\(\int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1258]

Optimal result

Integrand size = 29, antiderivative size = 287 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {x}{b^2}-\frac {2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^2 d}+\frac {4 \left (a^6-3 a^2 b^4+2 b^6\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^5 b^2 \sqrt {a^2-b^2} d}+\frac {b \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {2 b \left (3 a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{a^5 d}-\frac {\cot (c+d x)}{a^2 d}+\frac {3 \left (a^2-b^2\right ) \cot (c+d x)}{a^4 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {b \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a^4 b d (a+b \sin (c+d x))} \] Output:

-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^3/b^2/d+4*(a^6-3*a^2*b^4+2*b^6)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2 
)^(1/2))/a^5/b^2/(a^2-b^2)^(1/2)/d+b*arctanh(cos(d*x+c))/a^3/d-2*b*(3*a^2- 
2*b^2)*arctanh(cos(d*x+c))/a^5/d-cot(d*x+c)/a^2/d+3*(a^2-b^2)*cot(d*x+c)/a 
^4/d-1/3*cot(d*x+c)^3/a^2/d+b*cot(d*x+c)*csc(d*x+c)/a^3/d-(a^2-b^2)^2*cos( 
d*x+c)/a^4/b/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 6.83 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.49 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {c+d x}{b^2 d}+\frac {2 \left (a^2-b^2\right )^{3/2} \left (a^2+4 b^2\right ) \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^5 b^2 d}+\frac {\left (7 a^2 \cos \left (\frac {1}{2} (c+d x)\right )-9 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right ) \csc \left (\frac {1}{2} (c+d x)\right )}{6 a^4 d}+\frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{4 a^3 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 a^2 d}+\frac {\left (-5 a^2 b+4 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {\left (5 a^2 b-4 b^3\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}-\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{4 a^3 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (-7 a^2 \sin \left (\frac {1}{2} (c+d x)\right )+9 b^2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^4 d}+\frac {-a^4 \cos (c+d x)+2 a^2 b^2 \cos (c+d x)-b^4 \cos (c+d x)}{a^4 b d (a+b \sin (c+d x))}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{24 a^2 d} \] Input:

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + b*Sin[c + d*x])^2,x]
 

Output:

-((c + d*x)/(b^2*d)) + (2*(a^2 - b^2)^(3/2)*(a^2 + 4*b^2)*ArcTan[(Sec[(c + 
 d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(a^5 
*b^2*d) + ((7*a^2*Cos[(c + d*x)/2] - 9*b^2*Cos[(c + d*x)/2])*Csc[(c + d*x) 
/2])/(6*a^4*d) + (b*Csc[(c + d*x)/2]^2)/(4*a^3*d) - (Cot[(c + d*x)/2]*Csc[ 
(c + d*x)/2]^2)/(24*a^2*d) + ((-5*a^2*b + 4*b^3)*Log[Cos[(c + d*x)/2]])/(a 
^5*d) + ((5*a^2*b - 4*b^3)*Log[Sin[(c + d*x)/2]])/(a^5*d) - (b*Sec[(c + d* 
x)/2]^2)/(4*a^3*d) + (Sec[(c + d*x)/2]*(-7*a^2*Sin[(c + d*x)/2] + 9*b^2*Si 
n[(c + d*x)/2]))/(6*a^4*d) + (-(a^4*Cos[c + d*x]) + 2*a^2*b^2*Cos[c + d*x] 
 - b^4*Cos[c + d*x])/(a^4*b*d*(a + b*Sin[c + d*x])) + (Sec[(c + d*x)/2]^2* 
Tan[(c + d*x)/2])/(24*a^2*d)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^6}{\sin (c+d x)^4 (a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3376

\(\displaystyle \int \left (-\frac {2 b \csc ^3(c+d x)}{a^3}+\frac {\csc ^4(c+d x)}{a^2}+\frac {2 \left (3 a^2 b-2 b^3\right ) \csc (c+d x)}{a^5}-\frac {\left (a^2-b^2\right )^3}{a^4 b^2 (a+b \sin (c+d x))^2}-\frac {3 \left (a^2-b^2\right ) \csc ^2(c+d x)}{a^4}+\frac {2 \left (a^6-3 a^2 b^4+2 b^6\right )}{a^5 b^2 (a+b \sin (c+d x))}-\frac {1}{b^2}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + b*Sin[c + d*x])^2,x]
 

Output:

-(x/b^2) - (2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - 
 b^2]])/(a^3*b^2*d) + (4*(a^6 - 3*a^2*b^4 + 2*b^6)*ArcTan[(b + a*Tan[(c + 
d*x)/2])/Sqrt[a^2 - b^2]])/(a^5*b^2*Sqrt[a^2 - b^2]*d) + (b*ArcTanh[Cos[c 
+ d*x]])/(a^3*d) - (2*b*(3*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(a^5*d) - C 
ot[c + d*x]/(a^2*d) + (3*(a^2 - b^2)*Cot[c + d*x])/(a^4*d) - Cot[c + d*x]^ 
3/(3*a^2*d) + (b*Cot[c + d*x]*Csc[c + d*x])/(a^3*d) - ((a^2 - b^2)^2*Cos[c 
 + d*x])/(a^4*b*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 = 1.59 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}+12 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{4}}-\frac {1}{24 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-9 a^{2}+12 b^{2}}{8 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{4 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \left (5 a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{5}}+\frac {\frac {2 \left (-b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-b a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (a^{6}+2 a^{4} b^{2}-7 a^{2} b^{4}+4 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{5} b^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(327\)
default	\(\frac {\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-2 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}+12 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{4}}-\frac {1}{24 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-9 a^{2}+12 b^{2}}{8 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{4 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \left (5 a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{5}}+\frac {\frac {2 \left (-b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-b a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (a^{6}+2 a^{4} b^{2}-7 a^{2} b^{4}+4 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{5} b^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(327\)
risch	\(-\frac {x}{b^{2}}+\frac {2 i \left (-43 i a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-9 i a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}+12 i b^{5} {\mathrm e}^{6 i \left (d x +c \right )}+3 a^{5} {\mathrm e}^{7 i \left (d x +c \right )}-6 a^{3} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+6 a \,b^{4} {\mathrm e}^{7 i \left (d x +c \right )}-3 i a^{4} b +39 i b^{3} a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+13 i a^{2} b^{3}-9 a^{5} {\mathrm e}^{5 i \left (d x +c \right )}+36 a^{3} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-30 a \,b^{4} {\mathrm e}^{5 i \left (d x +c \right )}+3 i a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}-36 i b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+9 i a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a^{5} {\mathrm e}^{3 i \left (d x +c \right )}-42 a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+42 a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}-9 i a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-12 i b^{5}+36 i b^{5} {\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} a^{5}+20 a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}-18 a \,b^{4} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} b^{2} \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 ) d \,a^{4}}+\frac {5 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}-\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{5}}-\frac {5 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}+\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{5}}+\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} a}+\frac {3 \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^{3}}-\frac {4 \sqrt {-a^{2}+b^{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^{5}}-\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} a}-\frac {3 \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^{3}}+\frac {4 \sqrt {-a^{2}+b^{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^{5}}\)	\(824\)

Input:

int(cos(d*x+c)^2*cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/8/a^4*(1/3*a^2*tan(1/2*d*x+1/2*c)^3-2*a*b*tan(1/2*d*x+1/2*c)^2-9*ta 
n(1/2*d*x+1/2*c)*a^2+12*b^2*tan(1/2*d*x+1/2*c))-1/24/a^2/tan(1/2*d*x+1/2*c 
)^3-1/8*(-9*a^2+12*b^2)/a^4/tan(1/2*d*x+1/2*c)+1/4/a^3*b/tan(1/2*d*x+1/2*c 
)^2+1/a^5*b*(5*a^2-4*b^2)*ln(tan(1/2*d*x+1/2*c))+2/a^5/b^2*((-b^2*(a^4-2*a 
^2*b^2+b^4)*tan(1/2*d*x+1/2*c)-b*a*(a^4-2*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)+(a^6+2*a^4*b^2-7*a^2*b^4+4*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)))-2/b^2*arct 
an(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. 677 vs. \(2 (275) = 550\).

Time = 0.37 (sec) , antiderivative size = 1437, normalized size of antiderivative = 5.01 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

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

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(cos(d*x+c)**2*cot(d*x+c)**4/(a+b*sin(d*x+c))**2,x)
 

Output:

Integral(cos(c + d*x)**2*cot(c + d*x)**4/(a + b*sin(c + d*x))**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^2*cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

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 = 0.23 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.39 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {24 \, {\left (d x + c\right )}}{b^{2}} - \frac {24 \, {\left (5 \, a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{5}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}} - \frac {48 \, {\left (a^{6} + 2 \, a^{4} b^{2} - 7 \, a^{2} b^{4} + 4 \, b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{5} b^{2}} + \frac {48 \, {\left (a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a^{5} b} + \frac {220 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 176 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \] Input:

integrate(cos(d*x+c)^2*cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

-1/24*(24*(d*x + c)/b^2 - 24*(5*a^2*b - 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c 
)))/a^5 - (a^4*tan(1/2*d*x + 1/2*c)^3 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^2 - 2 
7*a^4*tan(1/2*d*x + 1/2*c) + 36*a^2*b^2*tan(1/2*d*x + 1/2*c))/a^6 - 48*(a^ 
6 + 2*a^4*b^2 - 7*a^2*b^4 + 4*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^5*b^2) + 48*(a^4*b*tan(1/2*d*x + 1/2*c) - 2*a^2*b^3*tan(1/2*d*x + 1/2*c 
) + b^5*tan(1/2*d*x + 1/2*c) + a^5 - 2*a^3*b^2 + a*b^4)/((a*tan(1/2*d*x + 
1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*a^5*b) + (220*a^2*b*tan(1/2*d*x + 
 1/2*c)^3 - 176*b^3*tan(1/2*d*x + 1/2*c)^3 - 27*a^3*tan(1/2*d*x + 1/2*c)^2 
 + 36*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/( 
a^5*tan(1/2*d*x + 1/2*c)^3))/d
 

Mupad [B] (verification not implemented)

Time = 24.09 (sec) , antiderivative size = 4740, normalized size of antiderivative = 16.52 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^2*cot(c + d*x)^4)/(a + b*sin(c + d*x))^2,x)
 

Output:

tan(c/2 + (d*x)/2)^3/(24*a^2*d) - (tan(c/2 + (d*x)/2)*((24*a^2 + 32*b^2)/( 
64*a^4) + 3/(4*a^2) - (2*b^2)/a^4))/d - (tan(c/2 + (d*x)/2)^2*(8*a*b^2 - ( 
26*a^3)/3) + a^3/3 - (4*a^2*b*tan(c/2 + (d*x)/2))/3 + (tan(c/2 + (d*x)/2)^ 
4*(7*a^4 + 16*b^4 - 20*a^2*b^2))/a + (4*tan(c/2 + (d*x)/2)^3*(4*a^4 + 10*b 
^4 - 13*a^2*b^2))/b)/(d*(8*a^5*tan(c/2 + (d*x)/2)^3 + 8*a^5*tan(c/2 + (d*x 
)/2)^5 + 16*a^4*b*tan(c/2 + (d*x)/2)^4)) + (2*atan((((((((((32*(4*a^14*b^7 
 - 3*a^16*b^5))/(a^12*b^2) + (32*tan(c/2 + (d*x)/2)*(16*a^13*b^10 - 17*a^1 
5*b^8 + 2*a^17*b^6))/(a^12*b^4))*1i)/b^2 - (32*(32*a^9*b^10 - 64*a^11*b^8 
+ 30*a^13*b^6 + a^15*b^4))/(a^12*b^2) + (32*tan(c/2 + (d*x)/2)*(148*a^10*b 
^11 - 64*a^8*b^13 - 97*a^12*b^9 + 10*a^14*b^7 + 4*a^16*b^5))/(a^12*b^4))*1 
i)/b^2 - (32*(3*a^16*b - 64*a^4*b^13 + 208*a^6*b^11 - 220*a^8*b^9 + 71*a^1 
0*b^7 + 5*a^12*b^5 - 4*a^14*b^3))/(a^12*b^2) + (32*tan(c/2 + (d*x)/2)*(16* 
a^5*b^14 - 40*a^7*b^12 + 5*a^9*b^10 + 44*a^11*b^8 - 6*a^13*b^6 - 20*a^15*b 
^4 + 2*a^17*b^2))/(a^12*b^4))*1i)/b^2 - (32*(6*a^15 + 16*a^5*b^10 - 56*a^7 
*b^8 + 97*a^9*b^6 - 84*a^11*b^4 + 20*a^13*b^2))/(a^12*b^2) + (32*tan(c/2 + 
 (d*x)/2)*(4*a^16*b - 64*b^17 + 304*a^2*b^15 - 540*a^4*b^13 + 405*a^6*b^11 
 - 124*a^8*b^9 + 78*a^10*b^7 - 72*a^12*b^5 + 10*a^14*b^3))/(a^12*b^4))/b^2 
 - ((((((((32*(4*a^14*b^7 - 3*a^16*b^5))/(a^12*b^2) + (32*tan(c/2 + (d*x)/ 
2)*(16*a^13*b^10 - 17*a^15*b^8 + 2*a^17*b^6))/(a^12*b^4))*1i)/b^2 + (32*(3 
2*a^9*b^10 - 64*a^11*b^8 + 30*a^13*b^6 + a^15*b^4))/(a^12*b^2) - (32*ta...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 610, normalized size of antiderivative = 2.13 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^2*cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x)
 

Output:

(6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( 
c + d*x)**4*a**4*b + 18*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sq 
rt(a**2 - b**2))*sin(c + d*x)**4*a**2*b**3 - 24*sqrt(a**2 - b**2)*atan((ta 
n((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**4*b**5 + 6*sqrt(a** 
2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**3 
*a**5 + 18*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b** 
2))*sin(c + d*x)**3*a**3*b**2 - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2 
)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**3*a*b**4 - 3*cos(c + d*x)*sin(c 
+ d*x)**3*a**5*b + 13*cos(c + d*x)*sin(c + d*x)**3*a**3*b**3 - 12*cos(c + 
d*x)*sin(c + d*x)**3*a*b**5 + 7*cos(c + d*x)*sin(c + d*x)**2*a**4*b**2 - 6 
*cos(c + d*x)*sin(c + d*x)**2*a**2*b**4 + 2*cos(c + d*x)*sin(c + d*x)*a**3 
*b**3 - cos(c + d*x)*a**4*b**2 + 15*log(tan((c + d*x)/2))*sin(c + d*x)**4* 
a**2*b**4 - 12*log(tan((c + d*x)/2))*sin(c + d*x)**4*b**6 + 15*log(tan((c 
+ d*x)/2))*sin(c + d*x)**3*a**3*b**3 - 12*log(tan((c + d*x)/2))*sin(c + d* 
x)**3*a*b**5 - 3*sin(c + d*x)**4*a**5*b*d*x - 3*sin(c + d*x)**3*a**6*d*x)/ 
(3*sin(c + d*x)**3*a**5*b**2*d*(sin(c + d*x)*b + a))