Integrand size = 29, antiderivative size = 122 \[ \int \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 B \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A+4 C) \tan ^3(c+d x)}{15 d} \]
3/8*B*arctanh(sin(d*x+c))/d+1/5*(5*A+4*C)*tan(d*x+c)/d+3/8*B*sec(d*x+c)*ta n(d*x+c)/d+1/4*B*sec(d*x+c)^3*tan(d*x+c)/d+1/5*C*sec(d*x+c)^4*tan(d*x+c)/d +1/15*(5*A+4*C)*tan(d*x+c)^3/d
Time = 0.49 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 B \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (45 B \sec (c+d x)+30 B \sec ^3(c+d x)+8 \left (15 (A+C)+5 (A+2 C) \tan ^2(c+d x)+3 C \tan ^4(c+d x)\right )\right )}{120 d} \]
(45*B*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(45*B*Sec[c + d*x] + 30*B*Sec[c + d*x]^3 + 8*(15*(A + C) + 5*(A + 2*C)*Tan[c + d*x]^2 + 3*C*Tan[c + d*x]^ 4)))/(120*d)
Time = 0.67 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 4535, 3042, 4255, 3042, 4255, 3042, 4257, 4534, 3042, 4254, 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.
| \(\displaystyle \int \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \int \sec ^4(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \sec ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} (5 A+4 C) \int \sec ^4(c+d x)dx+B \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+B \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {(5 A+4 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{5 d}+B \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {(5 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{5 d}+B \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
(C*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) - ((5*A + 4*C)*(-Tan[c + d*x] - Tan[ c + d*x]^3/3))/(5*d) + B*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(ArcTan h[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)
3.1.54.3.1 Defintions of rubi rules used
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Time = 0.47 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.85
| method | result | size |
| derivativedivides | \(\frac {-A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(104\) |
| default | \(\frac {-A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(104\) |
| parts | \(-\frac {A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(109\) |
| risch | \(-\frac {i \left (45 B \,{\mathrm e}^{9 i \left (d x +c \right )}+210 B \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A \,{\mathrm e}^{6 i \left (d x +c \right )}-560 A \,{\mathrm e}^{4 i \left (d x +c \right )}-640 C \,{\mathrm e}^{4 i \left (d x +c \right )}-210 B \,{\mathrm e}^{3 i \left (d x +c \right )}-400 A \,{\mathrm e}^{2 i \left (d x +c \right )}-320 C \,{\mathrm e}^{2 i \left (d x +c \right )}-45 B \,{\mathrm e}^{i \left (d x +c \right )}-80 A -64 C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(174\) |
| norman | \(\frac {-\frac {4 \left (25 A +29 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {\left (8 A -5 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (8 A +5 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (32 A -3 B +16 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (32 A +3 B +16 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {3 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(180\) |
| parallelrisch | \(\frac {-450 B \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+450 B \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (400 A +320 C \right ) \sin \left (3 d x +3 c \right )+\left (80 A +64 C \right ) \sin \left (5 d x +5 c \right )+420 B \sin \left (2 d x +2 c \right )+90 B \sin \left (4 d x +4 c \right )+320 \sin \left (d x +c \right ) \left (A +2 C \right )}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(197\) |
1/d*(-A*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+B*(-(-1/4*sec(d*x+c)^3-3/8*sec( d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-C*(-8/15-1/5*sec(d*x+c)^ 4-4/15*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00 \[ \int \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, B \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, B \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} + 45 \, B \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 30 \, B \cos \left (d x + c\right ) + 24 \, C\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
1/240*(45*B*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*B*cos(d*x + c)^5*log (-sin(d*x + c) + 1) + 2*(16*(5*A + 4*C)*cos(d*x + c)^4 + 45*B*cos(d*x + c) ^3 + 8*(5*A + 4*C)*cos(d*x + c)^2 + 30*B*cos(d*x + c) + 24*C)*sin(d*x + c) )/(d*cos(d*x + c)^5)
\[ \int \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.04 \[ \int \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C - 15 \, B {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A + 16*(3*tan(d*x + c)^5 + 10* tan(d*x + c)^3 + 15*tan(d*x + c))*C - 15*B*(2*(3*sin(d*x + c)^3 - 5*sin(d* x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (110) = 220\).
Time = 0.31 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.02 \[ \int \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 320 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 320 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
1/120*(45*B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*B*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*tan(1/2*d*x + 1/2*c)^9 - 75*B*tan(1/2*d*x + 1/2* c)^9 + 120*C*tan(1/2*d*x + 1/2*c)^9 - 320*A*tan(1/2*d*x + 1/2*c)^7 + 30*B* tan(1/2*d*x + 1/2*c)^7 - 160*C*tan(1/2*d*x + 1/2*c)^7 + 400*A*tan(1/2*d*x + 1/2*c)^5 + 464*C*tan(1/2*d*x + 1/2*c)^5 - 320*A*tan(1/2*d*x + 1/2*c)^3 - 30*B*tan(1/2*d*x + 1/2*c)^3 - 160*C*tan(1/2*d*x + 1/2*c)^3 + 120*A*tan(1/ 2*d*x + 1/2*c) + 75*B*tan(1/2*d*x + 1/2*c) + 120*C*tan(1/2*d*x + 1/2*c))/( tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 17.67 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.61 \[ \int \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\left (2\,A-\frac {5\,B}{4}+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B}{2}-\frac {16\,A}{3}-\frac {8\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A}{3}+\frac {116\,C}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,A}{3}-\frac {B}{2}-\frac {8\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A+\frac {5\,B}{4}+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(3*B*atanh(tan(c/2 + (d*x)/2)))/(4*d) - (tan(c/2 + (d*x)/2)^5*((20*A)/3 + (116*C)/15) + tan(c/2 + (d*x)/2)*(2*A + (5*B)/4 + 2*C) + tan(c/2 + (d*x)/2 )^9*(2*A - (5*B)/4 + 2*C) - tan(c/2 + (d*x)/2)^3*((16*A)/3 + B/2 + (8*C)/3 ) - tan(c/2 + (d*x)/2)^7*((16*A)/3 - B/2 + (8*C)/3))/(d*(5*tan(c/2 + (d*x) /2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d *x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))