\(\int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1077]

Optimal result

Integrand size = 25, antiderivative size = 92 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))} \] Output:

-2*b*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(1/2)/ 
d-arctanh(cos(d*x+c))/a^2/d+cos(d*x+c)/a/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a \cos (c+d x)}{a+b \sin (c+d x)}}{a^2 d} \] Input:

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

Output:

((-2*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 
 Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]] + (a*Cos[c + d*x])/(a + b*S 
in[c + d*x]))/(a^2*d)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3368, 3042, 3535, 27, 3042, 3226, 3042, 3139, 1083, 217, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{\sin (c+d x) (a+b \sin (c+d x))}dx}{a}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3226

\(\displaystyle \frac {\frac {\int \csc (c+d x)dx}{a}-\frac {b \int \frac {1}{a+b \sin (c+d x)}dx}{a}}{a}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \csc (c+d x)dx}{a}-\frac {b \int \frac {1}{a+b \sin (c+d x)}dx}{a}}{a}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a d \sqrt {a^2-b^2}}-\frac {\text {arctanh}(\cos (c+d x))}{a d}}{a}+\frac {\cos (c+d x)}{a d (a+b \sin (c+d x))}\)

Input:

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

Output:

((-2*b*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(a*Sqrt[a 
^2 - b^2]*d) - ArcTanh[Cos[c + d*x]]/(a*d))/a + Cos[c + d*x]/(a*d*(a + b*S 
in[c + d*x]))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3226

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), 
 x], x] - Simp[d/(b*c - a*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 3535

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[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m 
+ 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin 
[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) || 
 !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 
 0])))
 

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 = 0.66 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {4 \left (\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2}}}{d}\)	\(116\)
default	\(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {4 \left (\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2}}}{d}\)	\(116\)
risch	\(\frac {2 i \left (-i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{a b d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {i b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}-\frac {i b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}\)	\(240\)

Input:

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

Output:

1/d*(1/a^2*ln(tan(1/2*d*x+1/2*c))-4/a^2*((-1/2*b*tan(1/2*d*x+1/2*c)-1/2*a) 
/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)+1/2*b/(a^2-b^2)^(1/2)*a 
rctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (87) = 174\).

Time = 0.14 (sec) , antiderivative size = 483, normalized size of antiderivative = 5.25 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [-\frac {{\left (b^{2} \sin \left (d x + c\right ) + a b\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) + {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{4} b - a^{2} b^{3}\right )} d \sin \left (d x + c\right ) + {\left (a^{5} - a^{3} b^{2}\right )} d\right )}}, \frac {2 \, {\left (b^{2} \sin \left (d x + c\right ) + a b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) - {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{4} b - a^{2} b^{3}\right )} d \sin \left (d x + c\right ) + {\left (a^{5} - a^{3} b^{2}\right )} d\right )}}\right ] \] Input:

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

Output:

[-1/2*((b^2*sin(d*x + c) + a*b)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d 
*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c 
) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x 
+ c) - a^2 - b^2)) - 2*(a^3 - a*b^2)*cos(d*x + c) + (a^3 - a*b^2 + (a^2*b 
- b^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - (a^3 - a*b^2 + (a^2*b - 
 b^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^4*b - a^2*b^3)*d*sin 
(d*x + c) + (a^5 - a^3*b^2)*d), 1/2*(2*(b^2*sin(d*x + c) + a*b)*sqrt(a^2 - 
 b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 2*(a^ 
3 - a*b^2)*cos(d*x + c) - (a^3 - a*b^2 + (a^2*b - b^3)*sin(d*x + c))*log(1 
/2*cos(d*x + c) + 1/2) + (a^3 - a*b^2 + (a^2*b - b^3)*sin(d*x + c))*log(-1 
/2*cos(d*x + c) + 1/2))/((a^4*b - a^2*b^3)*d*sin(d*x + c) + (a^5 - a^3*b^2 
)*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.41 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{\sqrt {a^{2} - b^{2}} a^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a^{2}}}{d} \] Input:

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

Output:

-(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2 
*c) + b)/sqrt(a^2 - b^2)))*b/(sqrt(a^2 - b^2)*a^2) - log(abs(tan(1/2*d*x + 
 1/2*c)))/a^2 - 2*(b*tan(1/2*d*x + 1/2*c) + a)/((a*tan(1/2*d*x + 1/2*c)^2 
+ 2*b*tan(1/2*d*x + 1/2*c) + a)*a^2))/d
 

Mupad [B] (verification not implemented)

Time = 37.11 (sec) , antiderivative size = 523, normalized size of antiderivative = 5.68 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {a^3\,\cos \left (c+d\,x\right )-a\,b^2-b^3\,\sin \left (c+d\,x\right )+a^3+a^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-a\,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 )-b^3\,\sin \left (c+d\,x\right )\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-a\,b^2\,\cos \left (c+d\,x\right )+a^2\,b\,\sin \left (c+d\,x\right )+a^2\,b\,\sin \left (c+d\,x\right )\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+b^2\,\sin \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}+a\,b\,\mathrm {atan}\left (\frac {-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}}{a^2\,d\,\left (a^2-b^2\right )\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.45 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) b^{2}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a b +\cos \left (d x +c \right ) a^{3}-\cos \left (d x +c \right ) a \,b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}}{a^{2} d \left (\sin \left (d x +c \right ) a^{2} b -\sin \left (d x +c \right ) b^{3}+a^{3}-a \,b^{2}\right )} \] Input:

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

Output:

( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*s 
in(c + d*x)*b**2 - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt( 
a**2 - b**2))*a*b + cos(c + d*x)*a**3 - cos(c + d*x)*a*b**2 + log(tan((c + 
 d*x)/2))*sin(c + d*x)*a**2*b - log(tan((c + d*x)/2))*sin(c + d*x)*b**3 + 
log(tan((c + d*x)/2))*a**3 - log(tan((c + d*x)/2))*a*b**2)/(a**2*d*(sin(c 
+ d*x)*a**2*b - sin(c + d*x)*b**3 + a**3 - a*b**2))