\(\int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1128]

Optimal result

Integrand size = 27, antiderivative size = 158 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {6 b \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 d}+\frac {3 \left (a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 a^4 d}-\frac {\cos (c+d x)}{2 a^2 d \left (1-\cos ^2(c+d x)\right )}+\frac {2 b \cot (c+d x)}{a^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a^3 d (a+b \sin (c+d x))} \] Output:

6*b*(a^2-b^2)^(1/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^4/d 
+3/2*(a^2-2*b^2)*arctanh(cos(d*x+c))/a^4/d-1/2*cos(d*x+c)/a^2/d/(1-cos(d*x 
+c)^2)+2*b*cot(d*x+c)/a^3/d-(a^2-b^2)*cos(d*x+c)/a^3/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 4.75 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.21 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {48 b \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+8 a b \cot \left (\frac {1}{2} (c+d x)\right )-a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )+12 \left (a^2-2 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-12 \left (a^2-2 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {8 a \left (-a^2+b^2\right ) \cos (c+d x)}{a+b \sin (c+d x)}-8 a b \tan \left (\frac {1}{2} (c+d x)\right )}{8 a^4 d} \] Input:

Integrate[(Cos[c + d*x]*Cot[c + d*x]^3)/(a + b*Sin[c + d*x])^2,x]
 

Output:

(48*b*Sqrt[a^2 - b^2]*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 8 
*a*b*Cot[(c + d*x)/2] - a^2*Csc[(c + d*x)/2]^2 + 12*(a^2 - 2*b^2)*Log[Cos[ 
(c + d*x)/2]] - 12*(a^2 - 2*b^2)*Log[Sin[(c + d*x)/2]] + a^2*Sec[(c + d*x) 
/2]^2 + (8*a*(-a^2 + b^2)*Cos[c + d*x])/(a + b*Sin[c + d*x]) - 8*a*b*Tan[( 
c + d*x)/2])/(8*a^4*d)
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.27, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3369, 3042, 3534, 27, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x)^3 (a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3369

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle \frac {-\frac {3 \left (\frac {8 b^2 \left (a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d}+\frac {b \left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}\right )}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a^2 b}+\frac {\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

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

Input:

Int[(Cos[c + d*x]*Cot[c + d*x]^3)/(a + b*Sin[c + d*x])^2,x]
 

Output:

((-3*((-4*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) - (b*(a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(a*d)))/a - 
(2*(a^2 - 3*b^2)*Cot[c + d*x])/(a*d))/(2*a^2*b) + ((2*a^2 - 3*b^2)*Cot[c + 
 d*x])/(2*a^2*b*d*(a + b*Sin[c + d*x])) - (Cot[c + d*x]*Csc[c + d*x])/(2*a 
*d*(a + b*Sin[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 3369

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]*(d*Sin[ 
e + f*x])^(n + 1)*((a + b*Sin[e + f*x])^(m + 1)/(a*d*f*(n + 1))), x] + (-Si 
mp[(a^2*(n + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(( 
a + b*Sin[e + f*x])^(m + 1)/(a^2*b*d^2*f*(n + 1)*(m + 1))), x] + Simp[1/(a^ 
2*b*d*(n + 1)*(m + 1))   Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^ 
(m + 1)*Simp[a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 1 
)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 4))*Sin[e 
+ f*x]^2, x], x], x]) /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && 
IntegersQ[2*m, 2*n] && LtQ[m, -1] && LtQ[n, -1]
 

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 3534

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[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[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, 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 = 1.25 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}-4 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{3}}-\frac {1}{8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-6 a^{2}+12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{4}}+\frac {b}{a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\frac {4 \left (-\frac {b \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a^{3}}{2}+\frac {a \,b^{2}}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+6 b \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4}}}{d}\)	\(209\)
default	\(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}-4 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{3}}-\frac {1}{8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-6 a^{2}+12 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{4}}+\frac {b}{a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\frac {4 \left (-\frac {b \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a^{3}}{2}+\frac {a \,b^{2}}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+6 b \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4}}}{d}\)	\(209\)
risch	\(\frac {i \left (-3 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+6 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-9 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+6 b^{3}+2 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-4 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-2 a^{2} b \right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} b \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right ) d \,a^{3}}-\frac {3 i \sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,a^{4}}+\frac {3 i \sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,a^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{a^{4} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{a^{4} d}\)	\(395\)

Input:

int(cos(d*x+c)*cot(d*x+c)^3/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/4/a^3*(1/2*tan(1/2*d*x+1/2*c)^2*a-4*b*tan(1/2*d*x+1/2*c))-1/8/a^2/t 
an(1/2*d*x+1/2*c)^2+1/4/a^4*(-6*a^2+12*b^2)*ln(tan(1/2*d*x+1/2*c))+b/a^3/t 
an(1/2*d*x+1/2*c)+4/a^4*((-1/2*b*(a^2-b^2)*tan(1/2*d*x+1/2*c)-1/2*a^3+1/2* 
a*b^2)/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)+3/2*b*(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))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (147) = 294\).

Time = 0.15 (sec) , antiderivative size = 804, normalized size of antiderivative = 5.09 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="fricas" 
)
 

Output:

[-1/4*(6*a^2*b*cos(d*x + c)*sin(d*x + c) + 4*(a^3 - 3*a*b^2)*cos(d*x + c)^ 
3 - 6*(a*b*cos(d*x + c)^2 - a*b + (b^2*cos(d*x + c)^2 - 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)) - 6*(a^3 - 2 
*a*b^2)*cos(d*x + c) + 3*(a^3 - 2*a*b^2 - (a^3 - 2*a*b^2)*cos(d*x + c)^2 + 
 (a^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*co 
s(d*x + c) + 1/2) - 3*(a^3 - 2*a*b^2 - (a^3 - 2*a*b^2)*cos(d*x + c)^2 + (a 
^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos( 
d*x + c) + 1/2))/(a^5*d*cos(d*x + c)^2 - a^5*d + (a^4*b*d*cos(d*x + c)^2 - 
 a^4*b*d)*sin(d*x + c)), -1/4*(6*a^2*b*cos(d*x + c)*sin(d*x + c) + 4*(a^3 
- 3*a*b^2)*cos(d*x + c)^3 + 12*(a*b*cos(d*x + c)^2 - a*b + (b^2*cos(d*x + 
c)^2 - b^2)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sq 
rt(a^2 - b^2)*cos(d*x + c))) - 6*(a^3 - 2*a*b^2)*cos(d*x + c) + 3*(a^3 - 2 
*a*b^2 - (a^3 - 2*a*b^2)*cos(d*x + c)^2 + (a^2*b - 2*b^3 - (a^2*b - 2*b^3) 
*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*(a^3 - 2*a* 
b^2 - (a^3 - 2*a*b^2)*cos(d*x + c)^2 + (a^2*b - 2*b^3 - (a^2*b - 2*b^3)*co 
s(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/(a^5*d*cos(d*x + 
 c)^2 - a^5*d + (a^4*b*d*cos(d*x + c)^2 - a^4*b*d)*sin(d*x + c))]
 

Sympy [F]

\[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cos {\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(cos(d*x+c)*cot(d*x+c)**3/(a+b*sin(d*x+c))**2,x)
 

Output:

Integral(cos(c + d*x)*cot(c + d*x)**3/(a + b*sin(c + d*x))**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)^3/(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 = 0.18 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.74 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, {\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {48 \, {\left (a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} + \frac {16 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} - a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a^{4}} - \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:

integrate(cos(d*x+c)*cot(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")
 

Output:

-1/8*(12*(a^2 - 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - 48*(a^2*b - b^ 
3)*(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)))/(sqrt(a^2 - b^2)*a^4) - (a^2*tan(1/2*d*x + 1/2*c 
)^2 - 8*a*b*tan(1/2*d*x + 1/2*c))/a^4 + 16*(a^2*b*tan(1/2*d*x + 1/2*c) - b 
^3*tan(1/2*d*x + 1/2*c) + a^3 - a*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*ta 
n(1/2*d*x + 1/2*c) + a)*a^4) - (18*a^2*tan(1/2*d*x + 1/2*c)^2 - 36*b^2*tan 
(1/2*d*x + 1/2*c)^2 + 8*a*b*tan(1/2*d*x + 1/2*c) - a^2)/(a^4*tan(1/2*d*x + 
 1/2*c)^2))/d
 

Mupad [B] (verification not implemented)

Time = 33.79 (sec) , antiderivative size = 675, normalized size of antiderivative = 4.27 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int((cos(c + d*x)*cot(c + d*x)^3)/(a + b*sin(c + d*x))^2,x)
 

Output:

tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (tan(c/2 + (d*x)/2)^2*((17*a^2)/2 - 16*b^ 
2) + a^2/2 + (4*tan(c/2 + (d*x)/2)^3*(a^2*b - 2*b^3))/a - 3*a*b*tan(c/2 + 
(d*x)/2))/(d*(4*a^4*tan(c/2 + (d*x)/2)^2 + 4*a^4*tan(c/2 + (d*x)/2)^4 + 8* 
a^3*b*tan(c/2 + (d*x)/2)^3)) - (b*tan(c/2 + (d*x)/2))/(a^3*d) - (log(tan(c 
/2 + (d*x)/2))*(3*a^2 - 6*b^2))/(2*a^4*d) - (6*b*atanh((72*b^4*(b^2 - a^2) 
^(1/2))/(18*a^4*b + 72*b^5 - 90*a^2*b^3 - 216*a*b^4*tan(c/2 + (d*x)/2) + 7 
2*a^3*b^2*tan(c/2 + (d*x)/2) + (144*b^6*tan(c/2 + (d*x)/2))/a) - (54*b^2*( 
b^2 - a^2)^(1/2))/(18*a^2*b - 90*b^3 + (72*b^5)/a^2 + 72*a*b^2*tan(c/2 + ( 
d*x)/2) - (216*b^4*tan(c/2 + (d*x)/2))/a + (144*b^6*tan(c/2 + (d*x)/2))/a^ 
3) + (18*b*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(18*a*b - (90*b^3)/a + (7 
2*b^5)/a^3 + 72*b^2*tan(c/2 + (d*x)/2) - (216*b^4*tan(c/2 + (d*x)/2))/a^2 
+ (144*b^6*tan(c/2 + (d*x)/2))/a^4) - (144*b^3*tan(c/2 + (d*x)/2)*(b^2 - a 
^2)^(1/2))/(18*a^3*b - 90*a*b^3 + (72*b^5)/a - 216*b^4*tan(c/2 + (d*x)/2) 
+ 72*a^2*b^2*tan(c/2 + (d*x)/2) + (144*b^6*tan(c/2 + (d*x)/2))/a^2) + (144 
*b^5*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(72*a*b^5 + 18*a^5*b - 90*a^3*b 
^3 + 144*b^6*tan(c/2 + (d*x)/2) - 216*a^2*b^4*tan(c/2 + (d*x)/2) + 72*a^4* 
b^2*tan(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2))/(a^4*d)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.84 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{3} b^{2}+12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} a b -2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -\cos \left (d x +c \right ) a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a^{2} b +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} b^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a \,b^{2}}{2 \sin \left (d x +c \right )^{2} a^{4} d \left (\sin \left (d x +c \right ) b +a \right )} \] Input:

int(cos(d*x+c)*cot(d*x+c)^3/(a+b*sin(d*x+c))^2,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)**3*b**2 + 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqr 
t(a**2 - b**2))*sin(c + d*x)**2*a*b - 2*cos(c + d*x)*sin(c + d*x)**2*a**3 
+ 6*cos(c + d*x)*sin(c + d*x)**2*a*b**2 + 3*cos(c + d*x)*sin(c + d*x)*a**2 
*b - cos(c + d*x)*a**3 - 3*log(tan((c + d*x)/2))*sin(c + d*x)**3*a**2*b + 
6*log(tan((c + d*x)/2))*sin(c + d*x)**3*b**3 - 3*log(tan((c + d*x)/2))*sin 
(c + d*x)**2*a**3 + 6*log(tan((c + d*x)/2))*sin(c + d*x)**2*a*b**2)/(2*sin 
(c + d*x)**2*a**4*d*(sin(c + d*x)*b + a))