\(\int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [1084]

Optimal result

Integrand size = 25, antiderivative size = 154 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {b \left (3 a^2-2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))} \] Output:

-b*(3*a^2-2*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/(a^2 
-b^2)^(3/2)/d-arctanh(cos(d*x+c))/a^3/d+1/2*cos(d*x+c)/a/d/(a+b*sin(d*x+c) 
)^2+1/2*(a^2-2*b^2)*cos(d*x+c)/a^2/(a^2-b^2)/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {2 b \left (-3 a^2+2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a \cos (c+d x) \left (2 a^3-3 a b^2+b \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{(a-b) (a+b) (a+b \sin (c+d x))^2}}{2 a^3 d} \] Input:

Integrate[(Cos[c + d*x]*Cot[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

((2*b*(-3*a^2 + 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/( 
a^2 - b^2)^(3/2) - 2*Log[Cos[(c + d*x)/2]] + 2*Log[Sin[(c + d*x)/2]] + (a* 
Cos[c + d*x]*(2*a^3 - 3*a*b^2 + b*(a^2 - 2*b^2)*Sin[c + d*x]))/((a - b)*(a 
 + b)*(a + b*Sin[c + d*x])^2))/(2*a^3*d)
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.25, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3368, 3042, 3535, 3042, 3535, 3042, 3480, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

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

Input:

Int[(Cos[c + d*x]*Cot[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

Cos[c + d*x]/(2*a*d*(a + b*Sin[c + d*x])^2) + (((-2*b*(3*a^2 - 2*b^2)*Sqrt 
[a^2 - b^2]*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(a*d 
) - (2*(a^2 - b^2)^2*ArcTanh[Cos[c + d*x]])/(a*d))/(a*(a^2 - b^2)) + ((a^2 
 - 2*b^2)*Cos[c + d*x])/(a*d*(a + b*Sin[c + d*x])))/(2*a*(a^2 - b^2))
 

Defintions of rubi rules used

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 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 3480

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b 
- a*B)/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ 
(b*c - a*d)   Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

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 = 1.22 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.64

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {-\frac {a b \left (3 a^{2}-4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}-b^{2}\right )}-\frac {\left (2 a^{4}+a^{2} b^{2}-6 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{2}-b^{2}\right )}-\frac {a b \left (5 a^{2}-8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-b^{2}\right )}-\frac {a^{2} \left (2 a^{2}-3 b^{2}\right )}{2 \left (a^{2}-b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {b \left (3 a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\)	\(253\)
default	\(\frac {-\frac {2 \left (\frac {-\frac {a b \left (3 a^{2}-4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}-b^{2}\right )}-\frac {\left (2 a^{4}+a^{2} b^{2}-6 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{2}-b^{2}\right )}-\frac {a b \left (5 a^{2}-8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-b^{2}\right )}-\frac {a^{2} \left (2 a^{2}-3 b^{2}\right )}{2 \left (a^{2}-b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {b \left (3 a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\)	\(253\)
risch	\(-\frac {i \left (i a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+4 i a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-7 i a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+2 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-3 b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-a^{2} b^{2}+2 b^{4}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} a^{2} \left (a^{2}-b^{2}\right ) d b}+\frac {3 i b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}-\frac {i b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}-\frac {3 i b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}+\frac {i b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\)	\(539\)

Input:

int(cos(d*x+c)*cot(d*x+c)/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2/a^3*((-1/2*a*b*(3*a^2-4*b^2)/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3-1/2*(2 
*a^4+a^2*b^2-6*b^4)/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2-1/2*a*b*(5*a^2-8*b^2)/( 
a^2-b^2)*tan(1/2*d*x+1/2*c)-1/2*a^2*(2*a^2-3*b^2)/(a^2-b^2))/(tan(1/2*d*x+ 
1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)^2+1/2*b*(3*a^2-2*b^2)/(a^2-b^2)^(3/2) 
*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2)))+1/a^3*ln(tan(1/ 
2*d*x+1/2*c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (145) = 290\).

Time = 0.34 (sec) , antiderivative size = 996, normalized size of antiderivative = 6.47 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

[-1/4*(2*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)*sin(d*x + c) - (3*a^4* 
b + a^2*b^3 - 2*b^5 - (3*a^2*b^3 - 2*b^5)*cos(d*x + c)^2 + 2*(3*a^3*b^2 - 
2*a*b^4)*sin(d*x + c))*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*(2*a^6 - 5*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c) - 2*(a^6 - a^4* 
b^2 - a^2*b^4 + b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2*(a^5* 
b - 2*a^3*b^3 + a*b^5)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 2*(a^6 
- a^4*b^2 - a^2*b^4 + b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2 
*(a^5*b - 2*a^3*b^3 + a*b^5)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/( 
(a^7*b^2 - 2*a^5*b^4 + a^3*b^6)*d*cos(d*x + c)^2 - 2*(a^8*b - 2*a^6*b^3 + 
a^4*b^5)*d*sin(d*x + c) - (a^9 - a^7*b^2 - a^5*b^4 + a^3*b^6)*d), -1/2*((a 
^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)*sin(d*x + c) + (3*a^4*b + a^2*b^3 
 - 2*b^5 - (3*a^2*b^3 - 2*b^5)*cos(d*x + c)^2 + 2*(3*a^3*b^2 - 2*a*b^4)*si 
n(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)* 
cos(d*x + c))) + (2*a^6 - 5*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c) - (a^6 - a^4 
*b^2 - a^2*b^4 + b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2*(a^5 
*b - 2*a^3*b^3 + a*b^5)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^6 - 
 a^4*b^2 - a^2*b^4 + b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2* 
(a^5*b - 2*a^3*b^3 + a*b^5)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))...
 

Sympy [F]

\[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\cos {\left (c + d x \right )} \cot {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(cos(d*x+c)*cot(d*x+c)/(a+b*sin(d*x+c))**3,x)
 

Output:

Integral(cos(c + d*x)*cot(c + d*x)/(a + b*sin(c + d*x))**3, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)/(a+b*sin(d*x+c))^3,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 = 0.16 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.80 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {{\left (3 \, a^{2} b - 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{4} - 3 \, a^{2} b^{2}}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}}}{d} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

-((3*a^2*b - 2*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*t 
an(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^5 - a^3*b^2)*sqrt(a^2 - b^2 
)) - (3*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 4*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 2* 
a^4*tan(1/2*d*x + 1/2*c)^2 + a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*b^4*tan(1/ 
2*d*x + 1/2*c)^2 + 5*a^3*b*tan(1/2*d*x + 1/2*c) - 8*a*b^3*tan(1/2*d*x + 1/ 
2*c) + 2*a^4 - 3*a^2*b^2)/((a^5 - a^3*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b 
*tan(1/2*d*x + 1/2*c) + a)^2) - log(abs(tan(1/2*d*x + 1/2*c)))/a^3)/d
 

Mupad [B] (verification not implemented)

Time = 36.61 (sec) , antiderivative size = 1610, normalized size of antiderivative = 10.45 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)*cot(c + d*x))/(a + b*sin(c + d*x))^3,x)
 

Output:

log(tan(c/2 + (d*x)/2))/(a^3*d) + ((2*a^2 - 3*b^2)/(a*(a^2 - b^2)) + (tan( 
c/2 + (d*x)/2)^2*(2*a^4 - 6*b^4 + a^2*b^2))/(a^3*(a^2 - b^2)) + (tan(c/2 + 
 (d*x)/2)*(5*a^2*b - 8*b^3))/(a^2*(a^2 - b^2)) + (b*tan(c/2 + (d*x)/2)^3*( 
3*a^2 - 4*b^2))/(a^2*(a^2 - b^2)))/(d*(tan(c/2 + (d*x)/2)^2*(2*a^2 + 4*b^2 
) + a^2*tan(c/2 + (d*x)/2)^4 + a^2 + 4*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*ta 
n(c/2 + (d*x)/2))) + (b*atan(((b*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1 
/2)*((5*a^5*b - 4*a^3*b^3)/(a^6 - a^4*b^2) + (tan(c/2 + (d*x)/2)*(8*a*b^6 
- 2*a^7 - 20*a^3*b^4 + 14*a^5*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (b*((2*a 
^8*b - 2*a^6*b^3)/(a^6 - a^4*b^2) - (tan(c/2 + (d*x)/2)*(6*a^10 - 8*a^4*b^ 
6 + 22*a^6*b^4 - 20*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2))*(3*a^2 - 2*b^2) 
*(-(a + b)^3*(a - b)^3)^(1/2))/(2*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)) 
)*1i)/(2*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)) + (b*(3*a^2 - 2*b^2)*(-( 
a + b)^3*(a - b)^3)^(1/2)*((5*a^5*b - 4*a^3*b^3)/(a^6 - a^4*b^2) + (tan(c/ 
2 + (d*x)/2)*(8*a*b^6 - 2*a^7 - 20*a^3*b^4 + 14*a^5*b^2))/(a^7 + a^3*b^4 - 
 2*a^5*b^2) + (b*((2*a^8*b - 2*a^6*b^3)/(a^6 - a^4*b^2) - (tan(c/2 + (d*x) 
/2)*(6*a^10 - 8*a^4*b^6 + 22*a^6*b^4 - 20*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5 
*b^2))*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(2*(a^9 - a^3*b^6 + 3 
*a^5*b^4 - 3*a^7*b^2)))*1i)/(2*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))/( 
(2*(3*a^2*b - 2*b^3))/(a^6 - a^4*b^2) + (2*tan(c/2 + (d*x)/2)*(2*b^4 - 3*a 
^2*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (b*(3*a^2 - 2*b^2)*(-(a + b)^3*(...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 792, normalized size of antiderivative = 5.14 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)*cot(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Output:

( - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* 
sin(c + d*x)**2*a**2*b**3 + 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + 
 b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*b**5 - 24*sqrt(a**2 - b**2)*atan((t 
an((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**3*b**2 + 16*sqrt 
(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x 
)*a*b**4 - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - 
b**2))*a**4*b + 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a** 
2 - b**2))*a**2*b**3 + 2*cos(c + d*x)*sin(c + d*x)*a**5*b - 6*cos(c + d*x) 
*sin(c + d*x)*a**3*b**3 + 4*cos(c + d*x)*sin(c + d*x)*a*b**5 + 4*cos(c + d 
*x)*a**6 - 10*cos(c + d*x)*a**4*b**2 + 6*cos(c + d*x)*a**2*b**4 + 4*log(ta 
n((c + d*x)/2))*sin(c + d*x)**2*a**4*b**2 - 8*log(tan((c + d*x)/2))*sin(c 
+ d*x)**2*a**2*b**4 + 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**6 + 8*log 
(tan((c + d*x)/2))*sin(c + d*x)*a**5*b - 16*log(tan((c + d*x)/2))*sin(c + 
d*x)*a**3*b**3 + 8*log(tan((c + d*x)/2))*sin(c + d*x)*a*b**5 + 4*log(tan(( 
c + d*x)/2))*a**6 - 8*log(tan((c + d*x)/2))*a**4*b**2 + 4*log(tan((c + d*x 
)/2))*a**2*b**4 + sin(c + d*x)**2*a**4*b**2 - 3*sin(c + d*x)**2*a**2*b**4 
+ 2*sin(c + d*x)**2*b**6 + 2*sin(c + d*x)*a**5*b - 6*sin(c + d*x)*a**3*b** 
3 + 4*sin(c + d*x)*a*b**5 + a**6 - 3*a**4*b**2 + 2*a**2*b**4)/(4*a**3*d*(s 
in(c + d*x)**2*a**4*b**2 - 2*sin(c + d*x)**2*a**2*b**4 + sin(c + d*x)**2*b 
**6 + 2*sin(c + d*x)*a**5*b - 4*sin(c + d*x)*a**3*b**3 + 2*sin(c + d*x)...