Integrand size = 21, antiderivative size = 154 \[ \int \frac {\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {63 \text {arctanh}(\sin (c+d x))}{256 a d}-\frac {63 \sec (c+d x) \tan (c+d x)}{256 a d}+\frac {21 \sec (c+d x) \tan ^3(c+d x)}{128 a d}-\frac {21 \sec (c+d x) \tan ^5(c+d x)}{160 a d}+\frac {9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac {\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac {\tan ^{10}(c+d x)}{10 a d} \]
63/256*arctanh(sin(d*x+c))/a/d-63/256*sec(d*x+c)*tan(d*x+c)/a/d+21/128*sec (d*x+c)*tan(d*x+c)^3/a/d-21/160*sec(d*x+c)*tan(d*x+c)^5/a/d+9/80*sec(d*x+c )*tan(d*x+c)^7/a/d-1/10*sec(d*x+c)*tan(d*x+c)^9/a/d+1/10*tan(d*x+c)^10/a/d
Time = 1.55 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {630 \text {arctanh}(\sin (c+d x))+\frac {2 \left (128-187 \sin (c+d x)-827 \sin ^2(c+d x)+643 \sin ^3(c+d x)+1923 \sin ^4(c+d x)-765 \sin ^5(c+d x)-2045 \sin ^6(c+d x)+325 \sin ^7(c+d x)+965 \sin ^8(c+d x)\right )}{(-1+\sin (c+d x))^4 (1+\sin (c+d x))^5}}{2560 a d} \]
(630*ArcTanh[Sin[c + d*x]] + (2*(128 - 187*Sin[c + d*x] - 827*Sin[c + d*x] ^2 + 643*Sin[c + d*x]^3 + 1923*Sin[c + d*x]^4 - 765*Sin[c + d*x]^5 - 2045* Sin[c + d*x]^6 + 325*Sin[c + d*x]^7 + 965*Sin[c + d*x]^8))/((-1 + Sin[c + d*x])^4*(1 + Sin[c + d*x])^5))/(2560*a*d)
Time = 0.85 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {3042, 3185, 3042, 3087, 15, 3091, 3042, 3091, 3042, 3091, 3042, 3091, 3042, 3091, 3042, 4257}
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.
| \(\displaystyle \int \frac {\tan ^9(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^9}{a \sin (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3185 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) \tan ^9(c+d x)dx}{a}-\frac {\int \sec (c+d x) \tan ^{10}(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^9dx}{a}-\frac {\int \sec (c+d x) \tan (c+d x)^{10}dx}{a}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\int \tan ^9(c+d x)d\tan (c+d x)}{a d}-\frac {\int \sec (c+d x) \tan (c+d x)^{10}dx}{a}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\int \sec (c+d x) \tan (c+d x)^{10}dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \int \sec (c+d x) \tan ^8(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \int \sec (c+d x) \tan (c+d x)^8dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \left (\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \int \sec (c+d x) \tan ^6(c+d x)dx\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \left (\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \int \sec (c+d x) \tan (c+d x)^6dx\right )}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \left (\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \int \sec (c+d x) \tan ^4(c+d x)dx\right )\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \left (\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \int \sec (c+d x) \tan (c+d x)^4dx\right )\right )}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \left (\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan ^2(c+d x)dx\right )\right )\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \left (\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan (c+d x)^2dx\right )\right )\right )}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \left (\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \sec (c+d x)dx\right )\right )\right )\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \left (\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )\right )\right )}{a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\frac {\tan ^9(c+d x) \sec (c+d x)}{10 d}-\frac {9}{10} \left (\frac {\tan ^7(c+d x) \sec (c+d x)}{8 d}-\frac {7}{8} \left (\frac {\tan ^5(c+d x) \sec (c+d x)}{6 d}-\frac {5}{6} \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {\text {arctanh}(\sin (c+d x))}{2 d}\right )\right )\right )\right )}{a}\) |
Tan[c + d*x]^10/(10*a*d) - ((Sec[c + d*x]*Tan[c + d*x]^9)/(10*d) - (9*((Se c[c + d*x]*Tan[c + d*x]^7)/(8*d) - (7*((Sec[c + d*x]*Tan[c + d*x]^5)/(6*d) - (5*((Sec[c + d*x]*Tan[c + d*x]^3)/(4*d) - (3*(-1/2*ArcTanh[Sin[c + d*x] ]/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/6))/8))/10)/a
3.9.98.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.81 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.90
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {57}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {65}{256 \left (\sin \left (d x +c \right )-1\right )}-\frac {63 \ln \left (\sin \left (d x +c \right )-1\right )}{512}+\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {13}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {23}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {187}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{2 \sin \left (d x +c \right )+2}+\frac {63 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(139\) |
| default | \(\frac {\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {57}{512 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {65}{256 \left (\sin \left (d x +c \right )-1\right )}-\frac {63 \ln \left (\sin \left (d x +c \right )-1\right )}{512}+\frac {1}{160 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {13}{256 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {23}{128 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {187}{512 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{2 \sin \left (d x +c \right )+2}+\frac {63 \ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d a}\) | \(139\) |
| risch | \(\frac {i \left (8708 \,{\mathrm e}^{5 i \left (d x +c \right )}+1570 i {\mathrm e}^{14 i \left (d x +c \right )}+650 i {\mathrm e}^{16 i \left (d x +c \right )}-658 i {\mathrm e}^{8 i \left (d x +c \right )}-3626 i {\mathrm e}^{6 i \left (d x +c \right )}+3626 i {\mathrm e}^{12 i \left (d x +c \right )}+658 i {\mathrm e}^{10 i \left (d x +c \right )}+15470 \,{\mathrm e}^{9 i \left (d x +c \right )}+965 \,{\mathrm e}^{i \left (d x +c \right )}+8708 \,{\mathrm e}^{13 i \left (d x +c \right )}-1484 \,{\mathrm e}^{11 i \left (d x +c \right )}+965 \,{\mathrm e}^{17 i \left (d x +c \right )}+460 \,{\mathrm e}^{15 i \left (d x +c \right )}-1570 i {\mathrm e}^{4 i \left (d x +c \right )}-650 i {\mathrm e}^{2 i \left (d x +c \right )}+460 \,{\mathrm e}^{3 i \left (d x +c \right )}-1484 \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{640 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d a}-\frac {63 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 d a}+\frac {63 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 a d}\) | \(277\) |
| parallelrisch | \(\frac {\left (-8820 \sin \left (3 d x +3 c \right )-6300 \sin \left (5 d x +5 c \right )-2205 \sin \left (7 d x +7 c \right )-315 \sin \left (9 d x +9 c \right )-35280 \cos \left (2 d x +2 c \right )-17640 \cos \left (4 d x +4 c \right )-5040 \cos \left (6 d x +6 c \right )-630 \cos \left (8 d x +8 c \right )-4410 \sin \left (d x +c \right )-22050\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (8820 \sin \left (3 d x +3 c \right )+6300 \sin \left (5 d x +5 c \right )+2205 \sin \left (7 d x +7 c \right )+315 \sin \left (9 d x +9 c \right )+35280 \cos \left (2 d x +2 c \right )+17640 \cos \left (4 d x +4 c \right )+5040 \cos \left (6 d x +6 c \right )+630 \cos \left (8 d x +8 c \right )+4410 \sin \left (d x +c \right )+22050\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2016 \sin \left (3 d x +3 c \right )+600 \sin \left (5 d x +5 c \right )+9 \sin \left (7 d x +7 c \right )+187 \sin \left (9 d x +9 c \right )+17976 \cos \left (2 d x +2 c \right )+27888 \cos \left (4 d x +4 c \right )+3912 \cos \left (6 d x +6 c \right )+2304 \cos \left (8 d x +8 c \right )+1302 \sin \left (d x +c \right )+28560}{1280 a d \left (70+\sin \left (9 d x +9 c \right )+7 \sin \left (7 d x +7 c \right )+20 \sin \left (5 d x +5 c \right )+28 \sin \left (3 d x +3 c \right )+14 \sin \left (d x +c \right )+2 \cos \left (8 d x +8 c \right )+16 \cos \left (6 d x +6 c \right )+56 \cos \left (4 d x +4 c \right )+112 \cos \left (2 d x +2 c \right )\right )}\) | \(427\) |
1/d/a*(1/256/(sin(d*x+c)-1)^4+1/32/(sin(d*x+c)-1)^3+57/512/(sin(d*x+c)-1)^ 2+65/256/(sin(d*x+c)-1)-63/512*ln(sin(d*x+c)-1)+1/160/(1+sin(d*x+c))^5-13/ 256/(1+sin(d*x+c))^4+23/128/(1+sin(d*x+c))^3-187/512/(1+sin(d*x+c))^2+1/2/ (1+sin(d*x+c))+63/512*ln(1+sin(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21 \[ \int \frac {\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {1930 \, \cos \left (d x + c\right )^{8} - 3630 \, \cos \left (d x + c\right )^{6} + 3156 \, \cos \left (d x + c\right )^{4} - 1488 \, \cos \left (d x + c\right )^{2} + 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (325 \, \cos \left (d x + c\right )^{6} - 210 \, \cos \left (d x + c\right )^{4} + 88 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 288}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]
1/2560*(1930*cos(d*x + c)^8 - 3630*cos(d*x + c)^6 + 3156*cos(d*x + c)^4 - 1488*cos(d*x + c)^2 + 315*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*l og(sin(d*x + c) + 1) - 315*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)* log(-sin(d*x + c) + 1) - 2*(325*cos(d*x + c)^6 - 210*cos(d*x + c)^4 + 88*c os(d*x + c)^2 - 16)*sin(d*x + c) + 288)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)
Timed out. \[ \int \frac {\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.39 \[ \int \frac {\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (965 \, \sin \left (d x + c\right )^{8} + 325 \, \sin \left (d x + c\right )^{7} - 2045 \, \sin \left (d x + c\right )^{6} - 765 \, \sin \left (d x + c\right )^{5} + 1923 \, \sin \left (d x + c\right )^{4} + 643 \, \sin \left (d x + c\right )^{3} - 827 \, \sin \left (d x + c\right )^{2} - 187 \, \sin \left (d x + c\right ) + 128\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {315 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {315 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \]
1/2560*(2*(965*sin(d*x + c)^8 + 325*sin(d*x + c)^7 - 2045*sin(d*x + c)^6 - 765*sin(d*x + c)^5 + 1923*sin(d*x + c)^4 + 643*sin(d*x + c)^3 - 827*sin(d *x + c)^2 - 187*sin(d*x + c) + 128)/(a*sin(d*x + c)^9 + a*sin(d*x + c)^8 - 4*a*sin(d*x + c)^7 - 4*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 + 6*a*sin(d* x + c)^4 - 4*a*sin(d*x + c)^3 - 4*a*sin(d*x + c)^2 + a*sin(d*x + c) + a) + 315*log(sin(d*x + c) + 1)/a - 315*log(sin(d*x + c) - 1)/a)/d
Time = 0.43 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.01 \[ \int \frac {\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {1260 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {1260 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (525 \, \sin \left (d x + c\right )^{4} - 1580 \, \sin \left (d x + c\right )^{3} + 1818 \, \sin \left (d x + c\right )^{2} - 932 \, \sin \left (d x + c\right ) + 177\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {2877 \, \sin \left (d x + c\right )^{5} + 9265 \, \sin \left (d x + c\right )^{4} + 12030 \, \sin \left (d x + c\right )^{3} + 7430 \, \sin \left (d x + c\right )^{2} + 1965 \, \sin \left (d x + c\right ) + 113}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \]
1/10240*(1260*log(abs(sin(d*x + c) + 1))/a - 1260*log(abs(sin(d*x + c) - 1 ))/a + 5*(525*sin(d*x + c)^4 - 1580*sin(d*x + c)^3 + 1818*sin(d*x + c)^2 - 932*sin(d*x + c) + 177)/(a*(sin(d*x + c) - 1)^4) - (2877*sin(d*x + c)^5 + 9265*sin(d*x + c)^4 + 12030*sin(d*x + c)^3 + 7430*sin(d*x + c)^2 + 1965*s in(d*x + c) + 113)/(a*(sin(d*x + c) + 1)^5))/d
Time = 17.98 (sec) , antiderivative size = 497, normalized size of antiderivative = 3.23 \[ \int \frac {\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {63\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{128\,a\,d}-\frac {\frac {63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{128}+\frac {63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{64}-\frac {105\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{32}-\frac {483\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{64}+\frac {1407\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{160}+\frac {8043\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{320}-\frac {1779\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{160}-\frac {15159\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{320}+\frac {245\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}-\frac {15159\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{320}-\frac {1779\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{160}+\frac {8043\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{320}+\frac {1407\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160}-\frac {483\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64}-\frac {105\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32}+\frac {63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64}+\frac {63\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}-7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+56\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-112\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+14\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+140\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+14\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-112\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+56\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \]
(63*atanh(tan(c/2 + (d*x)/2)))/(128*a*d) - ((63*tan(c/2 + (d*x)/2))/128 + (63*tan(c/2 + (d*x)/2)^2)/64 - (105*tan(c/2 + (d*x)/2)^3)/32 - (483*tan(c/ 2 + (d*x)/2)^4)/64 + (1407*tan(c/2 + (d*x)/2)^5)/160 + (8043*tan(c/2 + (d* x)/2)^6)/320 - (1779*tan(c/2 + (d*x)/2)^7)/160 - (15159*tan(c/2 + (d*x)/2) ^8)/320 + (245*tan(c/2 + (d*x)/2)^9)/64 - (15159*tan(c/2 + (d*x)/2)^10)/32 0 - (1779*tan(c/2 + (d*x)/2)^11)/160 + (8043*tan(c/2 + (d*x)/2)^12)/320 + (1407*tan(c/2 + (d*x)/2)^13)/160 - (483*tan(c/2 + (d*x)/2)^14)/64 - (105*t an(c/2 + (d*x)/2)^15)/32 + (63*tan(c/2 + (d*x)/2)^16)/64 + (63*tan(c/2 + ( d*x)/2)^17)/128)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 7*a*tan(c/2 + (d*x)/2)^2 - 16*a*tan(c/2 + (d*x)/2)^3 + 20*a*tan(c/2 + (d*x)/2)^4 + 56*a*tan(c/2 + (d*x)/2)^5 - 28*a*tan(c/2 + (d*x)/2)^6 - 112*a*tan(c/2 + (d*x)/2)^7 + 14*a *tan(c/2 + (d*x)/2)^8 + 140*a*tan(c/2 + (d*x)/2)^9 + 14*a*tan(c/2 + (d*x)/ 2)^10 - 112*a*tan(c/2 + (d*x)/2)^11 - 28*a*tan(c/2 + (d*x)/2)^12 + 56*a*ta n(c/2 + (d*x)/2)^13 + 20*a*tan(c/2 + (d*x)/2)^14 - 16*a*tan(c/2 + (d*x)/2) ^15 - 7*a*tan(c/2 + (d*x)/2)^16 + 2*a*tan(c/2 + (d*x)/2)^17 + a*tan(c/2 + (d*x)/2)^18))