\(\int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [1265]

Optimal result

Integrand size = 27, antiderivative size = 399 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {x}{2 b^3}-\frac {3 \left (2 a^2-b^2\right ) x}{b^5}+\frac {\sqrt {a^2-b^2} \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a b^5 d}-\frac {2 \sqrt {a^2-b^2} \left (5 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a b^5 d}+\frac {2 \left (10 a^6-9 a^4 b^2-b^6\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^5 \sqrt {a^2-b^2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {3 a \cos (c+d x)}{b^4 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}+\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{2 b^4 d (a+b \sin (c+d x))}-\frac {\left (a^2-b^2\right ) \left (5 a^2+b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))} \] Output:

-1/2*x/b^3-3*(2*a^2-b^2)*x/b^5+(a^2-b^2)^(1/2)*(2*a^2+b^2)*arctan((b+a*tan 
(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a/b^5/d-2*(a^2-b^2)^(1/2)*(5*a^2+b^2)*ar 
ctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a/b^5/d+2*(10*a^6-9*a^4*b^2 
-b^6)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/b^5/(a^2-b^2)^( 
1/2)/d-arctanh(cos(d*x+c))/a^3/d-3*a*cos(d*x+c)/b^4/d+1/2*cos(d*x+c)*sin(d 
*x+c)/b^3/d+1/2*(a^2-b^2)^2*cos(d*x+c)/a/b^4/d/(a+b*sin(d*x+c))^2+3/2*(a^2 
-b^2)*cos(d*x+c)/b^4/d/(a+b*sin(d*x+c))-(a^2-b^2)*(5*a^2+b^2)*cos(d*x+c)/a 
^2/b^4/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 2.97 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {2 \left (-12 a^2+5 b^2\right ) (c+d x)}{b^5}+\frac {4 \left (12 a^6-11 a^4 b^2+a^2 b^4-2 b^6\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^5 \sqrt {a^2-b^2}}-\frac {12 a \cos (c+d x)}{b^4}-\frac {4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {2 \left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 (a+b \sin (c+d x))^2}+\frac {2 \left (-7 a^4+5 a^2 b^2+2 b^4\right ) \cos (c+d x)}{a^2 b^4 (a+b \sin (c+d x))}+\frac {\sin (2 (c+d x))}{b^3}}{4 d} \] Input:

Integrate[(Cos[c + d*x]^5*Cot[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

((2*(-12*a^2 + 5*b^2)*(c + d*x))/b^5 + (4*(12*a^6 - 11*a^4*b^2 + a^2*b^4 - 
 2*b^6)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^5*Sqrt[a^ 
2 - b^2]) - (12*a*Cos[c + d*x])/b^4 - (4*Log[Cos[(c + d*x)/2]])/a^3 + (4*L 
og[Sin[(c + d*x)/2]])/a^3 + (2*(a^2 - b^2)^2*Cos[c + d*x])/(a*b^4*(a + b*S 
in[c + d*x])^2) + (2*(-7*a^4 + 5*a^2*b^2 + 2*b^4)*Cos[c + d*x])/(a^2*b^4*( 
a + b*Sin[c + d*x])) + Sin[2*(c + d*x)]/b^3)/(4*d)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 3376, 2009}

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.

\(\displaystyle \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^6}{\sin (c+d x) (a+b \sin (c+d x))^3}dx\)

\(\Big \downarrow \) 3376

\(\displaystyle \int \left (\frac {\csc (c+d x)}{a^3}+\frac {\left (a^2-b^2\right )^3}{a b^5 (a+b \sin (c+d x))^3}-\frac {\left (5 a^2+b^2\right ) \left (a^2-b^2\right )^2}{a^2 b^5 (a+b \sin (c+d x))^2}+\frac {3 \left (b^2-2 a^2\right )}{b^5}+\frac {10 a^6-9 a^4 b^2-b^6}{a^3 b^5 (a+b \sin (c+d x))}+\frac {3 a \sin (c+d x)}{b^4}-\frac {\sin ^2(c+d x)}{b^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {\left (2 a^2+b^2\right ) \sqrt {a^2-b^2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a b^5 d}-\frac {2 \left (5 a^2+b^2\right ) \sqrt {a^2-b^2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a b^5 d}-\frac {3 x \left (2 a^2-b^2\right )}{b^5}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}-\frac {\left (5 a^2+b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))}+\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{2 b^4 d (a+b \sin (c+d x))}+\frac {2 \left (10 a^6-9 a^4 b^2-b^6\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 b^5 d \sqrt {a^2-b^2}}-\frac {3 a \cos (c+d x)}{b^4 d}+\frac {\sin (c+d x) \cos (c+d x)}{2 b^3 d}-\frac {x}{2 b^3}\)

Input:

Int[(Cos[c + d*x]^5*Cot[c + d*x])/(a + b*Sin[c + d*x])^3,x]
 

Output:

-1/2*x/b^3 - (3*(2*a^2 - b^2)*x)/b^5 + (Sqrt[a^2 - b^2]*(2*a^2 + b^2)*ArcT 
an[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*b^5*d) - (2*Sqrt[a^2 - b^ 
2]*(5*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*b^5* 
d) + (2*(10*a^6 - 9*a^4*b^2 - b^6)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^ 
2 - b^2]])/(a^3*b^5*Sqrt[a^2 - b^2]*d) - ArcTanh[Cos[c + d*x]]/(a^3*d) - ( 
3*a*Cos[c + d*x])/(b^4*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*b^3*d) + ((a^2 
- b^2)^2*Cos[c + d*x])/(2*a*b^4*d*(a + b*Sin[c + d*x])^2) + (3*(a^2 - b^2) 
*Cos[c + d*x])/(2*b^4*d*(a + b*Sin[c + d*x])) - ((a^2 - b^2)*(5*a^2 + b^2) 
*Cos[c + d*x])/(a^2*b^4*d*(a + b*Sin[c + d*x]))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3376

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) 
 + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ 
e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr 
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( 
LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
 

Maple [A] (verified)

Time = 15.85 (sec) , antiderivative size = 359, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+3 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+3 a b}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (12 a^{2}-5 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{5}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {\frac {2 \left (\left (-\frac {5}{2} a^{5} b^{2}+\frac {1}{2} a^{3} b^{4}+2 a \,b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {3 b \left (2 a^{6}+3 a^{4} b^{2}-3 a^{2} b^{4}-2 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {a \,b^{2} \left (19 a^{4}-11 a^{2} b^{2}-8 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {3 a^{2} b \left (2 a^{4}-a^{2} b^{2}-b^{4}\right )}{2}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {\left (12 a^{6}-11 a^{4} b^{2}+a^{2} b^{4}-2 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{3} b^{5}}}{d}\)	\(359\)
default	\(\frac {-\frac {2 \left (\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+3 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+3 a b}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (12 a^{2}-5 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{5}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {\frac {2 \left (\left (-\frac {5}{2} a^{5} b^{2}+\frac {1}{2} a^{3} b^{4}+2 a \,b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {3 b \left (2 a^{6}+3 a^{4} b^{2}-3 a^{2} b^{4}-2 b^{6}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {a \,b^{2} \left (19 a^{4}-11 a^{2} b^{2}-8 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {3 a^{2} b \left (2 a^{4}-a^{2} b^{2}-b^{4}\right )}{2}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {\left (12 a^{6}-11 a^{4} b^{2}+a^{2} b^{4}-2 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{3} b^{5}}}{d}\)	\(359\)
risch	\(-\frac {6 x \,a^{2}}{b^{5}}+\frac {5 x}{2 b^{3}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{3} d}-\frac {3 a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{4} d}-\frac {3 a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{4} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{3} d}+\frac {i \left (-8 i a^{5} b \,{\mathrm e}^{3 i \left (d x +c \right )}+7 i a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+i a \,b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+20 i a^{5} b \,{\mathrm e}^{i \left (d x +c \right )}-13 i a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}-7 i a \,b^{5} {\mathrm e}^{i \left (d x +c \right )}+14 a^{6} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{6} {\mathrm e}^{2 i \left (d x +c \right )}-7 a^{4} b^{2}+5 a^{2} b^{4}+2 b^{6}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} a^{2} d \,b^{5}}+\frac {6 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^{5}}+\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 )}{2 d \,b^{3} a}+\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^{3}}-\frac {6 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^{5}}-\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 )}{2 d \,b^{3} a}-\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^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\)	\(676\)

Input:

int(cos(d*x+c)^5*cot(d*x+c)/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2/b^5*((1/2*tan(1/2*d*x+1/2*c)^3*b^2+3*a*b*tan(1/2*d*x+1/2*c)^2-1/2* 
b^2*tan(1/2*d*x+1/2*c)+3*a*b)/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(12*a^2-5*b^2 
)*arctan(tan(1/2*d*x+1/2*c)))+1/a^3*ln(tan(1/2*d*x+1/2*c))+2/a^3/b^5*(((-5 
/2*a^5*b^2+1/2*a^3*b^4+2*a*b^6)*tan(1/2*d*x+1/2*c)^3-3/2*b*(2*a^6+3*a^4*b^ 
2-3*a^2*b^4-2*b^6)*tan(1/2*d*x+1/2*c)^2-1/2*a*b^2*(19*a^4-11*a^2*b^2-8*b^4 
)*tan(1/2*d*x+1/2*c)-3/2*a^2*b*(2*a^4-a^2*b^2-b^4))/(tan(1/2*d*x+1/2*c)^2* 
a+2*b*tan(1/2*d*x+1/2*c)+a)^2+1/2*(12*a^6-11*a^4*b^2+a^2*b^4-2*b^6)/(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 = 0.35 (sec) , antiderivative size = 1007, normalized size of antiderivative = 2.52 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas" 
)
 

Output:

[-1/4*(8*a^4*b^3*cos(d*x + c)^3 + 2*(12*a^5*b^2 - 5*a^3*b^4)*d*x*cos(d*x + 
 c)^2 - 2*(12*a^7 + 7*a^5*b^2 - 5*a^3*b^4)*d*x + (12*a^6 + 13*a^4*b^2 + 3* 
a^2*b^4 + 2*b^6 - (12*a^4*b^2 + a^2*b^4 + 2*b^6)*cos(d*x + c)^2 + 2*(12*a^ 
5*b + a^3*b^3 + 2*a*b^5)*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*s 
in(d*x + c) - a^2 - b^2)) - 2*(12*a^6*b + a^4*b^3 - 3*a^2*b^5)*cos(d*x + c 
) + 2*(b^7*cos(d*x + c)^2 - 2*a*b^6*sin(d*x + c) - a^2*b^5 - b^7)*log(1/2* 
cos(d*x + c) + 1/2) - 2*(b^7*cos(d*x + c)^2 - 2*a*b^6*sin(d*x + c) - a^2*b 
^5 - b^7)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^3*b^4*cos(d*x + c)^3 + 2*(12 
*a^6*b - 5*a^4*b^3)*d*x + 2*(9*a^5*b^2 - 3*a^3*b^4 - a*b^6)*cos(d*x + c))* 
sin(d*x + c))/(a^3*b^7*d*cos(d*x + c)^2 - 2*a^4*b^6*d*sin(d*x + c) - (a^5* 
b^5 + a^3*b^7)*d), -1/2*(4*a^4*b^3*cos(d*x + c)^3 + (12*a^5*b^2 - 5*a^3*b^ 
4)*d*x*cos(d*x + c)^2 - (12*a^7 + 7*a^5*b^2 - 5*a^3*b^4)*d*x - (12*a^6 + 1 
3*a^4*b^2 + 3*a^2*b^4 + 2*b^6 - (12*a^4*b^2 + a^2*b^4 + 2*b^6)*cos(d*x + c 
)^2 + 2*(12*a^5*b + a^3*b^3 + 2*a*b^5)*sin(d*x + c))*sqrt(a^2 - b^2)*arcta 
n(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - (12*a^6*b + a^4* 
b^3 - 3*a^2*b^5)*cos(d*x + c) + (b^7*cos(d*x + c)^2 - 2*a*b^6*sin(d*x + c) 
 - a^2*b^5 - b^7)*log(1/2*cos(d*x + c) + 1/2) - (b^7*cos(d*x + c)^2 - 2*a* 
b^6*sin(d*x + c) - a^2*b^5 - b^7)*log(-1/2*cos(d*x + c) + 1/2) - (a^3*b...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*cot(d*x+c)/(a+b*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)/(a+b*sin(d*x+c))^3,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 = 0.27 (sec) , antiderivative size = 635, normalized size of antiderivative = 1.59 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^5*cot(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

1/2*(2*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (12*a^2 - 5*b^2)*(d*x + c)/b^5 
 + 2*(12*a^6 - 11*a^4*b^2 + a^2*b^4 - 2*b^6)*(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^5) - 2*(6*a^5*b*tan(1/2*d*x + 1/2*c)^7 - a^3*b^3*tan(1/2* 
d*x + 1/2*c)^7 - 4*a*b^5*tan(1/2*d*x + 1/2*c)^7 + 12*a^6*tan(1/2*d*x + 1/2 
*c)^6 + 13*a^4*b^2*tan(1/2*d*x + 1/2*c)^6 - 9*a^2*b^4*tan(1/2*d*x + 1/2*c) 
^6 - 6*b^6*tan(1/2*d*x + 1/2*c)^6 + 54*a^5*b*tan(1/2*d*x + 1/2*c)^5 - 9*a^ 
3*b^3*tan(1/2*d*x + 1/2*c)^5 - 16*a*b^5*tan(1/2*d*x + 1/2*c)^5 + 36*a^6*ta 
n(1/2*d*x + 1/2*c)^4 + 39*a^4*b^2*tan(1/2*d*x + 1/2*c)^4 - 21*a^2*b^4*tan( 
1/2*d*x + 1/2*c)^4 - 12*b^6*tan(1/2*d*x + 1/2*c)^4 + 90*a^5*b*tan(1/2*d*x 
+ 1/2*c)^3 - 27*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 20*a*b^5*tan(1/2*d*x + 1/ 
2*c)^3 + 36*a^6*tan(1/2*d*x + 1/2*c)^2 + 23*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 
 - 15*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 - 6*b^6*tan(1/2*d*x + 1/2*c)^2 + 42*a 
^5*b*tan(1/2*d*x + 1/2*c) - 11*a^3*b^3*tan(1/2*d*x + 1/2*c) - 8*a*b^5*tan( 
1/2*d*x + 1/2*c) + 12*a^6 - 3*a^4*b^2 - 3*a^2*b^4)/((a*tan(1/2*d*x + 1/2*c 
)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/ 
2*d*x + 1/2*c) + a)^2*a^3*b^4))/d
 

Mupad [B] (verification not implemented)

Time = 22.66 (sec) , antiderivative size = 5354, normalized size of antiderivative = 13.42 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^5*cot(c + d*x))/(a + b*sin(c + d*x))^3,x)
 

Output:

log(tan(c/2 + (d*x)/2))/(a^3*d) + ((3*(b^4 - 4*a^4 + a^2*b^2))/(a*b^4) + ( 
tan(c/2 + (d*x)/2)*(8*b^4 - 42*a^4 + 11*a^2*b^2))/(a^2*b^3) - (3*tan(c/2 + 
 (d*x)/2)^4*(12*a^6 - 4*b^6 - 7*a^2*b^4 + 13*a^4*b^2))/(a^3*b^4) - (tan(c/ 
2 + (d*x)/2)^6*(12*a^6 - 6*b^6 - 9*a^2*b^4 + 13*a^4*b^2))/(a^3*b^4) - (tan 
(c/2 + (d*x)/2)^2*(36*a^6 - 6*b^6 - 15*a^2*b^4 + 23*a^4*b^2))/(a^3*b^4) + 
(tan(c/2 + (d*x)/2)^7*(4*b^4 - 6*a^4 + a^2*b^2))/(a^2*b^3) + (tan(c/2 + (d 
*x)/2)^5*(16*b^4 - 54*a^4 + 9*a^2*b^2))/(a^2*b^3) + (tan(c/2 + (d*x)/2)^3* 
(20*b^4 - 90*a^4 + 27*a^2*b^2))/(a^2*b^3))/(d*(tan(c/2 + (d*x)/2)^2*(4*a^2 
 + 4*b^2) + tan(c/2 + (d*x)/2)^6*(4*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^4*(6 
*a^2 + 8*b^2) + a^2*tan(c/2 + (d*x)/2)^8 + a^2 + 12*a*b*tan(c/2 + (d*x)/2) 
^3 + 12*a*b*tan(c/2 + (d*x)/2)^5 + 4*a*b*tan(c/2 + (d*x)/2)^7 + 4*a*b*tan( 
c/2 + (d*x)/2))) - (atan((((a^2*12i - b^2*5i)*((4*(20*a^3*b^12 - 6048*a^15 
 + 332*a^5*b^10 - 1947*a^7*b^8 + 3979*a^9*b^6 - 7038*a^11*b^4 + 10800*a^13 
*b^2))/(a^6*b^11) - ((a^2*12i - b^2*5i)*((4*(32*a^2*b^16 - 24*a^4*b^14 + 1 
60*a^6*b^12 + 32*a^8*b^10 - 1110*a^10*b^8 + 1872*a^12*b^6 - 864*a^14*b^4)) 
/(a^6*b^11) - ((a^2*12i - b^2*5i)*((4*(64*a^5*b^16 - 48*a^7*b^14 + 160*a^9 
*b^12 - 168*a^11*b^10))/(a^6*b^11) - (((4*(32*a^8*b^16 - 24*a^10*b^14))/(a 
^6*b^11) + (8*tan(c/2 + (d*x)/2)*(64*a^7*b^22 - 68*a^9*b^20 + 8*a^11*b^18) 
)/(a^6*b^16))*(a^2*12i - b^2*5i))/(2*b^5) + (8*tan(c/2 + (d*x)/2)*(64*a^4* 
b^22 - 68*a^6*b^20 + 192*a^8*b^18 - 280*a^10*b^16 + 96*a^12*b^14))/(a^6...
 

Reduce [B] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 872, normalized size of antiderivative = 2.19 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^5*cot(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Output:

(48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin 
(c + d*x)**2*a**4*b**2 + 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b) 
/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**2*b**4 + 8*sqrt(a**2 - b**2)*atan(( 
tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*b**6 + 96*sqrt( 
a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x) 
*a**5*b + 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b* 
*2))*sin(c + d*x)*a**3*b**3 + 16*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)* 
a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a*b**5 + 48*sqrt(a**2 - b**2)*atan( 
(tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**6 + 4*sqrt(a**2 - b**2)*ata 
n((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**4*b**2 + 8*sqrt(a**2 - b* 
*2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**2*b**4 + 2*cos(c + 
 d*x)*sin(c + d*x)**3*a**3*b**4 - 8*cos(c + d*x)*sin(c + d*x)**2*a**4*b**3 
 - 36*cos(c + d*x)*sin(c + d*x)*a**5*b**2 + 10*cos(c + d*x)*sin(c + d*x)*a 
**3*b**4 + 4*cos(c + d*x)*sin(c + d*x)*a*b**6 - 24*cos(c + d*x)*a**6*b + 6 
*cos(c + d*x)*a**4*b**3 + 6*cos(c + d*x)*a**2*b**5 + 4*log(tan((c + d*x)/2 
))*sin(c + d*x)**2*b**7 + 8*log(tan((c + d*x)/2))*sin(c + d*x)*a*b**6 + 4* 
log(tan((c + d*x)/2))*a**2*b**5 - 24*sin(c + d*x)**2*a**5*b**2*d*x - 18*si 
n(c + d*x)**2*a**4*b**3 + 10*sin(c + d*x)**2*a**3*b**4*d*x + 5*sin(c + d*x 
)**2*a**2*b**5 + 2*sin(c + d*x)**2*b**7 - 48*sin(c + d*x)*a**6*b*d*x - 36* 
sin(c + d*x)*a**5*b**2 + 20*sin(c + d*x)*a**4*b**3*d*x + 10*sin(c + d*x...