\(\int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx\) [60]

Optimal result

Integrand size = 21, antiderivative size = 169 \[ \int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\left (a^2+b^2\right )^2 \cot (c+d x)}{a^5 d}+\frac {b \left (2 a^2+b^2\right ) \cot ^2(c+d x)}{2 a^4 d}-\frac {\left (2 a^2+b^2\right ) \cot ^3(c+d x)}{3 a^3 d}+\frac {b \cot ^4(c+d x)}{4 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {b \left (a^2+b^2\right )^2 \log (\tan (c+d x))}{a^6 d}+\frac {b \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))}{a^6 d} \] Output:

-(a^2+b^2)^2*cot(d*x+c)/a^5/d+1/2*b*(2*a^2+b^2)*cot(d*x+c)^2/a^4/d-1/3*(2* 
a^2+b^2)*cot(d*x+c)^3/a^3/d+1/4*b*cot(d*x+c)^4/a^2/d-1/5*cot(d*x+c)^5/a/d- 
b*(a^2+b^2)^2*ln(tan(d*x+c))/a^6/d+b*(a^2+b^2)^2*ln(a+b*tan(d*x+c))/a^6/d
 

Mathematica [A] (verified)

Time = 5.10 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-4 \cot (c+d x) \left (8 a^5+25 a^3 b^2+15 a b^4+a^3 \left (4 a^2+5 b^2\right ) \csc ^2(c+d x)+3 a^5 \csc ^4(c+d x)\right )+15 b \left (2 a^2 \left (a^2+b^2\right ) \csc ^2(c+d x)+a^4 \csc ^4(c+d x)-4 \left (a^2+b^2\right )^2 (\log (\sin (c+d x))-\log (a \cos (c+d x)+b \sin (c+d x)))\right )}{60 a^6 d} \] Input:

Integrate[Csc[c + d*x]^6/(a + b*Tan[c + d*x]),x]
 

Output:

(-4*Cot[c + d*x]*(8*a^5 + 25*a^3*b^2 + 15*a*b^4 + a^3*(4*a^2 + 5*b^2)*Csc[ 
c + d*x]^2 + 3*a^5*Csc[c + d*x]^4) + 15*b*(2*a^2*(a^2 + b^2)*Csc[c + d*x]^ 
2 + a^4*Csc[c + d*x]^4 - 4*(a^2 + b^2)^2*(Log[Sin[c + d*x]] - Log[a*Cos[c 
+ d*x] + b*Sin[c + d*x]])))/(60*a^6*d)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3999, 522, 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.

\(\displaystyle \int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (c+d x)^6 (a+b \tan (c+d x))}dx\)

\(\Big \downarrow \) 3999

\(\displaystyle \frac {b \int \frac {\cot ^6(c+d x) \left (\tan ^2(c+d x) b^2+b^2\right )^2}{b^6 (a+b \tan (c+d x))}d(b \tan (c+d x))}{d}\)

\(\Big \downarrow \) 522

\(\displaystyle \frac {b \int \left (\frac {\cot ^6(c+d x)}{a b^2}-\frac {\cot ^5(c+d x)}{a^2 b}+\frac {\left (b^4+2 a^2 b^2\right ) \cot ^4(c+d x)}{a^3 b^4}+\frac {\left (-2 a^2-b^2\right ) \cot ^3(c+d x)}{a^4 b}+\frac {\left (a^2+b^2\right )^2 \cot ^2(c+d x)}{a^5 b^2}-\frac {\left (a^2+b^2\right )^2 \cot (c+d x)}{a^6 b}+\frac {\left (a^2+b^2\right )^2}{a^6 (a+b \tan (c+d x))}\right )d(b \tan (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (\frac {\cot ^4(c+d x)}{4 a^2}-\frac {\left (a^2+b^2\right )^2 \log (b \tan (c+d x))}{a^6}+\frac {\left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))}{a^6}-\frac {\left (a^2+b^2\right )^2 \cot (c+d x)}{a^5 b}+\frac {\left (2 a^2+b^2\right ) \cot ^2(c+d x)}{2 a^4}-\frac {\left (2 a^2+b^2\right ) \cot ^3(c+d x)}{3 a^3 b}-\frac {\cot ^5(c+d x)}{5 a b}\right )}{d}\)

Input:

Int[Csc[c + d*x]^6/(a + b*Tan[c + d*x]),x]
 

Output:

(b*(-(((a^2 + b^2)^2*Cot[c + d*x])/(a^5*b)) + ((2*a^2 + b^2)*Cot[c + d*x]^ 
2)/(2*a^4) - ((2*a^2 + b^2)*Cot[c + d*x]^3)/(3*a^3*b) + Cot[c + d*x]^4/(4* 
a^2) - Cot[c + d*x]^5/(5*a*b) - ((a^2 + b^2)^2*Log[b*Tan[c + d*x]])/a^6 + 
((a^2 + b^2)^2*Log[a + b*Tan[c + d*x]])/a^6))/d
 

Defintions of rubi rules used

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

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 3999

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[b/f   Subst[Int[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), 
 x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/2]
 

Maple [A] (verified)

Time = 6.94 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{6}}-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {2 a^{2}+b^{2}}{3 a^{3} \tan \left (d x +c \right )^{3}}-\frac {a^{4}+2 b^{2} a^{2}+b^{4}}{a^{5} \tan \left (d x +c \right )}+\frac {b}{4 a^{2} \tan \left (d x +c \right )^{4}}+\frac {\left (2 a^{2}+b^{2}\right ) b}{2 a^{4} \tan \left (d x +c \right )^{2}}-\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) b \ln \left (\tan \left (d x +c \right )\right )}{a^{6}}}{d}\)	\(165\)
default	\(\frac {\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{6}}-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {2 a^{2}+b^{2}}{3 a^{3} \tan \left (d x +c \right )^{3}}-\frac {a^{4}+2 b^{2} a^{2}+b^{4}}{a^{5} \tan \left (d x +c \right )}+\frac {b}{4 a^{2} \tan \left (d x +c \right )^{4}}+\frac {\left (2 a^{2}+b^{2}\right ) b}{2 a^{4} \tan \left (d x +c \right )^{2}}-\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) b \ln \left (\tan \left (d x +c \right )\right )}{a^{6}}}{d}\)	\(165\)
risch	\(-\frac {2 i \left (75 i a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}+45 i a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+15 a^{2} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+15 b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-15 i a \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+15 i a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-90 a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-60 b^{4} {\mathrm e}^{6 i \left (d x +c \right )}-75 i a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}-15 i a^{3} b \,{\mathrm e}^{8 i \left (d x +c \right )}+80 a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+160 a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+90 b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+15 i a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-45 i a \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-40 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-110 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-60 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+8 a^{4}+25 b^{2} a^{2}+15 b^{4}\right )}{15 d \,a^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{2} d}+\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{4} d}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{6} d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}-\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}-\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{6} d}\)	\(506\)

Input:

int(csc(d*x+c)^6/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*((a^4+2*a^2*b^2+b^4)/a^6*b*ln(a+b*tan(d*x+c))-1/5/a/tan(d*x+c)^5-1/3*( 
2*a^2+b^2)/a^3/tan(d*x+c)^3-(a^4+2*a^2*b^2+b^4)/a^5/tan(d*x+c)+1/4*b/a^2/t 
an(d*x+c)^4+1/2*(2*a^2+b^2)/a^4*b/tan(d*x+c)^2-(a^4+2*a^2*b^2+b^4)/a^6*b*l 
n(tan(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (161) = 322\).

Time = 0.11 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.28 \[ \int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {4 \, {\left (8 \, a^{5} + 25 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} - 20 \, {\left (4 \, a^{5} + 11 \, a^{3} b^{2} + 6 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 30 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5} + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) \sin \left (d x + c\right ) + 30 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5} + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) + 60 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) - 15 \, {\left (3 \, a^{4} b + 2 \, a^{2} b^{3} - 2 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{6} d \cos \left (d x + c\right )^{4} - 2 \, a^{6} d \cos \left (d x + c\right )^{2} + a^{6} d\right )} \sin \left (d x + c\right )} \] Input:

integrate(csc(d*x+c)^6/(a+b*tan(d*x+c)),x, algorithm="fricas")
 

Output:

-1/60*(4*(8*a^5 + 25*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5 - 20*(4*a^5 + 11*a 
^3*b^2 + 6*a*b^4)*cos(d*x + c)^3 - 30*(a^4*b + 2*a^2*b^3 + b^5 + (a^4*b + 
2*a^2*b^3 + b^5)*cos(d*x + c)^4 - 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c) 
^2)*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2 
)*sin(d*x + c) + 30*(a^4*b + 2*a^2*b^3 + b^5 + (a^4*b + 2*a^2*b^3 + b^5)*c 
os(d*x + c)^4 - 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^2)*log(-1/4*cos(d 
*x + c)^2 + 1/4)*sin(d*x + c) + 60*(a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c) 
- 15*(3*a^4*b + 2*a^2*b^3 - 2*(a^4*b + a^2*b^3)*cos(d*x + c)^2)*sin(d*x + 
c))/((a^6*d*cos(d*x + c)^4 - 2*a^6*d*cos(d*x + c)^2 + a^6*d)*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\csc ^{6}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \] Input:

integrate(csc(d*x+c)**6/(a+b*tan(d*x+c)),x)
 

Output:

Integral(csc(c + d*x)**6/(a + b*tan(c + d*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.99 \[ \int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {60 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6}} - \frac {60 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{6}} + \frac {15 \, a^{3} b \tan \left (d x + c\right ) - 60 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{4} - 12 \, a^{4} + 30 \, {\left (2 \, a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )^{3} - 20 \, {\left (2 \, a^{4} + a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{5} \tan \left (d x + c\right )^{5}}}{60 \, d} \] Input:

integrate(csc(d*x+c)^6/(a+b*tan(d*x+c)),x, algorithm="maxima")
 

Output:

1/60*(60*(a^4*b + 2*a^2*b^3 + b^5)*log(b*tan(d*x + c) + a)/a^6 - 60*(a^4*b 
 + 2*a^2*b^3 + b^5)*log(tan(d*x + c))/a^6 + (15*a^3*b*tan(d*x + c) - 60*(a 
^4 + 2*a^2*b^2 + b^4)*tan(d*x + c)^4 - 12*a^4 + 30*(2*a^3*b + a*b^3)*tan(d 
*x + c)^3 - 20*(2*a^4 + a^2*b^2)*tan(d*x + c)^2)/(a^5*tan(d*x + c)^5))/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.08 \[ \int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{6} d} + \frac {{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b d} + \frac {15 \, a^{4} b \tan \left (d x + c\right ) - 12 \, a^{5} - 60 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \tan \left (d x + c\right )^{4} + 30 \, {\left (2 \, a^{4} b + a^{2} b^{3}\right )} \tan \left (d x + c\right )^{3} - 20 \, {\left (2 \, a^{5} + a^{3} b^{2}\right )} \tan \left (d x + c\right )^{2}}{60 \, a^{6} d \tan \left (d x + c\right )^{5}} \] Input:

integrate(csc(d*x+c)^6/(a+b*tan(d*x+c)),x, algorithm="giac")
 

Output:

-(a^4*b + 2*a^2*b^3 + b^5)*log(abs(tan(d*x + c)))/(a^6*d) + (a^4*b^2 + 2*a 
^2*b^4 + b^6)*log(abs(b*tan(d*x + c) + a))/(a^6*b*d) + 1/60*(15*a^4*b*tan( 
d*x + c) - 12*a^5 - 60*(a^5 + 2*a^3*b^2 + a*b^4)*tan(d*x + c)^4 + 30*(2*a^ 
4*b + a^2*b^3)*tan(d*x + c)^3 - 20*(2*a^5 + a^3*b^2)*tan(d*x + c)^2)/(a^6* 
d*tan(d*x + c)^5)
 

Mupad [B] (verification not implemented)

Time = 1.40 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.99 \[ \int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {b\,{\left (a^2+b^2\right )}^2\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,\left (a^4\,b+2\,a^2\,b^3+b^5\right )}\right )\,{\left (a^2+b^2\right )}^2}{a^6\,d}-\frac {\frac {1}{5\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+b^2\right )}{3\,a^3}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^4+2\,a^2\,b^2+b^4\right )}{a^5}-\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (2\,a^2+b^2\right )}{2\,a^4}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \] Input:

int(1/(sin(c + d*x)^6*(a + b*tan(c + d*x))),x)
 

Output:

(2*b*atanh((b*(a^2 + b^2)^2*(a + 2*b*tan(c + d*x)))/(a*(a^4*b + b^5 + 2*a^ 
2*b^3)))*(a^2 + b^2)^2)/(a^6*d) - (1/(5*a) + (tan(c + d*x)^2*(2*a^2 + b^2) 
)/(3*a^3) + (tan(c + d*x)^4*(a^4 + b^4 + 2*a^2*b^2))/a^5 - (b*tan(c + d*x) 
)/(4*a^2) - (b*tan(c + d*x)^3*(2*a^2 + b^2))/(2*a^4))/(d*tan(c + d*x)^5)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.41 \[ \int \frac {\csc ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-256 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{5}-800 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3} b^{2}-480 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{4}-128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{5}-160 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3} b^{2}-96 \cos \left (d x +c \right ) a^{5}+480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right ) \sin \left (d x +c \right )^{5} a^{4} b +960 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right ) \sin \left (d x +c \right )^{5} a^{2} b^{3}+480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right ) \sin \left (d x +c \right )^{5} b^{5}-480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} a^{4} b -960 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} a^{2} b^{3}-480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5} b^{5}-165 \sin \left (d x +c \right )^{5} a^{4} b -120 \sin \left (d x +c \right )^{5} a^{2} b^{3}+240 \sin \left (d x +c \right )^{3} a^{4} b +240 \sin \left (d x +c \right )^{3} a^{2} b^{3}+120 \sin \left (d x +c \right ) a^{4} b}{480 \sin \left (d x +c \right )^{5} a^{6} d} \] Input:

int(csc(d*x+c)^6/(a+b*tan(d*x+c)),x)
 

Output:

( - 256*cos(c + d*x)*sin(c + d*x)**4*a**5 - 800*cos(c + d*x)*sin(c + d*x)* 
*4*a**3*b**2 - 480*cos(c + d*x)*sin(c + d*x)**4*a*b**4 - 128*cos(c + d*x)* 
sin(c + d*x)**2*a**5 - 160*cos(c + d*x)*sin(c + d*x)**2*a**3*b**2 - 96*cos 
(c + d*x)*a**5 + 480*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a) 
*sin(c + d*x)**5*a**4*b + 960*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/ 
2)*b - a)*sin(c + d*x)**5*a**2*b**3 + 480*log(tan((c + d*x)/2)**2*a - 2*ta 
n((c + d*x)/2)*b - a)*sin(c + d*x)**5*b**5 - 480*log(tan((c + d*x)/2))*sin 
(c + d*x)**5*a**4*b - 960*log(tan((c + d*x)/2))*sin(c + d*x)**5*a**2*b**3 
- 480*log(tan((c + d*x)/2))*sin(c + d*x)**5*b**5 - 165*sin(c + d*x)**5*a** 
4*b - 120*sin(c + d*x)**5*a**2*b**3 + 240*sin(c + d*x)**3*a**4*b + 240*sin 
(c + d*x)**3*a**2*b**3 + 120*sin(c + d*x)*a**4*b)/(480*sin(c + d*x)**5*a** 
6*d)