Integrand size = 27, antiderivative size = 134 \[ \int \frac {\sec (c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {5 \text {arctanh}(\sin (c+d x))}{128 a d}-\frac {5 \sec (c+d x) \tan (c+d x)}{128 a d}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{64 a d}-\frac {5 \sec ^3(c+d x) \tan ^3(c+d x)}{48 a d}+\frac {\sec ^3(c+d x) \tan ^5(c+d x)}{8 a d}-\frac {\tan ^8(c+d x)}{8 a d} \]
-5/128*arctanh(sin(d*x+c))/a/d-5/128*sec(d*x+c)*tan(d*x+c)/a/d+5/64*sec(d* x+c)^3*tan(d*x+c)/a/d-5/48*sec(d*x+c)^3*tan(d*x+c)^3/a/d+1/8*sec(d*x+c)^3* tan(d*x+c)^5/a/d-1/8*tan(d*x+c)^8/a/d
Time = 0.65 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.75 \[ \int \frac {\sec (c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {15 \text {arctanh}(\sin (c+d x))+\frac {48+63 \sin (c+d x)-129 \sin ^2(c+d x)-184 \sin ^3(c+d x)+104 \sin ^4(c+d x)+177 \sin ^5(c+d x)-15 \sin ^6(c+d x)}{(-1+\sin (c+d x))^3 (1+\sin (c+d x))^4}}{384 a d} \]
-1/384*(15*ArcTanh[Sin[c + d*x]] + (48 + 63*Sin[c + d*x] - 129*Sin[c + d*x ]^2 - 184*Sin[c + d*x]^3 + 104*Sin[c + d*x]^4 + 177*Sin[c + d*x]^5 - 15*Si n[c + d*x]^6)/((-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^4))/(a*d)
Time = 0.80 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {3042, 3314, 3042, 3087, 15, 3091, 3042, 3091, 3042, 3091, 3042, 4255, 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.
| \(\displaystyle \int \frac {\tan ^6(c+d x) \sec (c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6}{\cos (c+d x)^7 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3314 |
| \(\displaystyle \frac {\int \sec ^3(c+d x) \tan ^6(c+d x)dx}{a}-\frac {\int \sec ^2(c+d x) \tan ^7(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^3 \tan (c+d x)^6dx}{a}-\frac {\int \sec (c+d x)^2 \tan (c+d x)^7dx}{a}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\int \sec (c+d x)^3 \tan (c+d x)^6dx}{a}-\frac {\int \tan ^7(c+d x)d\tan (c+d x)}{a d}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {\int \sec (c+d x)^3 \tan (c+d x)^6dx}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \int \sec ^3(c+d x) \tan ^4(c+d x)dx}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \int \sec (c+d x)^3 \tan (c+d x)^4dx}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {1}{2} \int \sec ^3(c+d x) \tan ^2(c+d x)dx\right )}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}-\frac {1}{2} \int \sec (c+d x)^3 \tan (c+d x)^2dx\right )}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \sec ^3(c+d x)dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\tan ^5(c+d x) \sec ^3(c+d x)}{8 d}-\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {\tan ^3(c+d x) \sec ^3(c+d x)}{6 d}\right )}{a}-\frac {\tan ^8(c+d x)}{8 a d}\) |
-1/8*Tan[c + d*x]^8/(a*d) + ((Sec[c + d*x]^3*Tan[c + d*x]^5)/(8*d) - (5*(( Sec[c + d*x]^3*Tan[c + d*x]^3)/(6*d) + (-1/4*(Sec[c + d*x]^3*Tan[c + d*x]) /d + (ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d))/4)/ 2))/8)/a
3.9.83.3.1 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[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.92 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {7}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {15}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256}-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {11}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}-\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) | \(115\) |
| default | \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {7}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {15}{128 \left (\sin \left (d x +c \right )-1\right )}+\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{256}-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {11}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}-\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) | \(115\) |
| risch | \(\frac {i \left (-354 i {\mathrm e}^{12 i \left (d x +c \right )}+15 \,{\mathrm e}^{13 i \left (d x +c \right )}+298 i {\mathrm e}^{10 i \left (d x +c \right )}+326 \,{\mathrm e}^{11 i \left (d x +c \right )}-1140 i {\mathrm e}^{8 i \left (d x +c \right )}+625 \,{\mathrm e}^{9 i \left (d x +c \right )}+1140 i {\mathrm e}^{6 i \left (d x +c \right )}+1140 \,{\mathrm e}^{7 i \left (d x +c \right )}-298 i {\mathrm e}^{4 i \left (d x +c \right )}+625 \,{\mathrm e}^{5 i \left (d x +c \right )}+354 i {\mathrm e}^{2 i \left (d x +c \right )}+326 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 d a}\) | \(231\) |
| parallelrisch | \(\frac {\left (840 \cos \left (2 d x +2 c \right )+420 \cos \left (4 d x +4 c \right )+120 \cos \left (6 d x +6 c \right )+15 \cos \left (8 d x +8 c \right )+525\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-840 \cos \left (2 d x +2 c \right )-420 \cos \left (4 d x +4 c \right )-120 \cos \left (6 d x +6 c \right )-15 \cos \left (8 d x +8 c \right )-525\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-1790 \sin \left (3 d x +3 c \right )+794 \sin \left (5 d x +5 c \right )-30 \sin \left (7 d x +7 c \right )+2688 \cos \left (2 d x +2 c \right )-1344 \cos \left (4 d x +4 c \right )+384 \cos \left (6 d x +6 c \right )-48 \cos \left (8 d x +8 c \right )+3530 \sin \left (d x +c \right )-1680}{384 a d \left (\cos \left (8 d x +8 c \right )+8 \cos \left (6 d x +6 c \right )+28 \cos \left (4 d x +4 c \right )+56 \cos \left (2 d x +2 c \right )+35\right )}\) | \(260\) |
| norman | \(\frac {\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {5 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {33 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {33 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {5 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}+\frac {289 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}-\frac {85 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}-\frac {85 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}-\frac {35 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}-\frac {35 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}+\frac {113 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}+\frac {113 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6}}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{128 a d}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{128 a d}\) | \(315\) |
1/d/a*(-1/96/(sin(d*x+c)-1)^3-7/128/(sin(d*x+c)-1)^2-15/128/(sin(d*x+c)-1) +5/256*ln(sin(d*x+c)-1)-1/64/(1+sin(d*x+c))^4+1/12/(1+sin(d*x+c))^3-11/64/ (1+sin(d*x+c))^2+5/32/(1+sin(d*x+c))-5/256*ln(1+sin(d*x+c)))
Time = 0.28 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.25 \[ \int \frac {\sec (c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{6} + 118 \, \cos \left (d x + c\right )^{4} - 68 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (177 \, \cos \left (d x + c\right )^{4} - 170 \, \cos \left (d x + c\right )^{2} + 56\right )} \sin \left (d x + c\right ) + 16}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]
1/768*(30*cos(d*x + c)^6 + 118*cos(d*x + c)^4 - 68*cos(d*x + c)^2 - 15*(co s(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 15*(co s(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) + 2*(17 7*cos(d*x + c)^4 - 170*cos(d*x + c)^2 + 56)*sin(d*x + c) + 16)/(a*d*cos(d* x + c)^6*sin(d*x + c) + a*d*cos(d*x + c)^6)
Timed out. \[ \int \frac {\sec (c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.31 \[ \int \frac {\sec (c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{6} - 177 \, \sin \left (d x + c\right )^{5} - 104 \, \sin \left (d x + c\right )^{4} + 184 \, \sin \left (d x + c\right )^{3} + 129 \, \sin \left (d x + c\right )^{2} - 63 \, \sin \left (d x + c\right ) - 48\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \]
1/768*(2*(15*sin(d*x + c)^6 - 177*sin(d*x + c)^5 - 104*sin(d*x + c)^4 + 18 4*sin(d*x + c)^3 + 129*sin(d*x + c)^2 - 63*sin(d*x + c) - 48)/(a*sin(d*x + c)^7 + a*sin(d*x + c)^6 - 3*a*sin(d*x + c)^5 - 3*a*sin(d*x + c)^4 + 3*a*s in(d*x + c)^3 + 3*a*sin(d*x + c)^2 - a*sin(d*x + c) - a) - 15*log(sin(d*x + c) + 1)/a + 15*log(sin(d*x + c) - 1)/a)/d
Time = 0.40 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01 \[ \int \frac {\sec (c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (55 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} - 111 \, \sin \left (d x + c\right ) + 57\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 980 \, \sin \left (d x + c\right )^{3} + 1662 \, \sin \left (d x + c\right )^{2} + 1140 \, \sin \left (d x + c\right ) + 285}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]
-1/3072*(60*log(abs(sin(d*x + c) + 1))/a - 60*log(abs(sin(d*x + c) - 1))/a + 2*(55*sin(d*x + c)^3 + 15*sin(d*x + c)^2 - 111*sin(d*x + c) + 57)/(a*(s in(d*x + c) - 1)^3) - (125*sin(d*x + c)^4 + 980*sin(d*x + c)^3 + 1662*sin( d*x + c)^2 + 1140*sin(d*x + c) + 285)/(a*(sin(d*x + c) + 1)^4))/d
Time = 21.58 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.90 \[ \int \frac {\sec (c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}-\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{96}+\frac {113\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {33\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{16}+\frac {289\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{16}+\frac {33\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{16}+\frac {113\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}-\frac {85\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{96}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-40\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,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 )}-\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a\,d} \]
((5*tan(c/2 + (d*x)/2))/64 + (5*tan(c/2 + (d*x)/2)^2)/32 - (35*tan(c/2 + ( d*x)/2)^3)/96 - (85*tan(c/2 + (d*x)/2)^4)/96 + (113*tan(c/2 + (d*x)/2)^5)/ 192 + (33*tan(c/2 + (d*x)/2)^6)/16 + (289*tan(c/2 + (d*x)/2)^7)/16 + (33*t an(c/2 + (d*x)/2)^8)/16 + (113*tan(c/2 + (d*x)/2)^9)/192 - (85*tan(c/2 + ( d*x)/2)^10)/96 - (35*tan(c/2 + (d*x)/2)^11)/96 + (5*tan(c/2 + (d*x)/2)^12) /32 + (5*tan(c/2 + (d*x)/2)^13)/64)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 5*a*t an(c/2 + (d*x)/2)^2 - 12*a*tan(c/2 + (d*x)/2)^3 + 9*a*tan(c/2 + (d*x)/2)^4 + 30*a*tan(c/2 + (d*x)/2)^5 - 5*a*tan(c/2 + (d*x)/2)^6 - 40*a*tan(c/2 + ( d*x)/2)^7 - 5*a*tan(c/2 + (d*x)/2)^8 + 30*a*tan(c/2 + (d*x)/2)^9 + 9*a*tan (c/2 + (d*x)/2)^10 - 12*a*tan(c/2 + (d*x)/2)^11 - 5*a*tan(c/2 + (d*x)/2)^1 2 + 2*a*tan(c/2 + (d*x)/2)^13 + a*tan(c/2 + (d*x)/2)^14)) - (5*atanh(tan(c /2 + (d*x)/2)))/(64*a*d)