\(\int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [1302]

Optimal result

Integrand size = 27, antiderivative size = 123 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {x}{b}-\frac {2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b d}+\frac {\left (3 a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 a^3 d}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \] Output:

x/b-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^3 
/b/d+1/2*(3*a^2-2*b^2)*arctanh(cos(d*x+c))/a^3/d+b*cot(d*x+c)/a^2/d-1/2*co 
t(d*x+c)*csc(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 1.94 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.66 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {8 a^3 c+8 a^3 d x-16 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+4 a b^2 \cot \left (\frac {1}{2} (c+d x)\right )-a^2 b \csc ^2\left (\frac {1}{2} (c+d x)\right )+12 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-12 a^2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a^2 b \sec ^2\left (\frac {1}{2} (c+d x)\right )-4 a b^2 \tan \left (\frac {1}{2} (c+d x)\right )}{8 a^3 b d} \] Input:

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

Output:

(8*a^3*c + 8*a^3*d*x - 16*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2] 
)/Sqrt[a^2 - b^2]] + 4*a*b^2*Cot[(c + d*x)/2] - a^2*b*Csc[(c + d*x)/2]^2 + 
 12*a^2*b*Log[Cos[(c + d*x)/2]] - 8*b^3*Log[Cos[(c + d*x)/2]] - 12*a^2*b*L 
og[Sin[(c + d*x)/2]] + 8*b^3*Log[Sin[(c + d*x)/2]] + a^2*b*Sec[(c + d*x)/2 
]^2 - 4*a*b^2*Tan[(c + d*x)/2])/(8*a^3*b*d)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3372, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3372

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3536

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

-1/2*((-2*a^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) - ((3*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]] 
)/(a*d))/a^2 + (b*Cot[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*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 3372

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[b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x] 
)^(n + 2)/(a^2*d^2*f*(n + 1)*(n + 2))), x] - Simp[1/(a^2*d^2*(n + 1)*(n + 2 
))   Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) 
 - b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) 
- b^2*(m + n + 2)*(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] && (LtQ[n, -2] || EqQ[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 = 0.66 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.46

method	result	size
derivativedivides	\(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-6 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\left (-8 a^{4}+16 a^{2} b^{2}-8 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{4 a^{3} b \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\)	\(180\)
default	\(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-6 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\left (-8 a^{4}+16 a^{2} b^{2}-8 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{4 a^{3} b \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\)	\(180\)
risch	\(\frac {x}{b}+\frac {i \left (-i a \,{\mathrm e}^{3 i \left (d x +c \right )}-i a \,{\mathrm e}^{i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{d \,a^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{d \,a^{3}}-\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d b a}+\frac {i \sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,a^{3}}+\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d b a}-\frac {i \sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,a^{3}}\)	\(368\)

Input:

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

Output:

1/d*(1/4/a^2*(1/2*tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c))-1/8/a/tan 
(1/2*d*x+1/2*c)^2+1/4/a^3*(-6*a^2+4*b^2)*ln(tan(1/2*d*x+1/2*c))+1/2/a^2*b/ 
tan(1/2*d*x+1/2*c)+1/4*(-8*a^4+16*a^2*b^2-8*b^4)/a^3/b/(a^2-b^2)^(1/2)*arc 
tan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+2/b*arctan(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. 244 vs. \(2 (114) = 228\).

Time = 0.20 (sec) , antiderivative size = 572, normalized size of antiderivative = 4.65 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {4 \, a^{3} d x \cos \left (d x + c\right )^{2} - 4 \, a^{3} d x - 4 \, 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 ) - 2 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 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 ) - {\left (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{3} b d \cos \left (d x + c\right )^{2} - a^{3} b d\right )}}, \frac {4 \, a^{3} d x \cos \left (d x + c\right )^{2} - 4 \, a^{3} d x - 4 \, 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 ) + 4 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}\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 (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{3} b d \cos \left (d x + c\right )^{2} - a^{3} b d\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(cos(d*x+c)*cot(d*x+c)^3/(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 = 0.15 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.76 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {8 \, {\left (d x + c\right )}}{b} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {4 \, {\left (3 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {16 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3} b} + \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 20.50 (sec) , antiderivative size = 2718, normalized size of antiderivative = 22.10 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*1i)/(a^3*b*d*1i - a^3*b*d* 
cos(2*c + 2*d*x)*1i) - (a^3*atan((2*a^3*cos(c/2 + (d*x)/2) + 2*b^3*sin(c/2 
 + (d*x)/2) - 3*a^2*b*sin(c/2 + (d*x)/2))/(2*a^3*sin(c/2 + (d*x)/2) - 2*b^ 
3*cos(c/2 + (d*x)/2) + 3*a^2*b*cos(c/2 + (d*x)/2)))*2i)/(a^3*b*d*1i - a^3* 
b*d*cos(2*c + 2*d*x)*1i) + (2*atan((32*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 
 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 14*a^12*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3* 
a^2*b^4 + 3*a^4*b^2)^(1/2) - 14*a^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2* 
b^4 + 3*a^4*b^2)^(3/2) - 36*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) + 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) + 8*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) - 24*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) - 82*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) + 63*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) + 2*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) - 11*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) + 56*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) - 106*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) + 72*a^10*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2* 
b^4 + 3*a^4*b^2)^(1/2) + 16*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) + 19*a^5*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b...
 

Reduce [B] (verification not implemented)

Time = 0.88 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.72 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} a^{2}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}-\cos \left (d x +c \right ) a^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2} b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{3}+2 \sin \left (d x +c \right )^{2} a^{3} d x}{2 \sin \left (d x +c \right )^{2} a^{3} b d} \] Input:

int(cos(d*x+c)*cot(d*x+c)^3/(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))*s 
in(c + d*x)**2*a**2 + 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sq 
rt(a**2 - b**2))*sin(c + d*x)**2*b**2 + 2*cos(c + d*x)*sin(c + d*x)*a*b**2 
 - cos(c + d*x)*a**2*b - 3*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2*b + 
2*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**3 + 2*sin(c + d*x)**2*a**3*d*x) 
/(2*sin(c + d*x)**2*a**3*b*d)