\(\int \cot ^{12}(c+d x) (a+a \sec (c+d x))^3 \, dx\) [55]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 213 \[ \int \cot ^{12}(c+d x) (a+a \sec (c+d x))^3 \, dx=a^3 x+\frac {a^3 \cot (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}+\frac {a^3 \cot ^5(c+d x)}{5 d}-\frac {a^3 \cot ^7(c+d x)}{7 d}+\frac {a^3 \cot ^9(c+d x)}{9 d}-\frac {4 a^3 \cot ^{11}(c+d x)}{11 d}+\frac {3 a^3 \csc (c+d x)}{d}-\frac {16 a^3 \csc ^3(c+d x)}{3 d}+\frac {34 a^3 \csc ^5(c+d x)}{5 d}-\frac {36 a^3 \csc ^7(c+d x)}{7 d}+\frac {19 a^3 \csc ^9(c+d x)}{9 d}-\frac {4 a^3 \csc ^{11}(c+d x)}{11 d} \] Output:

a^3*x+a^3*cot(d*x+c)/d-1/3*a^3*cot(d*x+c)^3/d+1/5*a^3*cot(d*x+c)^5/d-1/7*a 
^3*cot(d*x+c)^7/d+1/9*a^3*cot(d*x+c)^9/d-4/11*a^3*cot(d*x+c)^11/d+3*a^3*cs 
c(d*x+c)/d-16/3*a^3*csc(d*x+c)^3/d+34/5*a^3*csc(d*x+c)^5/d-36/7*a^3*csc(d* 
x+c)^7/d+19/9*a^3*csc(d*x+c)^9/d-4/11*a^3*csc(d*x+c)^11/d
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.75 \[ \int \cot ^{12}(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {3 a^3 \cot ^{11}(c+d x)}{11 d}+\frac {3 a^3 \csc (c+d x)}{d}-\frac {16 a^3 \csc ^3(c+d x)}{3 d}+\frac {34 a^3 \csc ^5(c+d x)}{5 d}-\frac {36 a^3 \csc ^7(c+d x)}{7 d}+\frac {19 a^3 \csc ^9(c+d x)}{9 d}-\frac {4 a^3 \csc ^{11}(c+d x)}{11 d}-\frac {a^3 \cot ^{11}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {11}{2},1,-\frac {9}{2},-\tan ^2(c+d x)\right )}{11 d} \] Input:

Integrate[Cot[c + d*x]^12*(a + a*Sec[c + d*x])^3,x]
 

Output:

(-3*a^3*Cot[c + d*x]^11)/(11*d) + (3*a^3*Csc[c + d*x])/d - (16*a^3*Csc[c + 
 d*x]^3)/(3*d) + (34*a^3*Csc[c + d*x]^5)/(5*d) - (36*a^3*Csc[c + d*x]^7)/( 
7*d) + (19*a^3*Csc[c + d*x]^9)/(9*d) - (4*a^3*Csc[c + d*x]^11)/(11*d) - (a 
^3*Cot[c + d*x]^11*Hypergeometric2F1[-11/2, 1, -9/2, -Tan[c + d*x]^2])/(11 
*d)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4374, 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.

\(\displaystyle \int \cot ^{12}(c+d x) (a \sec (c+d x)+a)^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}{\cot \left (c+d x+\frac {\pi }{2}\right )^{12}}dx\)

\(\Big \downarrow \) 4374

\(\displaystyle \int \left (a^3 \cot ^{12}(c+d x)+3 a^3 \cot ^{11}(c+d x) \csc (c+d x)+3 a^3 \cot ^{10}(c+d x) \csc ^2(c+d x)+a^3 \cot ^9(c+d x) \csc ^3(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 a^3 \cot ^{11}(c+d x)}{11 d}+\frac {a^3 \cot ^9(c+d x)}{9 d}-\frac {a^3 \cot ^7(c+d x)}{7 d}+\frac {a^3 \cot ^5(c+d x)}{5 d}-\frac {a^3 \cot ^3(c+d x)}{3 d}+\frac {a^3 \cot (c+d x)}{d}-\frac {4 a^3 \csc ^{11}(c+d x)}{11 d}+\frac {19 a^3 \csc ^9(c+d x)}{9 d}-\frac {36 a^3 \csc ^7(c+d x)}{7 d}+\frac {34 a^3 \csc ^5(c+d x)}{5 d}-\frac {16 a^3 \csc ^3(c+d x)}{3 d}+\frac {3 a^3 \csc (c+d x)}{d}+a^3 x\)

Input:

Int[Cot[c + d*x]^12*(a + a*Sec[c + d*x])^3,x]
 

Output:

a^3*x + (a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*x]^3)/(3*d) + (a^3*Cot[c + d 
*x]^5)/(5*d) - (a^3*Cot[c + d*x]^7)/(7*d) + (a^3*Cot[c + d*x]^9)/(9*d) - ( 
4*a^3*Cot[c + d*x]^11)/(11*d) + (3*a^3*Csc[c + d*x])/d - (16*a^3*Csc[c + d 
*x]^3)/(3*d) + (34*a^3*Csc[c + d*x]^5)/(5*d) - (36*a^3*Csc[c + d*x]^7)/(7* 
d) + (19*a^3*Csc[c + d*x]^9)/(9*d) - (4*a^3*Csc[c + d*x]^11)/(11*d)
 

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 4374
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ 
c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 23.78 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.98

method result size
risch \(a^{3} x +\frac {2 i a^{3} \left (10395 \,{\mathrm e}^{15 i \left (d x +c \right )}-31185 \,{\mathrm e}^{14 i \left (d x +c \right )}+1155 \,{\mathrm e}^{13 i \left (d x +c \right )}+148995 \,{\mathrm e}^{12 i \left (d x +c \right )}-190113 \,{\mathrm e}^{11 i \left (d x +c \right )}-117117 \,{\mathrm e}^{10 i \left (d x +c \right )}+434775 \,{\mathrm e}^{9 i \left (d x +c \right )}-138105 \,{\mathrm e}^{8 i \left (d x +c \right )}-385055 \,{\mathrm e}^{7 i \left (d x +c \right )}+374781 \,{\mathrm e}^{6 i \left (d x +c \right )}+63289 \,{\mathrm e}^{5 i \left (d x +c \right )}-223655 \,{\mathrm e}^{4 i \left (d x +c \right )}+75685 \,{\mathrm e}^{3 i \left (d x +c \right )}+43345 \,{\mathrm e}^{2 i \left (d x +c \right )}-34323 \,{\mathrm e}^{i \left (d x +c \right )}+7453\right )}{3465 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{11} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) \(209\)
derivativedivides \(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{10}}{11 \sin \left (d x +c \right )^{11}}-\frac {\cos \left (d x +c \right )^{10}}{99 \sin \left (d x +c \right )^{9}}+\frac {\cos \left (d x +c \right )^{10}}{693 \sin \left (d x +c \right )^{7}}-\frac {\cos \left (d x +c \right )^{10}}{1155 \sin \left (d x +c \right )^{5}}+\frac {\cos \left (d x +c \right )^{10}}{693 \sin \left (d x +c \right )^{3}}-\frac {\cos \left (d x +c \right )^{10}}{99 \sin \left (d x +c \right )}-\frac {\left (\frac {128}{35}+\cos \left (d x +c \right )^{8}+\frac {8 \cos \left (d x +c \right )^{6}}{7}+\frac {48 \cos \left (d x +c \right )^{4}}{35}+\frac {64 \cos \left (d x +c \right )^{2}}{35}\right ) \sin \left (d x +c \right )}{99}\right )-\frac {3 a^{3} \cos \left (d x +c \right )^{11}}{11 \sin \left (d x +c \right )^{11}}+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{12}}{11 \sin \left (d x +c \right )^{11}}+\frac {\cos \left (d x +c \right )^{12}}{99 \sin \left (d x +c \right )^{9}}-\frac {\cos \left (d x +c \right )^{12}}{231 \sin \left (d x +c \right )^{7}}+\frac {\cos \left (d x +c \right )^{12}}{231 \sin \left (d x +c \right )^{5}}-\frac {\cos \left (d x +c \right )^{12}}{99 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{12}}{11 \sin \left (d x +c \right )}+\frac {\left (\frac {256}{63}+\cos \left (d x +c \right )^{10}+\frac {10 \cos \left (d x +c \right )^{8}}{9}+\frac {80 \cos \left (d x +c \right )^{6}}{63}+\frac {32 \cos \left (d x +c \right )^{4}}{21}+\frac {128 \cos \left (d x +c \right )^{2}}{63}\right ) \sin \left (d x +c \right )}{11}\right )+a^{3} \left (-\frac {\cot \left (d x +c \right )^{11}}{11}+\frac {\cot \left (d x +c \right )^{9}}{9}-\frac {\cot \left (d x +c \right )^{7}}{7}+\frac {\cot \left (d x +c \right )^{5}}{5}-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\) \(425\)
default \(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{10}}{11 \sin \left (d x +c \right )^{11}}-\frac {\cos \left (d x +c \right )^{10}}{99 \sin \left (d x +c \right )^{9}}+\frac {\cos \left (d x +c \right )^{10}}{693 \sin \left (d x +c \right )^{7}}-\frac {\cos \left (d x +c \right )^{10}}{1155 \sin \left (d x +c \right )^{5}}+\frac {\cos \left (d x +c \right )^{10}}{693 \sin \left (d x +c \right )^{3}}-\frac {\cos \left (d x +c \right )^{10}}{99 \sin \left (d x +c \right )}-\frac {\left (\frac {128}{35}+\cos \left (d x +c \right )^{8}+\frac {8 \cos \left (d x +c \right )^{6}}{7}+\frac {48 \cos \left (d x +c \right )^{4}}{35}+\frac {64 \cos \left (d x +c \right )^{2}}{35}\right ) \sin \left (d x +c \right )}{99}\right )-\frac {3 a^{3} \cos \left (d x +c \right )^{11}}{11 \sin \left (d x +c \right )^{11}}+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{12}}{11 \sin \left (d x +c \right )^{11}}+\frac {\cos \left (d x +c \right )^{12}}{99 \sin \left (d x +c \right )^{9}}-\frac {\cos \left (d x +c \right )^{12}}{231 \sin \left (d x +c \right )^{7}}+\frac {\cos \left (d x +c \right )^{12}}{231 \sin \left (d x +c \right )^{5}}-\frac {\cos \left (d x +c \right )^{12}}{99 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{12}}{11 \sin \left (d x +c \right )}+\frac {\left (\frac {256}{63}+\cos \left (d x +c \right )^{10}+\frac {10 \cos \left (d x +c \right )^{8}}{9}+\frac {80 \cos \left (d x +c \right )^{6}}{63}+\frac {32 \cos \left (d x +c \right )^{4}}{21}+\frac {128 \cos \left (d x +c \right )^{2}}{63}\right ) \sin \left (d x +c \right )}{11}\right )+a^{3} \left (-\frac {\cot \left (d x +c \right )^{11}}{11}+\frac {\cot \left (d x +c \right )^{9}}{9}-\frac {\cot \left (d x +c \right )^{7}}{7}+\frac {\cot \left (d x +c \right )^{5}}{5}-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\) \(425\)

Input:

int(cot(d*x+c)^12*(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

a^3*x+2/3465*I*a^3*(10395*exp(15*I*(d*x+c))-31185*exp(14*I*(d*x+c))+1155*e 
xp(13*I*(d*x+c))+148995*exp(12*I*(d*x+c))-190113*exp(11*I*(d*x+c))-117117* 
exp(10*I*(d*x+c))+434775*exp(9*I*(d*x+c))-138105*exp(8*I*(d*x+c))-385055*e 
xp(7*I*(d*x+c))+374781*exp(6*I*(d*x+c))+63289*exp(5*I*(d*x+c))-223655*exp( 
4*I*(d*x+c))+75685*exp(3*I*(d*x+c))+43345*exp(2*I*(d*x+c))-34323*exp(I*(d* 
x+c))+7453)/d/(exp(I*(d*x+c))-1)^11/(exp(I*(d*x+c))+1)^5
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.47 \[ \int \cot ^{12}(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {7453 \, a^{3} \cos \left (d x + c\right )^{8} - 11964 \, a^{3} \cos \left (d x + c\right )^{7} - 11866 \, a^{3} \cos \left (d x + c\right )^{6} + 30542 \, a^{3} \cos \left (d x + c\right )^{5} + 90 \, a^{3} \cos \left (d x + c\right )^{4} - 26438 \, a^{3} \cos \left (d x + c\right )^{3} + 8539 \, a^{3} \cos \left (d x + c\right )^{2} + 7671 \, a^{3} \cos \left (d x + c\right ) - 3712 \, a^{3} + 3465 \, {\left (a^{3} d x \cos \left (d x + c\right )^{7} - 3 \, a^{3} d x \cos \left (d x + c\right )^{6} + a^{3} d x \cos \left (d x + c\right )^{5} + 5 \, a^{3} d x \cos \left (d x + c\right )^{4} - 5 \, a^{3} d x \cos \left (d x + c\right )^{3} - a^{3} d x \cos \left (d x + c\right )^{2} + 3 \, a^{3} d x \cos \left (d x + c\right ) - a^{3} d x\right )} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{7} - 3 \, d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5} + 5 \, d \cos \left (d x + c\right )^{4} - 5 \, d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2} + 3 \, d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )} \] Input:

integrate(cot(d*x+c)^12*(a+a*sec(d*x+c))^3,x, algorithm="fricas")
 

Output:

1/3465*(7453*a^3*cos(d*x + c)^8 - 11964*a^3*cos(d*x + c)^7 - 11866*a^3*cos 
(d*x + c)^6 + 30542*a^3*cos(d*x + c)^5 + 90*a^3*cos(d*x + c)^4 - 26438*a^3 
*cos(d*x + c)^3 + 8539*a^3*cos(d*x + c)^2 + 7671*a^3*cos(d*x + c) - 3712*a 
^3 + 3465*(a^3*d*x*cos(d*x + c)^7 - 3*a^3*d*x*cos(d*x + c)^6 + a^3*d*x*cos 
(d*x + c)^5 + 5*a^3*d*x*cos(d*x + c)^4 - 5*a^3*d*x*cos(d*x + c)^3 - a^3*d* 
x*cos(d*x + c)^2 + 3*a^3*d*x*cos(d*x + c) - a^3*d*x)*sin(d*x + c))/((d*cos 
(d*x + c)^7 - 3*d*cos(d*x + c)^6 + d*cos(d*x + c)^5 + 5*d*cos(d*x + c)^4 - 
 5*d*cos(d*x + c)^3 - d*cos(d*x + c)^2 + 3*d*cos(d*x + c) - d)*sin(d*x + c 
))
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^{12}(c+d x) (a+a \sec (c+d x))^3 \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**12*(a+a*sec(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00 \[ \int \cot ^{12}(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {{\left (3465 \, d x + 3465 \, c + \frac {3465 \, \tan \left (d x + c\right )^{10} - 1155 \, \tan \left (d x + c\right )^{8} + 693 \, \tan \left (d x + c\right )^{6} - 495 \, \tan \left (d x + c\right )^{4} + 385 \, \tan \left (d x + c\right )^{2} - 315}{\tan \left (d x + c\right )^{11}}\right )} a^{3} + \frac {15 \, {\left (693 \, \sin \left (d x + c\right )^{10} - 1155 \, \sin \left (d x + c\right )^{8} + 1386 \, \sin \left (d x + c\right )^{6} - 990 \, \sin \left (d x + c\right )^{4} + 385 \, \sin \left (d x + c\right )^{2} - 63\right )} a^{3}}{\sin \left (d x + c\right )^{11}} - \frac {{\left (1155 \, \sin \left (d x + c\right )^{8} - 2772 \, \sin \left (d x + c\right )^{6} + 2970 \, \sin \left (d x + c\right )^{4} - 1540 \, \sin \left (d x + c\right )^{2} + 315\right )} a^{3}}{\sin \left (d x + c\right )^{11}} - \frac {945 \, a^{3}}{\tan \left (d x + c\right )^{11}}}{3465 \, d} \] Input:

integrate(cot(d*x+c)^12*(a+a*sec(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/3465*((3465*d*x + 3465*c + (3465*tan(d*x + c)^10 - 1155*tan(d*x + c)^8 + 
 693*tan(d*x + c)^6 - 495*tan(d*x + c)^4 + 385*tan(d*x + c)^2 - 315)/tan(d 
*x + c)^11)*a^3 + 15*(693*sin(d*x + c)^10 - 1155*sin(d*x + c)^8 + 1386*sin 
(d*x + c)^6 - 990*sin(d*x + c)^4 + 385*sin(d*x + c)^2 - 63)*a^3/sin(d*x + 
c)^11 - (1155*sin(d*x + c)^8 - 2772*sin(d*x + c)^6 + 2970*sin(d*x + c)^4 - 
 1540*sin(d*x + c)^2 + 315)*a^3/sin(d*x + c)^11 - 945*a^3/tan(d*x + c)^11) 
/d
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.76 \[ \int \cot ^{12}(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {693 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 11550 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 887040 \, {\left (d x + c\right )} a^{3} + 159390 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {5 \, {\left (264726 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 59136 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 18018 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 4554 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 770 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 63 \, a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11}}}{887040 \, d} \] Input:

integrate(cot(d*x+c)^12*(a+a*sec(d*x+c))^3,x, algorithm="giac")
 

Output:

-1/887040*(693*a^3*tan(1/2*d*x + 1/2*c)^5 - 11550*a^3*tan(1/2*d*x + 1/2*c) 
^3 - 887040*(d*x + c)*a^3 + 159390*a^3*tan(1/2*d*x + 1/2*c) - 5*(264726*a^ 
3*tan(1/2*d*x + 1/2*c)^10 - 59136*a^3*tan(1/2*d*x + 1/2*c)^8 + 18018*a^3*t 
an(1/2*d*x + 1/2*c)^6 - 4554*a^3*tan(1/2*d*x + 1/2*c)^4 + 770*a^3*tan(1/2* 
d*x + 1/2*c)^2 - 63*a^3)/tan(1/2*d*x + 1/2*c)^11)/d
 

Mupad [B] (verification not implemented)

Time = 14.17 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.19 \[ \int \cot ^{12}(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {a^3\,\left (315\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+693\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-11550\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+159390\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-1323630\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+295680\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-90090\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+22770\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3850\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-887040\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (c+d\,x\right )\right )}{887040\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}} \] Input:

int(cot(c + d*x)^12*(a + a/cos(c + d*x))^3,x)
 

Output:

-(a^3*(315*cos(c/2 + (d*x)/2)^16 + 693*sin(c/2 + (d*x)/2)^16 - 11550*cos(c 
/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^14 + 159390*cos(c/2 + (d*x)/2)^4*sin(c/ 
2 + (d*x)/2)^12 - 1323630*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^10 + 295 
680*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^8 - 90090*cos(c/2 + (d*x)/2)^1 
0*sin(c/2 + (d*x)/2)^6 + 22770*cos(c/2 + (d*x)/2)^12*sin(c/2 + (d*x)/2)^4 
- 3850*cos(c/2 + (d*x)/2)^14*sin(c/2 + (d*x)/2)^2 - 887040*cos(c/2 + (d*x) 
/2)^5*sin(c/2 + (d*x)/2)^11*(c + d*x)))/(887040*d*cos(c/2 + (d*x)/2)^5*sin 
(c/2 + (d*x)/2)^11)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.66 \[ \int \cot ^{12}(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^{3} \left (-693 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+11550 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}-159390 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+887040 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11} d x +1323630 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-295680 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+90090 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-22770 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3850 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-315\right )}{887040 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11} d} \] Input:

int(cot(d*x+c)^12*(a+a*sec(d*x+c))^3,x)
 

Output:

(a**3*( - 693*tan((c + d*x)/2)**16 + 11550*tan((c + d*x)/2)**14 - 159390*t 
an((c + d*x)/2)**12 + 887040*tan((c + d*x)/2)**11*d*x + 1323630*tan((c + d 
*x)/2)**10 - 295680*tan((c + d*x)/2)**8 + 90090*tan((c + d*x)/2)**6 - 2277 
0*tan((c + d*x)/2)**4 + 3850*tan((c + d*x)/2)**2 - 315))/(887040*tan((c + 
d*x)/2)**11*d)