3.11.82 \(\int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1082]

3.11.82.1 Optimal result
3.11.82.2 Mathematica [A] (verified)
3.11.82.3 Rubi [A] (verified)
3.11.82.4 Maple [A] (verified)
3.11.82.5 Fricas [B] (verification not implemented)
3.11.82.6 Sympy [F]
3.11.82.7 Maxima [F(-2)]
3.11.82.8 Giac [A] (verification not implemented)
3.11.82.9 Mupad [B] (verification not implemented)

3.11.82.1 Optimal result

Integrand size = 21, antiderivative size = 115 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \left (a^2-2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \sqrt {a^2-b^2} d}+\frac {2 b \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))} \]

output
2*b*arctanh(cos(d*x+c))/a^3/d-2*cot(d*x+c)/a^2/d+cot(d*x+c)/a/d/(a+b*sin(d 
*x+c))-2*(a^2-2*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/ 
d/(a^2-b^2)^(1/2)
 
3.11.82.2 Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.21 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {4 \left (a^2-2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+a \cot \left (\frac {1}{2} (c+d x)\right )-4 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 a b \cos (c+d x)}{a+b \sin (c+d x)}-a \tan \left (\frac {1}{2} (c+d x)\right )}{2 a^3 d} \]

input
Integrate[Cot[c + d*x]^2/(a + b*Sin[c + d*x])^2,x]
 
output
-1/2*((4*(a^2 - 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/S 
qrt[a^2 - b^2] + a*Cot[(c + d*x)/2] - 4*b*Log[Cos[(c + d*x)/2]] + 4*b*Log[ 
Sin[(c + d*x)/2]] + (2*a*b*Cos[c + d*x])/(a + b*Sin[c + d*x]) - a*Tan[(c + 
 d*x)/2])/(a^3*d)
 
3.11.82.3 Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.40, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3202, 3042, 3535, 3042, 3535, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3202

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

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

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

3.11.82.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 3202
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^2, 
x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((1 - Sin[e + f*x]^2)/Sin[e + f*x]^ 
2), x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0]
 

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]
 
3.11.82.4 Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.36

method result size
derivativedivides \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}+a b}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{a^{3}}}{d}\) \(156\)
default \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}+a b}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{a^{3}}}{d}\) \(156\)
risch \(-\frac {2 \left (-3 a \,{\mathrm e}^{i \left (d x +c \right )}+2 i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i b +a \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right ) a^{2} d}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d a}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{3} d}\) \(417\)

input
int(cos(d*x+c)^2*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 
output
1/d*(1/2*tan(1/2*d*x+1/2*c)/a^2-1/2/a^2/tan(1/2*d*x+1/2*c)-2/a^3*b*ln(tan( 
1/2*d*x+1/2*c))-2/a^3*((tan(1/2*d*x+1/2*c)*b^2+a*b)/(tan(1/2*d*x+1/2*c)^2* 
a+2*b*tan(1/2*d*x+1/2*c)+a)+(a^2-2*b^2)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*ta 
n(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))
 
3.11.82.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (110) = 220\).

Time = 0.37 (sec) , antiderivative size = 768, normalized size of antiderivative = 6.68 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [\frac {4 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )\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^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right ) - 2 \, {\left (a^{2} b^{2} - b^{4} - {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a 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 (a^{2} b^{2} - b^{4} - {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a 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^{5} b - a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} - a^{4} b^{2}\right )} d \sin \left (d x + c\right ) - {\left (a^{5} b - a^{3} b^{3}\right )} d\right )}}, \frac {2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )\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 ) + {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right ) - {\left (a^{2} b^{2} - b^{4} - {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a 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^{2} b^{2} - b^{4} - {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a 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^{5} b - a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} - a^{4} b^{2}\right )} d \sin \left (d x + c\right ) - {\left (a^{5} b - a^{3} b^{3}\right )} d}\right ] \]

input
integrate(cos(d*x+c)^2*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="frica 
s")
 
output
[1/2*(4*(a^3*b - a*b^3)*cos(d*x + c)*sin(d*x + c) - (a^2*b - 2*b^3 - (a^2* 
b - 2*b^3)*cos(d*x + c)^2 + (a^3 - 2*a*b^2)*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*(a^4 - a^2*b^2)*cos(d*x + 
c) - 2*(a^2*b^2 - b^4 - (a^2*b^2 - b^4)*cos(d*x + c)^2 + (a^3*b - a*b^3)*s 
in(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 2*(a^2*b^2 - b^4 - (a^2*b^2 - b 
^4)*cos(d*x + c)^2 + (a^3*b - a*b^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 
 1/2))/((a^5*b - a^3*b^3)*d*cos(d*x + c)^2 - (a^6 - a^4*b^2)*d*sin(d*x + c 
) - (a^5*b - a^3*b^3)*d), (2*(a^3*b - a*b^3)*cos(d*x + c)*sin(d*x + c) - ( 
a^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2 + (a^3 - 2*a*b^2)*sin(d*x + 
 c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x 
 + c))) + (a^4 - a^2*b^2)*cos(d*x + c) - (a^2*b^2 - b^4 - (a^2*b^2 - b^4)* 
cos(d*x + c)^2 + (a^3*b - a*b^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) 
 + (a^2*b^2 - b^4 - (a^2*b^2 - b^4)*cos(d*x + c)^2 + (a^3*b - a*b^3)*sin(d 
*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^5*b - a^3*b^3)*d*cos(d*x + c)^2 
 - (a^6 - a^4*b^2)*d*sin(d*x + c) - (a^5*b - a^3*b^3)*d)]
 
3.11.82.6 Sympy [F]

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

input
integrate(cos(d*x+c)**2*csc(d*x+c)**2/(a+b*sin(d*x+c))**2,x)
 
output
Integral(cos(c + d*x)**2*csc(c + d*x)**2/(a + b*sin(c + d*x))**2, x)
 
3.11.82.7 Maxima [F(-2)]

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

input
integrate(cos(d*x+c)^2*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="maxim 
a")
 
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
 
3.11.82.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.90 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} + \frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a^{2} - 2 \, b^{2}\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} - \frac {4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 14 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{3}}}{6 \, d} \]

input
integrate(cos(d*x+c)^2*csc(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac" 
)
 
output
-1/6*(12*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 3*tan(1/2*d*x + 1/2*c)/a^2 
 + 12*(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)))*(a^2 - 2*b^2)/(sqrt(a^2 - b^2)*a^3) - (4*a*b* 
tan(1/2*d*x + 1/2*c)^3 - 3*a^2*tan(1/2*d*x + 1/2*c)^2 - 4*b^2*tan(1/2*d*x 
+ 1/2*c)^2 - 14*a*b*tan(1/2*d*x + 1/2*c) - 3*a^2)/((a*tan(1/2*d*x + 1/2*c) 
^3 + 2*b*tan(1/2*d*x + 1/2*c)^2 + a*tan(1/2*d*x + 1/2*c))*a^3))/d
 
3.11.82.9 Mupad [B] (verification not implemented)

Time = 11.61 (sec) , antiderivative size = 1616, normalized size of antiderivative = 14.05 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \]

input
int(cos(c + d*x)^2/(sin(c + d*x)^2*(a + b*sin(c + d*x))^2),x)
 
output
-(a^4*cos(c + d*x) - b^4/2 - b^4*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2) 
) + (a^2*b^2)/2 + (b^4*cos(2*c + 2*d*x))/2 - a^2*b^2*cos(c + d*x) - a*b^3* 
sin(2*c + 2*d*x) + a^3*b*sin(2*c + 2*d*x) + a^2*b^2*log(sin(c/2 + (d*x)/2) 
/cos(c/2 + (d*x)/2)) + b^4*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos( 
2*c + 2*d*x) + b^3*atan((a^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i - b^3 
*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i - a*b^2*cos(c/2 + (d*x)/2)*(b^2 - 
 a^2)^(1/2)*4i + a^2*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i)/(a^4*sin(c 
/2 + (d*x)/2) + 8*b^4*sin(c/2 + (d*x)/2) + 4*a*b^3*cos(c/2 + (d*x)/2) - 3* 
a^3*b*cos(c/2 + (d*x)/2) - 8*a^2*b^2*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2 
)*2i - (a^2*b^2*cos(2*c + 2*d*x))/2 - a*b^3*sin(c + d*x) + a^3*b*sin(c + d 
*x) - a^2*b*atan((a^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i - b^3*sin(c/ 
2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i - a*b^2*cos(c/2 + (d*x)/2)*(b^2 - a^2)^( 
1/2)*4i + a^2*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i)/(a^4*sin(c/2 + (d 
*x)/2) + 8*b^4*sin(c/2 + (d*x)/2) + 4*a*b^3*cos(c/2 + (d*x)/2) - 3*a^3*b*c 
os(c/2 + (d*x)/2) - 8*a^2*b^2*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2)*1i - 
a^3*sin(c + d*x)*atan((a^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i - b^3*s 
in(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i - a*b^2*cos(c/2 + (d*x)/2)*(b^2 - a 
^2)^(1/2)*4i + a^2*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i)/(a^4*sin(c/2 
 + (d*x)/2) + 8*b^4*sin(c/2 + (d*x)/2) + 4*a*b^3*cos(c/2 + (d*x)/2) - 3*a^ 
3*b*cos(c/2 + (d*x)/2) - 8*a^2*b^2*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/...