sin(c + d*x)**2*b**2 - 16*cos(c + d*x)*sin(c + d*x)*a*b - 
4*cos(c + d*x)*a**2 - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2 + 8*log 
(tan((c + d*x)/2))*sin(c + d*x)**2*b**2 + sin(c + d*x)**2*a**2 - 16*sin(c 
+ d*x)**2*a*b*d*x - 8*sin(c + d*x)**2*b**2)/(8*sin(c + d*x)**2*d)