Integrand size = 19, antiderivative size = 127 \[ \int (a+b \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {5 a x}{16}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d} \]
5/16*a*x+b*arctanh(sin(d*x+c))/d-b*sin(d*x+c)/d-5/16*a*cos(d*x+c)*sin(d*x+ c)/d-1/3*b*sin(d*x+c)^3/d-5/24*a*cos(d*x+c)*sin(d*x+c)^3/d-1/5*b*sin(d*x+c )^5/d-1/6*a*cos(d*x+c)*sin(d*x+c)^5/d
Time = 0.46 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.93 \[ \int (a+b \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {5 a (c+d x)}{16 d}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin ^5(c+d x)}{5 d}-\frac {15 a \sin (2 (c+d x))}{64 d}+\frac {3 a \sin (4 (c+d x))}{64 d}-\frac {a \sin (6 (c+d x))}{192 d} \]
(5*a*(c + d*x))/(16*d) + (b*ArcTanh[Sin[c + d*x]])/d - (b*Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/(3*d) - (b*Sin[c + d*x]^5)/(5*d) - (15*a*Sin[2*(c + d *x)])/(64*d) + (3*a*Sin[4*(c + d*x)])/(64*d) - (a*Sin[6*(c + d*x)])/(192*d )
Time = 0.61 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.842, Rules used = {3042, 4360, 25, 25, 3042, 3317, 3042, 3072, 254, 2009, 3115, 3042, 3115, 3042, 3115, 24}
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.
| \(\displaystyle \int \sin ^6(c+d x) (a+b \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^6 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\left (\sin ^5(c+d x) \tan (c+d x) (-a \cos (c+d x)-b)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\left ((b+a \cos (c+d x)) \sin ^5(c+d x) \tan (c+d x)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \sin ^5(c+d x) \tan (c+d x) (a \cos (c+d x)+b)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^6 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3317 |
| \(\displaystyle a \int \sin ^6(c+d x)dx+b \int \sin ^5(c+d x) \tan (c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sin (c+d x)^6dx+b \int \sin (c+d x)^5 \tan (c+d x)dx\) |
| \(\Big \downarrow \) 3072 |
| \(\displaystyle a \int \sin (c+d x)^6dx+\frac {b \int \frac {\sin ^6(c+d x)}{1-\sin ^2(c+d x)}d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle a \int \sin (c+d x)^6dx+\frac {b \int \left (-\sin ^4(c+d x)-\sin ^2(c+d x)+\frac {1}{1-\sin ^2(c+d x)}-1\right )d\sin (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \int \sin (c+d x)^6dx+\frac {b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a \left (\frac {5}{6} \int \sin ^4(c+d x)dx-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {5}{6} \int \sin (c+d x)^4dx-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \int \sin ^2(c+d x)dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \int \sin (c+d x)^2dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 d}\right )+\frac {b \left (\text {arctanh}(\sin (c+d x))-\frac {1}{5} \sin ^5(c+d x)-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
(b*(ArcTanh[Sin[c + d*x]] - Sin[c + d*x] - Sin[c + d*x]^3/3 - Sin[c + d*x] ^5/5))/d + a*(-1/6*(Cos[c + d*x]*Sin[c + d*x]^5)/d + (5*(-1/4*(Cos[c + d*x ]*Sin[c + d*x]^3)/d + (3*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/6)
3.2.68.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
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 = 1.89 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(\frac {a \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+b \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(96\) |
| default | \(\frac {a \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+b \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(96\) |
| parts | \(\frac {a \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}+\frac {b \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(98\) |
| parallelrisch | \(\frac {300 a x d -1320 \sin \left (d x +c \right ) b -5 a \sin \left (6 d x +6 c \right )+45 a \sin \left (4 d x +4 c \right )-225 a \sin \left (2 d x +2 c \right )-12 \sin \left (5 d x +5 c \right ) b +140 \sin \left (3 d x +3 c \right ) b -960 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+960 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{960 d}\) | \(111\) |
| risch | \(\frac {5 a x}{16}+\frac {11 i b \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {11 i b \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \sin \left (6 d x +6 c \right )}{192 d}-\frac {b \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 a \sin \left (4 d x +4 c \right )}{64 d}+\frac {7 b \sin \left (3 d x +3 c \right )}{48 d}-\frac {15 a \sin \left (2 d x +2 c \right )}{64 d}\) | \(150\) |
| norman | \(\frac {\frac {5 a x}{16}+\frac {15 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {75 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16}+\frac {25 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4}+\frac {75 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16}+\frac {15 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {5 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16}+\frac {\left (5 a -16 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}-\frac {\left (5 a +16 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (85 a -304 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}-\frac {\left (85 a +304 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}+\frac {\left (165 a -688 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 d}-\frac {\left (165 a +688 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(284\) |
1/d*(a*(-1/6*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/ 16*d*x+5/16*c)+b*(-1/5*sin(d*x+c)^5-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(sec(d*x +c)+tan(d*x+c))))
Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80 \[ \int (a+b \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {75 \, a d x + 120 \, b \log \left (\sin \left (d x + c\right ) + 1\right ) - 120 \, b \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (40 \, a \cos \left (d x + c\right )^{5} + 48 \, b \cos \left (d x + c\right )^{4} - 130 \, a \cos \left (d x + c\right )^{3} - 176 \, b \cos \left (d x + c\right )^{2} + 165 \, a \cos \left (d x + c\right ) + 368 \, b\right )} \sin \left (d x + c\right )}{240 \, d} \]
1/240*(75*a*d*x + 120*b*log(sin(d*x + c) + 1) - 120*b*log(-sin(d*x + c) + 1) - (40*a*cos(d*x + c)^5 + 48*b*cos(d*x + c)^4 - 130*a*cos(d*x + c)^3 - 1 76*b*cos(d*x + c)^2 + 165*a*cos(d*x + c) + 368*b)*sin(d*x + c))/d
\[ \int (a+b \sec (c+d x)) \sin ^6(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \sin ^{6}{\left (c + d x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int (a+b \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a - 32 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} b}{960 \, d} \]
1/960*(5*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*s in(2*d*x + 2*c))*a - 32*(6*sin(d*x + c)^5 + 10*sin(d*x + c)^3 - 15*log(sin (d*x + c) + 1) + 15*log(sin(d*x + c) - 1) + 30*sin(d*x + c))*b)/d
Time = 0.32 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.80 \[ \int (a+b \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {75 \, {\left (d x + c\right )} a + 240 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 240 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (75 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 240 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 425 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1520 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 990 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4128 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 990 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4128 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 425 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1520 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 240 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]
1/240*(75*(d*x + c)*a + 240*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 240*b*l og(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(75*a*tan(1/2*d*x + 1/2*c)^11 - 240* b*tan(1/2*d*x + 1/2*c)^11 + 425*a*tan(1/2*d*x + 1/2*c)^9 - 1520*b*tan(1/2* d*x + 1/2*c)^9 + 990*a*tan(1/2*d*x + 1/2*c)^7 - 4128*b*tan(1/2*d*x + 1/2*c )^7 - 990*a*tan(1/2*d*x + 1/2*c)^5 - 4128*b*tan(1/2*d*x + 1/2*c)^5 - 425*a *tan(1/2*d*x + 1/2*c)^3 - 1520*b*tan(1/2*d*x + 1/2*c)^3 - 75*a*tan(1/2*d*x + 1/2*c) - 240*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d
Time = 15.25 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.61 \[ \int (a+b \sec (c+d x)) \sin ^6(c+d x) \, dx=\frac {5\,a\,\mathrm {atan}\left (\frac {125\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,\left (\frac {125\,a^3}{64}+20\,a\,b^2\right )}+\frac {20\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {125\,a^3}{64}+20\,a\,b^2}\right )}{8\,d}+\frac {2\,b\,\mathrm {atanh}\left (\frac {64\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {25\,a^2\,b}{4}+64\,b^3}+\frac {25\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {25\,a^2\,b}{4}+64\,b^3\right )}\right )}{d}-\frac {\left (2\,b-\frac {5\,a}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {38\,b}{3}-\frac {85\,a}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {172\,b}{5}-\frac {33\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {33\,a}{4}+\frac {172\,b}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {85\,a}{24}+\frac {38\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a}{8}+2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(5*a*atan((125*a^3*tan(c/2 + (d*x)/2))/(64*(20*a*b^2 + (125*a^3)/64)) + (2 0*a*b^2*tan(c/2 + (d*x)/2))/(20*a*b^2 + (125*a^3)/64)))/(8*d) + (2*b*atanh ((64*b^3*tan(c/2 + (d*x)/2))/((25*a^2*b)/4 + 64*b^3) + (25*a^2*b*tan(c/2 + (d*x)/2))/(4*((25*a^2*b)/4 + 64*b^3))))/d - (tan(c/2 + (d*x)/2)*((5*a)/8 + 2*b) - tan(c/2 + (d*x)/2)^11*((5*a)/8 - 2*b) + tan(c/2 + (d*x)/2)^3*((85 *a)/24 + (38*b)/3) - tan(c/2 + (d*x)/2)^9*((85*a)/24 - (38*b)/3) + tan(c/2 + (d*x)/2)^5*((33*a)/4 + (172*b)/5) - tan(c/2 + (d*x)/2)^7*((33*a)/4 - (1 72*b)/5))/(d*(6*tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/ 2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c /2 + (d*x)/2)^12 + 1))