\(\int \frac {\cot ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [85]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
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 = 179 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {x}{a^2}-\frac {\cot (c+d x)}{a^2 d}+\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}+\frac {\cot ^7(c+d x)}{7 a^2 d}-\frac {2 \cot ^9(c+d x)}{9 a^2 d}+\frac {2 \csc (c+d x)}{a^2 d}-\frac {8 \csc ^3(c+d x)}{3 a^2 d}+\frac {12 \csc ^5(c+d x)}{5 a^2 d}-\frac {8 \csc ^7(c+d x)}{7 a^2 d}+\frac {2 \csc ^9(c+d x)}{9 a^2 d} \] Output:

-x/a^2-cot(d*x+c)/a^2/d+1/3*cot(d*x+c)^3/a^2/d-1/5*cot(d*x+c)^5/a^2/d+1/7* 
cot(d*x+c)^7/a^2/d-2/9*cot(d*x+c)^9/a^2/d+2*csc(d*x+c)/a^2/d-8/3*csc(d*x+c 
)^3/a^2/d+12/5*csc(d*x+c)^5/a^2/d-8/7*csc(d*x+c)^7/a^2/d+2/9*csc(d*x+c)^9/ 
a^2/d
 

Mathematica [A] (verified)

Time = 5.79 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.54 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-63 \cot ^2\left (\frac {c}{2}\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) \left (-17+\csc ^2\left (\frac {1}{2} (c+d x)\right )\right )-\frac {63}{8} \cot \left (\frac {c}{2}\right ) \left (5120 d x+(-543+736 \cos (c+d x)-201 \cos (2 (c+d x))) \csc \left (\frac {c}{2}\right ) \csc ^5\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )\right )+\frac {1}{256} \csc \left (\frac {c}{2}\right ) \sec ^9\left (\frac {1}{2} (c+d x)\right ) \left (4360986 \sin \left (\frac {d x}{2}\right )-3688020 \sin \left (c+\frac {d x}{2}\right )+3365964 \sin \left (c+\frac {3 d x}{2}\right )-2000040 \sin \left (2 c+\frac {3 d x}{2}\right )+1660896 \sin \left (2 c+\frac {5 d x}{2}\right )-638820 \sin \left (3 c+\frac {5 d x}{2}\right )+479484 \sin \left (3 c+\frac {7 d x}{2}\right )-95445 \sin \left (4 c+\frac {7 d x}{2}\right )+63881 \sin \left (4 c+\frac {9 d x}{2}\right )\right )\right ) \tan \left (\frac {c}{2}\right )}{10080 a^2 d (1+\sec (c+d x))^2} \] Input:

Integrate[Cot[c + d*x]^6/(a + a*Sec[c + d*x])^2,x]
 

Output:

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(-63*Cot[c/2]^2*Csc[(c + d*x)/2]^2*(-17 
 + Csc[(c + d*x)/2]^2) - (63*Cot[c/2]*(5120*d*x + (-543 + 736*Cos[c + d*x] 
 - 201*Cos[2*(c + d*x)])*Csc[c/2]*Csc[(c + d*x)/2]^5*Sin[(d*x)/2]))/8 + (C 
sc[c/2]*Sec[(c + d*x)/2]^9*(4360986*Sin[(d*x)/2] - 3688020*Sin[c + (d*x)/2 
] + 3365964*Sin[c + (3*d*x)/2] - 2000040*Sin[2*c + (3*d*x)/2] + 1660896*Si 
n[2*c + (5*d*x)/2] - 638820*Sin[3*c + (5*d*x)/2] + 479484*Sin[3*c + (7*d*x 
)/2] - 95445*Sin[4*c + (7*d*x)/2] + 63881*Sin[4*c + (9*d*x)/2]))/256)*Tan[ 
c/2])/(10080*a^2*d*(1 + Sec[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4376, 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.

\(\displaystyle \int \frac {\cot ^6(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^6 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\)

\(\Big \downarrow \) 4376

\(\displaystyle \frac {\int \cot ^{10}(c+d x) (a-a \sec (c+d x))^2dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}{\cot \left (c+d x+\frac {\pi }{2}\right )^{10}}dx}{a^4}\)

\(\Big \downarrow \) 4374

\(\displaystyle \frac {\int \left (a^2 \cot ^{10}(c+d x)-2 a^2 \csc (c+d x) \cot ^9(c+d x)+a^2 \csc ^2(c+d x) \cot ^8(c+d x)\right )dx}{a^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {2 a^2 \cot ^9(c+d x)}{9 d}+\frac {a^2 \cot ^7(c+d x)}{7 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x)}{d}+\frac {2 a^2 \csc ^9(c+d x)}{9 d}-\frac {8 a^2 \csc ^7(c+d x)}{7 d}+\frac {12 a^2 \csc ^5(c+d x)}{5 d}-\frac {8 a^2 \csc ^3(c+d x)}{3 d}+\frac {2 a^2 \csc (c+d x)}{d}-a^2 x}{a^4}\)

Input:

Int[Cot[c + d*x]^6/(a + a*Sec[c + d*x])^2,x]
 

Output:

(-(a^2*x) - (a^2*Cot[c + d*x])/d + (a^2*Cot[c + d*x]^3)/(3*d) - (a^2*Cot[c 
 + d*x]^5)/(5*d) + (a^2*Cot[c + d*x]^7)/(7*d) - (2*a^2*Cot[c + d*x]^9)/(9* 
d) + (2*a^2*Csc[c + d*x])/d - (8*a^2*Csc[c + d*x]^3)/(3*d) + (12*a^2*Csc[c 
 + d*x]^5)/(5*d) - (8*a^2*Csc[c + d*x]^7)/(7*d) + (2*a^2*Csc[c + d*x]^9)/( 
9*d))/a^4
 

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]
 

rule 4376
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Simp[a^(2*n)/e^(2*n)   Int[(e*Cot[c + d*x])^(m + 2* 
n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[a 
^2 - b^2, 0] && ILtQ[n, 0]
 
Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.69

method result size
derivativedivides \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {37 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+163 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {37}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-256 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d \,a^{2}}\) \(124\)
default \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {37 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+163 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {37}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-256 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d \,a^{2}}\) \(124\)
risch \(-\frac {x}{a^{2}}+\frac {4 i \left (315 \,{\mathrm e}^{13 i \left (d x +c \right )}+315 \,{\mathrm e}^{12 i \left (d x +c \right )}-1470 \,{\mathrm e}^{11 i \left (d x +c \right )}-3360 \,{\mathrm e}^{10 i \left (d x +c \right )}+1113 \,{\mathrm e}^{9 i \left (d x +c \right )}+6447 \,{\mathrm e}^{8 i \left (d x +c \right )}+2028 \,{\mathrm e}^{7 i \left (d x +c \right )}-7008 \,{\mathrm e}^{6 i \left (d x +c \right )}-4867 \,{\mathrm e}^{5 i \left (d x +c \right )}+2321 \,{\mathrm e}^{4 i \left (d x +c \right )}+3314 \,{\mathrm e}^{3 i \left (d x +c \right )}+16 \,{\mathrm e}^{2 i \left (d x +c \right )}-881 \,{\mathrm e}^{i \left (d x +c \right )}-299\right )}{315 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{5}}\) \(188\)

Input:

int(cot(d*x+c)^6/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/128/d/a^2*(1/9*tan(1/2*d*x+1/2*c)^9-9/7*tan(1/2*d*x+1/2*c)^7+37/5*tan(1/ 
2*d*x+1/2*c)^5-31*tan(1/2*d*x+1/2*c)^3+163*tan(1/2*d*x+1/2*c)-1/5/tan(1/2* 
d*x+1/2*c)^5+3/tan(1/2*d*x+1/2*c)^3-37/tan(1/2*d*x+1/2*c)-256*arctan(tan(1 
/2*d*x+1/2*c)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.40 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {598 \, \cos \left (d x + c\right )^{7} + 566 \, \cos \left (d x + c\right )^{6} - 1212 \, \cos \left (d x + c\right )^{5} - 1310 \, \cos \left (d x + c\right )^{4} + 860 \, \cos \left (d x + c\right )^{3} + 1014 \, \cos \left (d x + c\right )^{2} + 315 \, {\left (d x \cos \left (d x + c\right )^{6} + 2 \, d x \cos \left (d x + c\right )^{5} - d x \cos \left (d x + c\right )^{4} - 4 \, d x \cos \left (d x + c\right )^{3} - d x \cos \left (d x + c\right )^{2} + 2 \, d x \cos \left (d x + c\right ) + d x\right )} \sin \left (d x + c\right ) - 197 \, \cos \left (d x + c\right ) - 256}{315 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} + 2 \, a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{4} - 4 \, a^{2} d \cos \left (d x + c\right )^{3} - a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )} \] Input:

integrate(cot(d*x+c)^6/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/315*(598*cos(d*x + c)^7 + 566*cos(d*x + c)^6 - 1212*cos(d*x + c)^5 - 13 
10*cos(d*x + c)^4 + 860*cos(d*x + c)^3 + 1014*cos(d*x + c)^2 + 315*(d*x*co 
s(d*x + c)^6 + 2*d*x*cos(d*x + c)^5 - d*x*cos(d*x + c)^4 - 4*d*x*cos(d*x + 
 c)^3 - d*x*cos(d*x + c)^2 + 2*d*x*cos(d*x + c) + d*x)*sin(d*x + c) - 197* 
cos(d*x + c) - 256)/((a^2*d*cos(d*x + c)^6 + 2*a^2*d*cos(d*x + c)^5 - a^2* 
d*cos(d*x + c)^4 - 4*a^2*d*cos(d*x + c)^3 - a^2*d*cos(d*x + c)^2 + 2*a^2*d 
*cos(d*x + c) + a^2*d)*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {\cot ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\cot ^{6}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:

integrate(cot(d*x+c)**6/(a+a*sec(d*x+c))**2,x)
 

Output:

Integral(cot(c + d*x)**6/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.10 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {\frac {51345 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9765 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2331 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {405 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{2}} - \frac {80640 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {63 \, {\left (\frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {185 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{2} \sin \left (d x + c\right )^{5}}}{40320 \, d} \] Input:

integrate(cot(d*x+c)^6/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

1/40320*((51345*sin(d*x + c)/(cos(d*x + c) + 1) - 9765*sin(d*x + c)^3/(cos 
(d*x + c) + 1)^3 + 2331*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 405*sin(d*x 
+ c)^7/(cos(d*x + c) + 1)^7 + 35*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/a^2 
- 80640*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 63*(15*sin(d*x + c)^ 
2/(cos(d*x + c) + 1)^2 - 185*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1)*(cos 
(d*x + c) + 1)^5/(a^2*sin(d*x + c)^5))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.80 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {40320 \, {\left (d x + c\right )}}{a^{2}} + \frac {63 \, {\left (185 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - \frac {35 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 405 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2331 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9765 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 51345 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{18}}}{40320 \, d} \] Input:

integrate(cot(d*x+c)^6/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

-1/40320*(40320*(d*x + c)/a^2 + 63*(185*tan(1/2*d*x + 1/2*c)^4 - 15*tan(1/ 
2*d*x + 1/2*c)^2 + 1)/(a^2*tan(1/2*d*x + 1/2*c)^5) - (35*a^16*tan(1/2*d*x 
+ 1/2*c)^9 - 405*a^16*tan(1/2*d*x + 1/2*c)^7 + 2331*a^16*tan(1/2*d*x + 1/2 
*c)^5 - 9765*a^16*tan(1/2*d*x + 1/2*c)^3 + 51345*a^16*tan(1/2*d*x + 1/2*c) 
)/a^18)/d
 

Mupad [B] (verification not implemented)

Time = 12.74 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.28 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {63\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-35\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+405\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-2331\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+9765\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-51345\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+11655\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-945\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+40320\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (c+d\,x\right )}{40320\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:

int(cot(c + d*x)^6/(a + a/cos(c + d*x))^2,x)
 

Output:

-(63*cos(c/2 + (d*x)/2)^14 - 35*sin(c/2 + (d*x)/2)^14 + 405*cos(c/2 + (d*x 
)/2)^2*sin(c/2 + (d*x)/2)^12 - 2331*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2 
)^10 + 9765*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^8 - 51345*cos(c/2 + (d 
*x)/2)^8*sin(c/2 + (d*x)/2)^6 + 11655*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x 
)/2)^4 - 945*cos(c/2 + (d*x)/2)^12*sin(c/2 + (d*x)/2)^2 + 40320*cos(c/2 + 
(d*x)/2)^9*sin(c/2 + (d*x)/2)^5*(c + d*x))/(40320*a^2*d*cos(c/2 + (d*x)/2) 
^9*sin(c/2 + (d*x)/2)^5)
 

Reduce [B] (verification not implemented)

Time = 8.40 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.71 \[ \int \frac {\cot ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}-405 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+2331 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-9765 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+51345 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-40320 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} d x -11655 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+945 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-63}{40320 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a^{2} d} \] Input:

int(cot(d*x+c)^6/(a+a*sec(d*x+c))^2,x)
 

Output:

(35*tan((c + d*x)/2)**14 - 405*tan((c + d*x)/2)**12 + 2331*tan((c + d*x)/2 
)**10 - 9765*tan((c + d*x)/2)**8 + 51345*tan((c + d*x)/2)**6 - 40320*tan(( 
c + d*x)/2)**5*d*x - 11655*tan((c + d*x)/2)**4 + 945*tan((c + d*x)/2)**2 - 
 63)/(40320*tan((c + d*x)/2)**5*a**2*d)