\(\int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [344]

Optimal result

Integrand size = 29, antiderivative size = 104 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a x}{b^2}+\frac {2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^2 d}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d} \] Output:

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

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 a^3 c+2 a^3 d x-4 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+2 a^2 b \cos (c+d x)+a b^2 \cot \left (\frac {1}{2} (c+d x)\right )-2 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-a b^2 \tan \left (\frac {1}{2} (c+d x)\right )}{2 a^2 b^2 d} \] Input:

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

Output:

-1/2*(2*a^3*c + 2*a^3*d*x - 4*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x 
)/2])/Sqrt[a^2 - b^2]] + 2*a^2*b*Cos[c + d*x] + a*b^2*Cot[(c + d*x)/2] - 2 
*b^3*Log[Cos[(c + d*x)/2]] + 2*b^3*Log[Sin[(c + d*x)/2]] - a*b^2*Tan[(c + 
d*x)/2])/(a^2*b^2*d)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3373, 3042, 3536, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x)^2 (a+b \sin (c+d x))}dx\)

\(\Big \downarrow \) 3373

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3536

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

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 3373

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[Cos[e + f*x]*(a + b* 
Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 1)/(a*d*f*(n + 1))), x] + (-Si 
mp[Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 2)/(b*d 
^2*f*(m + n + 4))), x] + Simp[1/(a*b*d*(n + 1)*(m + n + 4))   Int[(a + b*Si 
n[e + f*x])^m*(d*Sin[e + f*x])^(n + 1)*Simp[a^2*(n + 1)*(n + 2) - b^2*(m + 
n + 2)*(m + n + 4) + a*b*(m + 3)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2* 
(m + n + 3)*(m + n + 4))*Sin[e + f*x]^2, x], x], x]) /; FreeQ[{a, b, d, e, 
f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n]) &&  ! 
m < -1 && LtQ[n, -1] && NeQ[m + n + 4, 0]
 

rule 3536

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) 
*(x_)])), x_Symbol] :> Simp[C*(x/(b*d)), x] + (Simp[(A*b^2 - a*b*B + a^2*C) 
/(b*(b*c - a*d))   Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[(c^2*C - B*c*d 
 + A*d^2)/(d*(b*c - a*d))   Int[1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a 
, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 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 = 2.29 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.49

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+\frac {\left (4 a^{4}-8 a^{2} b^{2}+4 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 a^{2} b^{2} \sqrt {a^{2}-b^{2}}}-\frac {2 \left (\frac {b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\)	\(155\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+\frac {\left (4 a^{4}-8 a^{2} b^{2}+4 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 a^{2} b^{2} \sqrt {a^{2}-b^{2}}}-\frac {2 \left (\frac {b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\)	\(155\)
risch	\(-\frac {a x}{b^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}-\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{2}}\)	\(306\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 396, normalized size of antiderivative = 3.81 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a b^{2} \cos \left (d x + c\right ) - {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{3} d x + a^{2} b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, a^{2} b^{2} d \sin \left (d x + c\right )}, \frac {b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a b^{2} \cos \left (d x + c\right ) - 2 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{3} d x + a^{2} b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, a^{2} b^{2} d \sin \left (d x + c\right )}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(cos(d*x+c)^2*cot(d*x+c)^2/(a+b*sin(d*x+c)),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (99) = 198\).

Time = 0.16 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.12 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {6 \, {\left (d x + c\right )} a}{b^{2}} + \frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2} b^{2}} - \frac {2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{2} b}}{6 \, d} \] Input:

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

Output:

-1/6*(6*(d*x + c)*a/b^2 + 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 3*tan(1 
/2*d*x + 1/2*c)/a - 12*(a^4 - 2*a^2*b^2 + b^4)*(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)))/(sqr 
t(a^2 - b^2)*a^2*b^2) - (2*b^2*tan(1/2*d*x + 1/2*c)^3 - 3*a*b*tan(1/2*d*x 
+ 1/2*c)^2 - 12*a^2*tan(1/2*d*x + 1/2*c) + 2*b^2*tan(1/2*d*x + 1/2*c) - 3* 
a*b)/((tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))*a^2*b))/d
 

Mupad [B] (verification not implemented)

Time = 41.42 (sec) , antiderivative size = 1167, normalized size of antiderivative = 11.22 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((16*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) 
 - 4*a^12*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 4 
*a^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 12*a^3 
*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + a^5*b^ 
7*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 4*a^7*b^5 
*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 6*a^9*b^3* 
cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 29*a^2*b^4* 
sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + 18*a^4*b^2* 
sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + a^2*b^10*si 
n(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 4*a^4*b^8*sin 
(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 22*a^6*b^6*sin 
(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 32*a^8*b^4*sin 
(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 18*a^10*b^2*si 
n(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 8*a*b^5*cos(c 
/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + 5*a^5*b*cos(c/2 
+ (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + 2*a^11*b*cos(c/2 + 
(d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(b^15*sin(c/2 + (d*x)/ 
2)*16i + a*b^14*cos(c/2 + (d*x)/2)*8i - a^14*b*sin(c/2 + (d*x)/2)*3i - a^3 
*b^12*cos(c/2 + (d*x)/2)*48i + a^5*b^10*cos(c/2 + (d*x)/2)*123i - a^7*b^8* 
cos(c/2 + (d*x)/2)*167i + a^9*b^6*cos(c/2 + (d*x)/2)*126i - a^11*b^4*co...
 

Reduce [B] (verification not implemented)

Time = 16.02 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.94 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a^{2}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -\cos \left (d x +c \right ) a \,b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) b^{3}-\sin \left (d x +c \right ) a^{3} c -\sin \left (d x +c \right ) a^{3} d x +\sin \left (d x +c \right ) a^{2} b}{\sin \left (d x +c \right ) a^{2} b^{2} d} \] Input:

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

Output:

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