Integrand size = 21, antiderivative size = 300 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^4} \, dx=-\frac {\left (a^4+20 a^2 b^2+35 b^4\right ) \cot (c+d x)}{a^8 d}+\frac {2 b \left (2 a^2+5 b^2\right ) \cot ^2(c+d x)}{a^7 d}-\frac {2 \left (a^2+5 b^2\right ) \cot ^3(c+d x)}{3 a^6 d}+\frac {b \cot ^4(c+d x)}{a^5 d}-\frac {\cot ^5(c+d x)}{5 a^4 d}-\frac {4 b \left (a^4+10 a^2 b^2+14 b^4\right ) \log (\tan (c+d x))}{a^9 d}+\frac {4 b \left (a^4+10 a^2 b^2+14 b^4\right ) \log (a+b \tan (c+d x))}{a^9 d}-\frac {b \left (a^2+b^2\right )^2}{3 a^6 d (a+b \tan (c+d x))^3}-\frac {b \left (a^2+b^2\right ) \left (a^2+3 b^2\right )}{a^7 d (a+b \tan (c+d x))^2}-\frac {b \left (3 a^4+20 a^2 b^2+21 b^4\right )}{a^8 d (a+b \tan (c+d x))} \]
-(a^4+20*a^2*b^2+35*b^4)*cot(d*x+c)/a^8/d+2*b*(2*a^2+5*b^2)*cot(d*x+c)^2/a ^7/d-2/3*(a^2+5*b^2)*cot(d*x+c)^3/a^6/d+b*cot(d*x+c)^4/a^5/d-1/5*cot(d*x+c )^5/a^4/d-4*b*(a^4+10*a^2*b^2+14*b^4)*ln(tan(d*x+c))/a^9/d+4*b*(a^4+10*a^2 *b^2+14*b^4)*ln(a+b*tan(d*x+c))/a^9/d-1/3*b*(a^2+b^2)^2/a^6/d/(a+b*tan(d*x +c))^3-b*(a^2+b^2)*(a^2+3*b^2)/a^7/d/(a+b*tan(d*x+c))^2-b*(3*a^4+20*a^2*b^ 2+21*b^4)/a^8/d/(a+b*tan(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(673\) vs. \(2(300)=600\).
Time = 3.10 (sec) , antiderivative size = 673, normalized size of antiderivative = 2.24 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (-7680 b \left (a^4+10 a^2 b^2+14 b^4\right ) \log (\sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^3+7680 b \left (a^4+10 a^2 b^2+14 b^4\right ) \log (a \cos (c+d x)+b \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^3+\csc ^5(c+d x) \left (-200 a^8+380 a^6 b^2+3070 a^4 b^4+11375 a^2 b^6+11025 b^8-4 \left (52 a^8+194 a^6 b^2+1510 a^4 b^4+5705 a^2 b^6+4410 b^8\right ) \cos (2 (c+d x))+4 \left (4 a^8-16 a^6 b^2+1010 a^4 b^4+4585 a^2 b^6+2205 b^8\right ) \cos (4 (c+d x))+16 a^8 \cos (6 (c+d x))+776 a^6 b^2 \cos (6 (c+d x))-1000 a^4 b^4 \cos (6 (c+d x))-8540 a^2 b^6 \cos (6 (c+d x))-2520 b^8 \cos (6 (c+d x))-8 a^8 \cos (8 (c+d x))-316 a^6 b^2 \cos (8 (c+d x))-70 a^4 b^4 \cos (8 (c+d x))+1645 a^2 b^6 \cos (8 (c+d x))+315 b^8 \cos (8 (c+d x))+264 a^7 b \sin (2 (c+d x))+372 a^5 b^3 \sin (2 (c+d x))+4830 a^3 b^5 \sin (2 (c+d x))+1470 a b^7 \sin (2 (c+d x))+144 a^7 b \sin (4 (c+d x))-2476 a^5 b^3 \sin (4 (c+d x))-9730 a^3 b^5 \sin (4 (c+d x))-1470 a b^7 \sin (4 (c+d x))-24 a^7 b \sin (6 (c+d x))+2756 a^5 b^3 \sin (6 (c+d x))+7670 a^3 b^5 \sin (6 (c+d x))+630 a b^7 \sin (6 (c+d x))-24 a^7 b \sin (8 (c+d x))-922 a^5 b^3 \sin (8 (c+d x))-2095 a^3 b^5 \sin (8 (c+d x))-105 a b^7 \sin (8 (c+d x))\right )\right )}{1920 a^9 d (a+b \tan (c+d x))^4} \]
(Sec[c + d*x]^4*(a*Cos[c + d*x] + b*Sin[c + d*x])*(-7680*b*(a^4 + 10*a^2*b ^2 + 14*b^4)*Log[Sin[c + d*x]]*(a*Cos[c + d*x] + b*Sin[c + d*x])^3 + 7680* b*(a^4 + 10*a^2*b^2 + 14*b^4)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]*(a*Cos[ c + d*x] + b*Sin[c + d*x])^3 + Csc[c + d*x]^5*(-200*a^8 + 380*a^6*b^2 + 30 70*a^4*b^4 + 11375*a^2*b^6 + 11025*b^8 - 4*(52*a^8 + 194*a^6*b^2 + 1510*a^ 4*b^4 + 5705*a^2*b^6 + 4410*b^8)*Cos[2*(c + d*x)] + 4*(4*a^8 - 16*a^6*b^2 + 1010*a^4*b^4 + 4585*a^2*b^6 + 2205*b^8)*Cos[4*(c + d*x)] + 16*a^8*Cos[6* (c + d*x)] + 776*a^6*b^2*Cos[6*(c + d*x)] - 1000*a^4*b^4*Cos[6*(c + d*x)] - 8540*a^2*b^6*Cos[6*(c + d*x)] - 2520*b^8*Cos[6*(c + d*x)] - 8*a^8*Cos[8* (c + d*x)] - 316*a^6*b^2*Cos[8*(c + d*x)] - 70*a^4*b^4*Cos[8*(c + d*x)] + 1645*a^2*b^6*Cos[8*(c + d*x)] + 315*b^8*Cos[8*(c + d*x)] + 264*a^7*b*Sin[2 *(c + d*x)] + 372*a^5*b^3*Sin[2*(c + d*x)] + 4830*a^3*b^5*Sin[2*(c + d*x)] + 1470*a*b^7*Sin[2*(c + d*x)] + 144*a^7*b*Sin[4*(c + d*x)] - 2476*a^5*b^3 *Sin[4*(c + d*x)] - 9730*a^3*b^5*Sin[4*(c + d*x)] - 1470*a*b^7*Sin[4*(c + d*x)] - 24*a^7*b*Sin[6*(c + d*x)] + 2756*a^5*b^3*Sin[6*(c + d*x)] + 7670*a ^3*b^5*Sin[6*(c + d*x)] + 630*a*b^7*Sin[6*(c + d*x)] - 24*a^7*b*Sin[8*(c + d*x)] - 922*a^5*b^3*Sin[8*(c + d*x)] - 2095*a^3*b^5*Sin[8*(c + d*x)] - 10 5*a*b^7*Sin[8*(c + d*x)])))/(1920*a^9*d*(a + b*Tan[c + d*x])^4)
Time = 0.54 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.93, 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.
\(\displaystyle \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^6 (a+b \tan (c+d x))^4}dx\) |
\(\Big \downarrow \) 3999 |
\(\displaystyle \frac {b \int \frac {\cot ^6(c+d x) \left (\tan ^2(c+d x) b^2+b^2\right )^2}{b^6 (a+b \tan (c+d x))^4}d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {b \int \left (\frac {\cot ^6(c+d x)}{a^4 b^2}-\frac {4 \cot ^5(c+d x)}{a^5 b}+\frac {2 \left (a^2+5 b^2\right ) \cot ^4(c+d x)}{a^6 b^2}-\frac {4 \left (5 b^4+2 a^2 b^2\right ) \cot ^3(c+d x)}{a^7 b^3}+\frac {\left (a^4+20 b^2 a^2+35 b^4\right ) \cot ^2(c+d x)}{a^8 b^2}-\frac {4 \left (a^4+10 b^2 a^2+14 b^4\right ) \cot (c+d x)}{a^9 b}+\frac {4 \left (a^4+10 b^2 a^2+14 b^4\right )}{a^9 (a+b \tan (c+d x))}+\frac {3 a^4+20 b^2 a^2+21 b^4}{a^8 (a+b \tan (c+d x))^2}+\frac {2 \left (a^4+4 b^2 a^2+3 b^4\right )}{a^7 (a+b \tan (c+d x))^3}+\frac {\left (a^2+b^2\right )^2}{a^6 (a+b \tan (c+d x))^4}\right )d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (\frac {\cot ^4(c+d x)}{a^5}-\frac {\cot ^5(c+d x)}{5 a^4 b}-\frac {\left (a^2+b^2\right ) \left (a^2+3 b^2\right )}{a^7 (a+b \tan (c+d x))^2}+\frac {2 \left (2 a^2+5 b^2\right ) \cot ^2(c+d x)}{a^7}-\frac {\left (a^2+b^2\right )^2}{3 a^6 (a+b \tan (c+d x))^3}-\frac {2 \left (a^2+5 b^2\right ) \cot ^3(c+d x)}{3 a^6 b}-\frac {4 \left (a^4+10 a^2 b^2+14 b^4\right ) \log (b \tan (c+d x))}{a^9}+\frac {4 \left (a^4+10 a^2 b^2+14 b^4\right ) \log (a+b \tan (c+d x))}{a^9}-\frac {3 a^4+20 a^2 b^2+21 b^4}{a^8 (a+b \tan (c+d x))}-\frac {\left (a^4+20 a^2 b^2+35 b^4\right ) \cot (c+d x)}{a^8 b}\right )}{d}\) |
(b*(-(((a^4 + 20*a^2*b^2 + 35*b^4)*Cot[c + d*x])/(a^8*b)) + (2*(2*a^2 + 5* b^2)*Cot[c + d*x]^2)/a^7 - (2*(a^2 + 5*b^2)*Cot[c + d*x]^3)/(3*a^6*b) + Co t[c + d*x]^4/a^5 - Cot[c + d*x]^5/(5*a^4*b) - (4*(a^4 + 10*a^2*b^2 + 14*b^ 4)*Log[b*Tan[c + d*x]])/a^9 + (4*(a^4 + 10*a^2*b^2 + 14*b^4)*Log[a + b*Tan [c + d*x]])/a^9 - (a^2 + b^2)^2/(3*a^6*(a + b*Tan[c + d*x])^3) - ((a^2 + b ^2)*(a^2 + 3*b^2))/(a^7*(a + b*Tan[c + d*x])^2) - (3*a^4 + 20*a^2*b^2 + 21 *b^4)/(a^8*(a + b*Tan[c + d*x]))))/d
3.1.77.3.1 Defintions of rubi rules used
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]
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]
Time = 7.96 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {-\frac {b \left (3 a^{4}+20 a^{2} b^{2}+21 b^{4}\right )}{a^{8} \left (a +b \tan \left (d x +c \right )\right )}-\frac {\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) b}{3 a^{6} \left (a +b \tan \left (d x +c \right )\right )^{3}}-\frac {b \left (a^{4}+4 a^{2} b^{2}+3 b^{4}\right )}{a^{7} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {4 b \left (a^{4}+10 a^{2} b^{2}+14 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{9}}-\frac {1}{5 a^{4} \tan \left (d x +c \right )^{5}}-\frac {2 a^{2}+10 b^{2}}{3 a^{6} \tan \left (d x +c \right )^{3}}-\frac {a^{4}+20 a^{2} b^{2}+35 b^{4}}{a^{8} \tan \left (d x +c \right )}+\frac {b}{a^{5} \tan \left (d x +c \right )^{4}}+\frac {2 b \left (2 a^{2}+5 b^{2}\right )}{a^{7} \tan \left (d x +c \right )^{2}}-\frac {4 b \left (a^{4}+10 a^{2} b^{2}+14 b^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{9}}}{d}\) | \(280\) |
default | \(\frac {-\frac {b \left (3 a^{4}+20 a^{2} b^{2}+21 b^{4}\right )}{a^{8} \left (a +b \tan \left (d x +c \right )\right )}-\frac {\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) b}{3 a^{6} \left (a +b \tan \left (d x +c \right )\right )^{3}}-\frac {b \left (a^{4}+4 a^{2} b^{2}+3 b^{4}\right )}{a^{7} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {4 b \left (a^{4}+10 a^{2} b^{2}+14 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{9}}-\frac {1}{5 a^{4} \tan \left (d x +c \right )^{5}}-\frac {2 a^{2}+10 b^{2}}{3 a^{6} \tan \left (d x +c \right )^{3}}-\frac {a^{4}+20 a^{2} b^{2}+35 b^{4}}{a^{8} \tan \left (d x +c \right )}+\frac {b}{a^{5} \tan \left (d x +c \right )^{4}}+\frac {2 b \left (2 a^{2}+5 b^{2}\right )}{a^{7} \tan \left (d x +c \right )^{2}}-\frac {4 b \left (a^{4}+10 a^{2} b^{2}+14 b^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{9}}}{d}\) | \(280\) |
risch | \(\text {Expression too large to display}\) | \(1317\) |
1/d*(-b*(3*a^4+20*a^2*b^2+21*b^4)/a^8/(a+b*tan(d*x+c))-1/3*(a^4+2*a^2*b^2+ b^4)*b/a^6/(a+b*tan(d*x+c))^3-b*(a^4+4*a^2*b^2+3*b^4)/a^7/(a+b*tan(d*x+c)) ^2+4*b*(a^4+10*a^2*b^2+14*b^4)/a^9*ln(a+b*tan(d*x+c))-1/5/a^4/tan(d*x+c)^5 -1/3*(2*a^2+10*b^2)/a^6/tan(d*x+c)^3-(a^4+20*a^2*b^2+35*b^4)/a^8/tan(d*x+c )+1/a^5*b/tan(d*x+c)^4+2*b*(2*a^2+5*b^2)/a^7/tan(d*x+c)^2-4*b*(a^4+10*a^2* b^2+14*b^4)/a^9*ln(tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 1536 vs. \(2 (294) = 588\).
Time = 0.37 (sec) , antiderivative size = 1536, normalized size of antiderivative = 5.12 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\text {Too large to display} \]
-1/15*(110*a^6*b^4 + 510*a^4*b^6 + 420*a^2*b^8 - 4*(2*a^10 + 81*a^8*b^2 + 29*a^6*b^4 - 660*a^4*b^6 - 630*a^2*b^8)*cos(d*x + c)^8 + 2*(10*a^10 + 423* a^8*b^2 - 47*a^6*b^4 - 4320*a^4*b^6 - 3990*a^2*b^8)*cos(d*x + c)^6 - 15*(a ^10 + 47*a^8*b^2 - 44*a^6*b^4 - 658*a^4*b^6 - 588*a^2*b^8)*cos(d*x + c)^4 + 20*(9*a^8*b^2 - 28*a^6*b^4 - 219*a^4*b^6 - 189*a^2*b^8)*cos(d*x + c)^2 + 30*(a^6*b^4 + 11*a^4*b^6 + 24*a^2*b^8 + 14*b^10 - (3*a^8*b^2 + 32*a^6*b^4 + 61*a^4*b^6 + 18*a^2*b^8 - 14*b^10)*cos(d*x + c)^8 + (9*a^8*b^2 + 95*a^6 *b^4 + 172*a^4*b^6 + 30*a^2*b^8 - 56*b^10)*cos(d*x + c)^6 - 3*(3*a^8*b^2 + 31*a^6*b^4 + 50*a^4*b^6 - 6*a^2*b^8 - 28*b^10)*cos(d*x + c)^4 + (3*a^8*b^ 2 + 29*a^6*b^4 + 28*a^4*b^6 - 54*a^2*b^8 - 56*b^10)*cos(d*x + c)^2 + ((a^9 *b + 8*a^7*b^3 - 9*a^5*b^5 - 58*a^3*b^7 - 42*a*b^9)*cos(d*x + c)^7 - (2*a^ 9*b + 13*a^7*b^3 - 51*a^5*b^5 - 188*a^3*b^7 - 126*a*b^9)*cos(d*x + c)^5 + (a^9*b + 2*a^7*b^3 - 75*a^5*b^5 - 202*a^3*b^7 - 126*a*b^9)*cos(d*x + c)^3 + 3*(a^7*b^3 + 11*a^5*b^5 + 24*a^3*b^7 + 14*a*b^9)*cos(d*x + c))*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^ 2) - 30*(a^6*b^4 + 11*a^4*b^6 + 24*a^2*b^8 + 14*b^10 - (3*a^8*b^2 + 32*a^6 *b^4 + 61*a^4*b^6 + 18*a^2*b^8 - 14*b^10)*cos(d*x + c)^8 + (9*a^8*b^2 + 95 *a^6*b^4 + 172*a^4*b^6 + 30*a^2*b^8 - 56*b^10)*cos(d*x + c)^6 - 3*(3*a^8*b ^2 + 31*a^6*b^4 + 50*a^4*b^6 - 6*a^2*b^8 - 28*b^10)*cos(d*x + c)^4 + (3*a^ 8*b^2 + 29*a^6*b^4 + 28*a^4*b^6 - 54*a^2*b^8 - 56*b^10)*cos(d*x + c)^2 ...
\[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\int \frac {\csc ^{6}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{4}}\, dx \]
Time = 0.65 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.08 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\frac {\frac {6 \, a^{6} b \tan \left (d x + c\right ) - 60 \, {\left (a^{4} b^{3} + 10 \, a^{2} b^{5} + 14 \, b^{7}\right )} \tan \left (d x + c\right )^{7} - 3 \, a^{7} - 150 \, {\left (a^{5} b^{2} + 10 \, a^{3} b^{4} + 14 \, a b^{6}\right )} \tan \left (d x + c\right )^{6} - 110 \, {\left (a^{6} b + 10 \, a^{4} b^{3} + 14 \, a^{2} b^{5}\right )} \tan \left (d x + c\right )^{5} - 15 \, {\left (a^{7} + 10 \, a^{5} b^{2} + 14 \, a^{3} b^{4}\right )} \tan \left (d x + c\right )^{4} + 6 \, {\left (5 \, a^{6} b + 7 \, a^{4} b^{3}\right )} \tan \left (d x + c\right )^{3} - 2 \, {\left (5 \, a^{7} + 7 \, a^{5} b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{8} b^{3} \tan \left (d x + c\right )^{8} + 3 \, a^{9} b^{2} \tan \left (d x + c\right )^{7} + 3 \, a^{10} b \tan \left (d x + c\right )^{6} + a^{11} \tan \left (d x + c\right )^{5}} + \frac {60 \, {\left (a^{4} b + 10 \, a^{2} b^{3} + 14 \, b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{9}} - \frac {60 \, {\left (a^{4} b + 10 \, a^{2} b^{3} + 14 \, b^{5}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{9}}}{15 \, d} \]
1/15*((6*a^6*b*tan(d*x + c) - 60*(a^4*b^3 + 10*a^2*b^5 + 14*b^7)*tan(d*x + c)^7 - 3*a^7 - 150*(a^5*b^2 + 10*a^3*b^4 + 14*a*b^6)*tan(d*x + c)^6 - 110 *(a^6*b + 10*a^4*b^3 + 14*a^2*b^5)*tan(d*x + c)^5 - 15*(a^7 + 10*a^5*b^2 + 14*a^3*b^4)*tan(d*x + c)^4 + 6*(5*a^6*b + 7*a^4*b^3)*tan(d*x + c)^3 - 2*( 5*a^7 + 7*a^5*b^2)*tan(d*x + c)^2)/(a^8*b^3*tan(d*x + c)^8 + 3*a^9*b^2*tan (d*x + c)^7 + 3*a^10*b*tan(d*x + c)^6 + a^11*tan(d*x + c)^5) + 60*(a^4*b + 10*a^2*b^3 + 14*b^5)*log(b*tan(d*x + c) + a)/a^9 - 60*(a^4*b + 10*a^2*b^3 + 14*b^5)*log(tan(d*x + c))/a^9)/d
Time = 0.76 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.43 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^4} \, dx=-\frac {\frac {60 \, {\left (a^{4} b + 10 \, a^{2} b^{3} + 14 \, b^{5}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{9}} - \frac {60 \, {\left (a^{4} b^{2} + 10 \, a^{2} b^{4} + 14 \, b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{9} b} + \frac {5 \, {\left (22 \, a^{4} b^{4} \tan \left (d x + c\right )^{3} + 220 \, a^{2} b^{6} \tan \left (d x + c\right )^{3} + 308 \, b^{8} \tan \left (d x + c\right )^{3} + 75 \, a^{5} b^{3} \tan \left (d x + c\right )^{2} + 720 \, a^{3} b^{5} \tan \left (d x + c\right )^{2} + 987 \, a b^{7} \tan \left (d x + c\right )^{2} + 87 \, a^{6} b^{2} \tan \left (d x + c\right ) + 792 \, a^{4} b^{4} \tan \left (d x + c\right ) + 1059 \, a^{2} b^{6} \tan \left (d x + c\right ) + 35 \, a^{7} b + 294 \, a^{5} b^{3} + 381 \, a^{3} b^{5}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3} a^{9}} - \frac {137 \, a^{4} b \tan \left (d x + c\right )^{5} + 1370 \, a^{2} b^{3} \tan \left (d x + c\right )^{5} + 1918 \, b^{5} \tan \left (d x + c\right )^{5} - 15 \, a^{5} \tan \left (d x + c\right )^{4} - 300 \, a^{3} b^{2} \tan \left (d x + c\right )^{4} - 525 \, a b^{4} \tan \left (d x + c\right )^{4} + 60 \, a^{4} b \tan \left (d x + c\right )^{3} + 150 \, a^{2} b^{3} \tan \left (d x + c\right )^{3} - 10 \, a^{5} \tan \left (d x + c\right )^{2} - 50 \, a^{3} b^{2} \tan \left (d x + c\right )^{2} + 15 \, a^{4} b \tan \left (d x + c\right ) - 3 \, a^{5}}{a^{9} \tan \left (d x + c\right )^{5}}}{15 \, d} \]
-1/15*(60*(a^4*b + 10*a^2*b^3 + 14*b^5)*log(abs(tan(d*x + c)))/a^9 - 60*(a ^4*b^2 + 10*a^2*b^4 + 14*b^6)*log(abs(b*tan(d*x + c) + a))/(a^9*b) + 5*(22 *a^4*b^4*tan(d*x + c)^3 + 220*a^2*b^6*tan(d*x + c)^3 + 308*b^8*tan(d*x + c )^3 + 75*a^5*b^3*tan(d*x + c)^2 + 720*a^3*b^5*tan(d*x + c)^2 + 987*a*b^7*t an(d*x + c)^2 + 87*a^6*b^2*tan(d*x + c) + 792*a^4*b^4*tan(d*x + c) + 1059* a^2*b^6*tan(d*x + c) + 35*a^7*b + 294*a^5*b^3 + 381*a^3*b^5)/((b*tan(d*x + c) + a)^3*a^9) - (137*a^4*b*tan(d*x + c)^5 + 1370*a^2*b^3*tan(d*x + c)^5 + 1918*b^5*tan(d*x + c)^5 - 15*a^5*tan(d*x + c)^4 - 300*a^3*b^2*tan(d*x + c)^4 - 525*a*b^4*tan(d*x + c)^4 + 60*a^4*b*tan(d*x + c)^3 + 150*a^2*b^3*ta n(d*x + c)^3 - 10*a^5*tan(d*x + c)^2 - 50*a^3*b^2*tan(d*x + c)^2 + 15*a^4* b*tan(d*x + c) - 3*a^5)/(a^9*tan(d*x + c)^5))/d
Time = 6.83 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\frac {8\,b\,\mathrm {atanh}\left (\frac {4\,b\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^4+10\,a^2\,b^2+14\,b^4\right )}{a\,\left (4\,a^4\,b+40\,a^2\,b^3+56\,b^5\right )}\right )\,\left (a^4+10\,a^2\,b^2+14\,b^4\right )}{a^9\,d}-\frac {\frac {1}{5\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^4+10\,a^2\,b^2+14\,b^4\right )}{a^5}+\frac {2\,{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (5\,a^2+7\,b^2\right )}{15\,a^3}-\frac {2\,b\,\mathrm {tan}\left (c+d\,x\right )}{5\,a^2}+\frac {22\,b\,{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (a^4+10\,a^2\,b^2+14\,b^4\right )}{3\,a^6}+\frac {10\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^6\,\left (a^4+10\,a^2\,b^2+14\,b^4\right )}{a^7}+\frac {4\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^7\,\left (a^4+10\,a^2\,b^2+14\,b^4\right )}{a^8}-\frac {2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (5\,a^2+7\,b^2\right )}{5\,a^4}}{d\,\left (a^3\,{\mathrm {tan}\left (c+d\,x\right )}^5+3\,a^2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^6+3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^7+b^3\,{\mathrm {tan}\left (c+d\,x\right )}^8\right )} \]
(8*b*atanh((4*b*(a + 2*b*tan(c + d*x))*(a^4 + 14*b^4 + 10*a^2*b^2))/(a*(4* a^4*b + 56*b^5 + 40*a^2*b^3)))*(a^4 + 14*b^4 + 10*a^2*b^2))/(a^9*d) - (1/( 5*a) + (tan(c + d*x)^4*(a^4 + 14*b^4 + 10*a^2*b^2))/a^5 + (2*tan(c + d*x)^ 2*(5*a^2 + 7*b^2))/(15*a^3) - (2*b*tan(c + d*x))/(5*a^2) + (22*b*tan(c + d *x)^5*(a^4 + 14*b^4 + 10*a^2*b^2))/(3*a^6) + (10*b^2*tan(c + d*x)^6*(a^4 + 14*b^4 + 10*a^2*b^2))/a^7 + (4*b^3*tan(c + d*x)^7*(a^4 + 14*b^4 + 10*a^2* b^2))/a^8 - (2*b*tan(c + d*x)^3*(5*a^2 + 7*b^2))/(5*a^4))/(d*(a^3*tan(c + d*x)^5 + b^3*tan(c + d*x)^8 + 3*a^2*b*tan(c + d*x)^6 + 3*a*b^2*tan(c + d*x )^7))