Integrand size = 21, antiderivative size = 192 \[ \int \csc ^8(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15 a^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {13 a^3 \cot (c+d x)}{d}-\frac {7 a^3 \cot ^3(c+d x)}{d}-\frac {3 a^3 \cot ^5(c+d x)}{d}-\frac {4 a^3 \cot ^7(c+d x)}{7 d}-\frac {15 a^3 \csc (c+d x)}{2 d}-\frac {5 a^3 \csc ^3(c+d x)}{2 d}-\frac {3 a^3 \csc ^5(c+d x)}{2 d}-\frac {15 a^3 \csc ^7(c+d x)}{14 d}+\frac {a^3 \csc ^7(c+d x) \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \tan (c+d x)}{d} \]
15/2*a^3*arctanh(sin(d*x+c))/d-13*a^3*cot(d*x+c)/d-7*a^3*cot(d*x+c)^3/d-3* a^3*cot(d*x+c)^5/d-4/7*a^3*cot(d*x+c)^7/d-15/2*a^3*csc(d*x+c)/d-5/2*a^3*cs c(d*x+c)^3/d-3/2*a^3*csc(d*x+c)^5/d-15/14*a^3*csc(d*x+c)^7/d+1/2*a^3*csc(d *x+c)^7*sec(d*x+c)^2/d+3*a^3*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(430\) vs. \(2(192)=384\).
Time = 3.79 (sec) , antiderivative size = 430, normalized size of antiderivative = 2.24 \[ \int \csc ^8(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^3 \cos (c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^3 \left (-860160 \cos ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+860160 \cos ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 \csc (2 c) \csc ^6\left (\frac {1}{2} (c+d x)\right ) \csc (c+d x) (5264 \sin (2 c)-9580 \sin (d x)+8480 \sin (2 d x)+2776 \sin (c-d x)-6080 \sin (c+d x)+8816 \sin (2 (c+d x))-7904 \sin (3 (c+d x))+4864 \sin (4 (c+d x))-1824 \sin (5 (c+d x))+304 \sin (6 (c+d x))-9580 \sin (2 c+d x)-10024 \sin (3 c+d x)+13891 \sin (c+2 d x)+7720 \sin (2 (c+2 d x))+13891 \sin (3 c+2 d x)+10080 \sin (4 c+2 d x)-10060 \sin (c+3 d x)-12454 \sin (2 c+3 d x)-12454 \sin (4 c+3 d x)-6580 \sin (5 c+3 d x)+7664 \sin (3 c+4 d x)+7664 \sin (5 c+4 d x)+2520 \sin (6 c+4 d x)-3420 \sin (3 c+5 d x)-2874 \sin (4 c+5 d x)-2874 \sin (6 c+5 d x)-420 \sin (7 c+5 d x)+640 \sin (4 c+6 d x)+479 \sin (5 c+6 d x)+479 \sin (7 c+6 d x))\right )}{917504 d} \]
(a^3*Cos[c + d*x]*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(-860160*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 860160*Cos[c + d*x]^2*L og[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 8*Csc[2*c]*Csc[(c + d*x)/2]^6*Cs c[c + d*x]*(5264*Sin[2*c] - 9580*Sin[d*x] + 8480*Sin[2*d*x] + 2776*Sin[c - d*x] - 6080*Sin[c + d*x] + 8816*Sin[2*(c + d*x)] - 7904*Sin[3*(c + d*x)] + 4864*Sin[4*(c + d*x)] - 1824*Sin[5*(c + d*x)] + 304*Sin[6*(c + d*x)] - 9 580*Sin[2*c + d*x] - 10024*Sin[3*c + d*x] + 13891*Sin[c + 2*d*x] + 7720*Si n[2*(c + 2*d*x)] + 13891*Sin[3*c + 2*d*x] + 10080*Sin[4*c + 2*d*x] - 10060 *Sin[c + 3*d*x] - 12454*Sin[2*c + 3*d*x] - 12454*Sin[4*c + 3*d*x] - 6580*S in[5*c + 3*d*x] + 7664*Sin[3*c + 4*d*x] + 7664*Sin[5*c + 4*d*x] + 2520*Sin [6*c + 4*d*x] - 3420*Sin[3*c + 5*d*x] - 2874*Sin[4*c + 5*d*x] - 2874*Sin[6 *c + 5*d*x] - 420*Sin[7*c + 5*d*x] + 640*Sin[4*c + 6*d*x] + 479*Sin[5*c + 6*d*x] + 479*Sin[7*c + 6*d*x])))/(917504*d)
Time = 0.97 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4360, 25, 25, 3042, 25, 3352, 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.
| \(\displaystyle \int \csc ^8(c+d x) (a \sec (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3}{\cos \left (c+d x-\frac {\pi }{2}\right )^8}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \csc ^8(c+d x) \sec ^3(c+d x) \left (-(a (-\cos (c+d x))-a)^3\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -(\cos (c+d x) a+a)^3 \csc ^8(c+d x) \sec ^3(c+d x)dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \csc ^8(c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}{\sin \left (c+d x-\frac {\pi }{2}\right )^3 \cos \left (c+d x-\frac {\pi }{2}\right )^8}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\left (a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^3}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^8 \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )^3}dx\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle -\int \left (-a^3 \sec ^8\left (\frac {1}{2} (2 c-\pi )+d x\right )-a^3 \sec ^3(c+d x) \sec ^8\left (\frac {1}{2} (2 c-\pi )+d x\right )-3 a^3 \sec ^2(c+d x) \sec ^8\left (\frac {1}{2} (2 c-\pi )+d x\right )-3 a^3 \sec (c+d x) \sec ^8\left (\frac {1}{2} (2 c-\pi )+d x\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {15 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}-\frac {4 a^3 \cot ^7(c+d x)}{7 d}-\frac {3 a^3 \cot ^5(c+d x)}{d}-\frac {7 a^3 \cot ^3(c+d x)}{d}-\frac {13 a^3 \cot (c+d x)}{d}-\frac {15 a^3 \csc ^7(c+d x)}{14 d}-\frac {3 a^3 \csc ^5(c+d x)}{2 d}-\frac {5 a^3 \csc ^3(c+d x)}{2 d}-\frac {15 a^3 \csc (c+d x)}{2 d}+\frac {a^3 \csc ^7(c+d x) \sec ^2(c+d x)}{2 d}\) |
(15*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (13*a^3*Cot[c + d*x])/d - (7*a^3*Co t[c + d*x]^3)/d - (3*a^3*Cot[c + d*x]^5)/d - (4*a^3*Cot[c + d*x]^7)/(7*d) - (15*a^3*Csc[c + d*x])/(2*d) - (5*a^3*Csc[c + d*x]^3)/(2*d) - (3*a^3*Csc[ c + d*x]^5)/(2*d) - (15*a^3*Csc[c + d*x]^7)/(14*d) + (a^3*Csc[c + d*x]^7*S ec[c + d*x]^2)/(2*d) + (3*a^3*Tan[c + d*x])/d
3.1.55.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 1.56 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00
| method | result | size |
| norman | \(\frac {-\frac {a^{3}}{112 d}-\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{28 d}-\frac {85 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{112 d}-\frac {15 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 d}+\frac {395 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16 d}-\frac {57 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4 d}-\frac {a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {15 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {15 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(192\) |
| risch | \(-\frac {i a^{3} \left (105 \,{\mathrm e}^{11 i \left (d x +c \right )}-630 \,{\mathrm e}^{10 i \left (d x +c \right )}+1645 \,{\mathrm e}^{9 i \left (d x +c \right )}-2520 \,{\mathrm e}^{8 i \left (d x +c \right )}+2506 \,{\mathrm e}^{7 i \left (d x +c \right )}-1316 \,{\mathrm e}^{6 i \left (d x +c \right )}-694 \,{\mathrm e}^{5 i \left (d x +c \right )}+2120 \,{\mathrm e}^{4 i \left (d x +c \right )}-2515 \,{\mathrm e}^{3 i \left (d x +c \right )}+1930 \,{\mathrm e}^{2 i \left (d x +c \right )}-855 \,{\mathrm e}^{i \left (d x +c \right )}+160\right )}{7 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{7} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(215\) |
| parallelrisch | \(\frac {5 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \left (-\frac {329}{40}+\frac {21 \left (\frac {\sin \left (6 d x +6 c \right )}{16}+\frac {29 \sin \left (2 d x +2 c \right )}{16}-\frac {3 \sin \left (5 d x +5 c \right )}{8}-\frac {5 \sin \left (d x +c \right )}{4}+\sin \left (4 d x +4 c \right )-\frac {13 \sin \left (3 d x +3 c \right )}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {21 \left (-\frac {\sin \left (6 d x +6 c \right )}{16}-\frac {29 \sin \left (2 d x +2 c \right )}{16}+\frac {3 \sin \left (5 d x +5 c \right )}{8}+\frac {5 \sin \left (d x +c \right )}{4}-\sin \left (4 d x +4 c \right )+\frac {13 \sin \left (3 d x +3 c \right )}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\cos \left (6 d x +6 c \right )+\frac {453 \cos \left (d x +c \right )}{40}-\frac {5 \cos \left (2 d x +2 c \right )}{2}-\frac {87 \cos \left (3 d x +3 c \right )}{16}+\frac {65 \cos \left (4 d x +4 c \right )}{8}-\frac {75 \cos \left (5 d x +5 c \right )}{16}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{112 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(259\) |
| derivativedivides | \(\frac {a^{3} \left (-\frac {1}{7 \sin \left (d x +c \right )^{7} \cos \left (d x +c \right )^{2}}-\frac {9}{35 \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{2}}-\frac {3}{5 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {3}{2 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {9}{2 \sin \left (d x +c \right )}+\frac {9 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (-\frac {1}{7 \sin \left (d x +c \right )^{7} \cos \left (d x +c \right )}-\frac {8}{35 \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )}-\frac {16}{35 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {64}{35 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {128 \cot \left (d x +c \right )}{35}\right )+3 a^{3} \left (-\frac {1}{7 \sin \left (d x +c \right )^{7}}-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {16}{35}-\frac {\csc \left (d x +c \right )^{6}}{7}-\frac {6 \csc \left (d x +c \right )^{4}}{35}-\frac {8 \csc \left (d x +c \right )^{2}}{35}\right ) \cot \left (d x +c \right )}{d}\) | \(297\) |
| default | \(\frac {a^{3} \left (-\frac {1}{7 \sin \left (d x +c \right )^{7} \cos \left (d x +c \right )^{2}}-\frac {9}{35 \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{2}}-\frac {3}{5 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {3}{2 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {9}{2 \sin \left (d x +c \right )}+\frac {9 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (-\frac {1}{7 \sin \left (d x +c \right )^{7} \cos \left (d x +c \right )}-\frac {8}{35 \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )}-\frac {16}{35 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {64}{35 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {128 \cot \left (d x +c \right )}{35}\right )+3 a^{3} \left (-\frac {1}{7 \sin \left (d x +c \right )^{7}}-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {16}{35}-\frac {\csc \left (d x +c \right )^{6}}{7}-\frac {6 \csc \left (d x +c \right )^{4}}{35}-\frac {8 \csc \left (d x +c \right )^{2}}{35}\right ) \cot \left (d x +c \right )}{d}\) | \(297\) |
(-1/112/d*a^3-3/28*a^3/d*tan(1/2*d*x+1/2*c)^2-85/112*a^3/d*tan(1/2*d*x+1/2 *c)^4-15/2*a^3/d*tan(1/2*d*x+1/2*c)^6+395/16*a^3/d*tan(1/2*d*x+1/2*c)^8-57 /4*a^3/d*tan(1/2*d*x+1/2*c)^10-1/16*a^3/d*tan(1/2*d*x+1/2*c)^12)/tan(1/2*d *x+1/2*c)^7/(-1+tan(1/2*d*x+1/2*c)^2)^2-15/2/d*a^3*ln(tan(1/2*d*x+1/2*c)-1 )+15/2/d*a^3*ln(tan(1/2*d*x+1/2*c)+1)
Time = 0.28 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.45 \[ \int \csc ^8(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {320 \, a^{3} \cos \left (d x + c\right )^{6} - 750 \, a^{3} \cos \left (d x + c\right )^{5} + 170 \, a^{3} \cos \left (d x + c\right )^{4} + 720 \, a^{3} \cos \left (d x + c\right )^{3} - 520 \, a^{3} \cos \left (d x + c\right )^{2} + 42 \, a^{3} \cos \left (d x + c\right ) + 14 \, a^{3} - 105 \, {\left (a^{3} \cos \left (d x + c\right )^{5} - 3 \, a^{3} \cos \left (d x + c\right )^{4} + 3 \, a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 105 \, {\left (a^{3} \cos \left (d x + c\right )^{5} - 3 \, a^{3} \cos \left (d x + c\right )^{4} + 3 \, a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{28 \, {\left (d \cos \left (d x + c\right )^{5} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )} \]
-1/28*(320*a^3*cos(d*x + c)^6 - 750*a^3*cos(d*x + c)^5 + 170*a^3*cos(d*x + c)^4 + 720*a^3*cos(d*x + c)^3 - 520*a^3*cos(d*x + c)^2 + 42*a^3*cos(d*x + c) + 14*a^3 - 105*(a^3*cos(d*x + c)^5 - 3*a^3*cos(d*x + c)^4 + 3*a^3*cos( d*x + c)^3 - a^3*cos(d*x + c)^2)*log(sin(d*x + c) + 1)*sin(d*x + c) + 105* (a^3*cos(d*x + c)^5 - 3*a^3*cos(d*x + c)^4 + 3*a^3*cos(d*x + c)^3 - a^3*co s(d*x + c)^2)*log(-sin(d*x + c) + 1)*sin(d*x + c))/((d*cos(d*x + c)^5 - 3* d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^3 - d*cos(d*x + c)^2)*sin(d*x + c))
Timed out. \[ \int \csc ^8(c+d x) (a+a \sec (c+d x))^3 \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.40 \[ \int \csc ^8(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {a^{3} {\left (\frac {2 \, {\left (315 \, \sin \left (d x + c\right )^{8} - 210 \, \sin \left (d x + c\right )^{6} - 42 \, \sin \left (d x + c\right )^{4} - 18 \, \sin \left (d x + c\right )^{2} - 10\right )}}{\sin \left (d x + c\right )^{9} - \sin \left (d x + c\right )^{7}} - 315 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 315 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{3} {\left (\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{6} + 35 \, \sin \left (d x + c\right )^{4} + 21 \, \sin \left (d x + c\right )^{2} + 15\right )}}{\sin \left (d x + c\right )^{7}} - 105 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} {\left (\frac {140 \, \tan \left (d x + c\right )^{6} + 70 \, \tan \left (d x + c\right )^{4} + 28 \, \tan \left (d x + c\right )^{2} + 5}{\tan \left (d x + c\right )^{7}} - 35 \, \tan \left (d x + c\right )\right )} + \frac {4 \, {\left (35 \, \tan \left (d x + c\right )^{6} + 35 \, \tan \left (d x + c\right )^{4} + 21 \, \tan \left (d x + c\right )^{2} + 5\right )} a^{3}}{\tan \left (d x + c\right )^{7}}}{140 \, d} \]
-1/140*(a^3*(2*(315*sin(d*x + c)^8 - 210*sin(d*x + c)^6 - 42*sin(d*x + c)^ 4 - 18*sin(d*x + c)^2 - 10)/(sin(d*x + c)^9 - sin(d*x + c)^7) - 315*log(si n(d*x + c) + 1) + 315*log(sin(d*x + c) - 1)) + 2*a^3*(2*(105*sin(d*x + c)^ 6 + 35*sin(d*x + c)^4 + 21*sin(d*x + c)^2 + 15)/sin(d*x + c)^7 - 105*log(s in(d*x + c) + 1) + 105*log(sin(d*x + c) - 1)) + 12*a^3*((140*tan(d*x + c)^ 6 + 70*tan(d*x + c)^4 + 28*tan(d*x + c)^2 + 5)/tan(d*x + c)^7 - 35*tan(d*x + c)) + 4*(35*tan(d*x + c)^6 + 35*tan(d*x + c)^4 + 21*tan(d*x + c)^2 + 5) *a^3/tan(d*x + c)^7)/d
Time = 0.43 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.88 \[ \int \csc ^8(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {840 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 840 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {112 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac {1050 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 112 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 14 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}}}{112 \, d} \]
1/112*(840*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 840*a^3*log(abs(tan(1/ 2*d*x + 1/2*c) - 1)) - 7*a^3*tan(1/2*d*x + 1/2*c) - 112*(5*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 - (1050*a^3*tan(1/2*d*x + 1/2*c)^6 + 112*a^3*tan(1/2*d*x + 1/2*c)^4 + 14*a^ 3*tan(1/2*d*x + 1/2*c)^2 + a^3)/tan(1/2*d*x + 1/2*c)^7)/d
Time = 16.14 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.88 \[ \int \csc ^8(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {230\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-396\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+120\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {85\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}+\frac {12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{7}+\frac {a^3}{7}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\right )}-\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d} \]
(15*a^3*atanh(tan(c/2 + (d*x)/2)))/d - ((12*a^3*tan(c/2 + (d*x)/2)^2)/7 + (85*a^3*tan(c/2 + (d*x)/2)^4)/7 + 120*a^3*tan(c/2 + (d*x)/2)^6 - 396*a^3*t an(c/2 + (d*x)/2)^8 + 230*a^3*tan(c/2 + (d*x)/2)^10 + a^3/7)/(d*(16*tan(c/ 2 + (d*x)/2)^7 - 32*tan(c/2 + (d*x)/2)^9 + 16*tan(c/2 + (d*x)/2)^11)) - (a ^3*tan(c/2 + (d*x)/2))/(16*d)