\(\int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) [1300]

Optimal result

Integrand size = 27, antiderivative size = 124 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (2 a^2-3 b^2\right ) x}{2 b^3}-\frac {2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {-4 a^3 c+6 a b^2 c-4 a^3 d x+6 a b^2 d x+8 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )-4 a^2 b \cos (c+d x)+4 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a b^2 \sin (2 (c+d x))}{4 a b^3 d} \] Input:

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

Output:

-1/4*(-4*a^3*c + 6*a*b^2*c - 4*a^3*d*x + 6*a*b^2*d*x + 8*(a^2 - b^2)^(3/2) 
*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] - 4*a^2*b*Cos[c + d*x] + 
 4*b^3*Log[Cos[(c + d*x)/2]] - 4*b^3*Log[Sin[(c + d*x)/2]] + a*b^2*Sin[2*( 
c + d*x)])/(a*b^3*d)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3374, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3374

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3536

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

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 3374

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a*(n + 3)*Cos[e + f* 
x]*(d*Sin[e + f*x])^(n + 1)*((a + b*Sin[e + f*x])^(m + 1)/(b^2*d*f*(m + n + 
 3)*(m + n + 4))), x] + (-Simp[Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*((a + 
b*Sin[e + f*x])^(m + 1)/(b*d^2*f*(m + n + 4))), x] - Simp[1/(b^2*(m + n + 3 
)*(m + n + 4))   Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*Simp[a^2*(n 
+ 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 
 2)*(n + 3) - b^2*(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x]) /; F 
reeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Integ 
ersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && 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 = 1.90 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.45

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+a b \right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 a^{2}-3 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (-2 a^{4}+4 a^{2} b^{2}-2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \,b^{3} \sqrt {a^{2}-b^{2}}}}{d}\)	\(180\)
default	\(\frac {\frac {\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+a b \right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 a^{2}-3 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (-2 a^{4}+4 a^{2} b^{2}-2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \,b^{3} \sqrt {a^{2}-b^{2}}}}{d}\)	\(180\)
risch	\(\frac {x \,a^{2}}{b^{3}}-\frac {3 x}{2 b}+\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {i \sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,b^{3}}-\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d b a}-\frac {i \sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,b^{3}}+\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d b a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}-\frac {\sin \left (2 d x +2 c \right )}{4 b d}\)	\(320\)

Input:

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

Output:

1/d*(2/b^3*((1/2*tan(1/2*d*x+1/2*c)^3*b^2+a*b*tan(1/2*d*x+1/2*c)^2-1/2*b^2 
*tan(1/2*d*x+1/2*c)+a*b)/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(2*a^2-3*b^2)*arct 
an(tan(1/2*d*x+1/2*c)))+1/a*ln(tan(1/2*d*x+1/2*c))+(-2*a^4+4*a^2*b^2-2*b^4 
)/a/b^3/(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 [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.82 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b \cos \left (d x + c\right ) + b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x - {\left (-a^{2} + b^{2}\right )}^{\frac {3}{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right )}{2 \, a b^{3} d}, -\frac {a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b \cos \left (d x + c\right ) + b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x - 2 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right )}{2 \, a b^{3} d}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(cos(d*x+c)^3*cot(d*x+c)/(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 [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {{\left (2 \, a^{2} - 3 \, b^{2}\right )} {\left (d x + c\right )}}{b^{3}} - \frac {4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a b^{3}} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \] Input:

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

Output:

1/2*(2*log(abs(tan(1/2*d*x + 1/2*c)))/a + (2*a^2 - 3*b^2)*(d*x + c)/b^3 - 
4*(a^4 - 2*a^2*b^2 + b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arcta 
n((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a*b^3) + 
 2*(b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x 
+ 1/2*c) + 2*a)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*b^2))/d
 

Mupad [B] (verification not implemented)

Time = 21.29 (sec) , antiderivative size = 1320, normalized size of antiderivative = 10.65 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(a*d) - sin(2*c + 2*d*x)/(4*b*d 
) - (3*atan((2*a^3*cos(c/2 + (d*x)/2) + 2*b^3*sin(c/2 + (d*x)/2) - 3*a*b^2 
*cos(c/2 + (d*x)/2))/(2*b^3*cos(c/2 + (d*x)/2) - 2*a^3*sin(c/2 + (d*x)/2) 
+ 3*a*b^2*sin(c/2 + (d*x)/2))))/(b*d) + (a*cos(c + d*x))/(b^2*d) + (2*a^2* 
atan((2*a^3*cos(c/2 + (d*x)/2) + 2*b^3*sin(c/2 + (d*x)/2) - 3*a*b^2*cos(c/ 
2 + (d*x)/2))/(2*b^3*cos(c/2 + (d*x)/2) - 2*a^3*sin(c/2 + (d*x)/2) + 3*a*b 
^2*sin(c/2 + (d*x)/2))))/(b^3*d) + (atan((b^6*sin(c/2 + (d*x)/2)*(b^6 - a^ 
6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)*64i - a^12*sin(c/2 + (d*x)/2)*(b^6 - a^6 
- 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*16i - a^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3 
*a^2*b^4 + 3*a^4*b^2)^(3/2)*16i - 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)*42i + a^3*b^9*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 
 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*66i - 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)*176i + 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)*178i - 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)*81i - 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)*116i + 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)*72i + a^2*b^10*sin(c/2 + (d*x)/2)*(b 
^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*148i - 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)*460i + 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)*577i - a^8*b^4*sin(c/2 + (d...
 

Reduce [B] (verification not implemented)

Time = 3.72 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.36 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}+2 \cos \left (d x +c \right ) a^{2} b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+2 a^{3} c +2 a^{3} d x -3 a \,b^{2} c -3 a \,b^{2} d x}{2 a \,b^{3} d} \] Input:

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

Output:

( - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a 
**2 + 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2)) 
*b**2 - cos(c + d*x)*sin(c + d*x)*a*b**2 + 2*cos(c + d*x)*a**2*b + 2*log(t 
an((c + d*x)/2))*b**3 + 2*a**3*c + 2*a**3*d*x - 3*a*b**2*c - 3*a*b**2*d*x) 
/(2*a*b**3*d)