Integrand size = 21, antiderivative size = 192 \[ \int (a+a \sec (c+d x))^2 \tan ^9(c+d x) \, dx=-\frac {a^2 \log (\cos (c+d x))}{d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {3 a^2 \sec ^2(c+d x)}{2 d}-\frac {8 a^2 \sec ^3(c+d x)}{3 d}+\frac {a^2 \sec ^4(c+d x)}{2 d}+\frac {12 a^2 \sec ^5(c+d x)}{5 d}+\frac {a^2 \sec ^6(c+d x)}{3 d}-\frac {8 a^2 \sec ^7(c+d x)}{7 d}-\frac {3 a^2 \sec ^8(c+d x)}{8 d}+\frac {2 a^2 \sec ^9(c+d x)}{9 d}+\frac {a^2 \sec ^{10}(c+d x)}{10 d} \]
-a^2*ln(cos(d*x+c))/d+2*a^2*sec(d*x+c)/d-3/2*a^2*sec(d*x+c)^2/d-8/3*a^2*se c(d*x+c)^3/d+1/2*a^2*sec(d*x+c)^4/d+12/5*a^2*sec(d*x+c)^5/d+1/3*a^2*sec(d* x+c)^6/d-8/7*a^2*sec(d*x+c)^7/d-3/8*a^2*sec(d*x+c)^8/d+2/9*a^2*sec(d*x+c)^ 9/d+1/10*a^2*sec(d*x+c)^10/d
Time = 1.19 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.01 \[ \int (a+a \sec (c+d x))^2 \tan ^9(c+d x) \, dx=\frac {a^2 (63080 \cos (c+d x)+39600 \cos (3 (c+d x))-1050 \cos (2 (c+d x)) (68+63 \log (\cos (c+d x)))-21 (1662-1072 \cos (5 (c+d x))+600 \cos (6 (c+d x))-220 \cos (7 (c+d x))+90 \cos (8 (c+d x))-60 \cos (9 (c+d x))+1890 \log (\cos (c+d x))+675 \cos (6 (c+d x)) \log (\cos (c+d x))+150 \cos (8 (c+d x)) \log (\cos (c+d x))+15 \cos (10 (c+d x)) \log (\cos (c+d x))+40 \cos (4 (c+d x)) (37+45 \log (\cos (c+d x))))) \sec ^{10}(c+d x)}{161280 d} \]
(a^2*(63080*Cos[c + d*x] + 39600*Cos[3*(c + d*x)] - 1050*Cos[2*(c + d*x)]* (68 + 63*Log[Cos[c + d*x]]) - 21*(1662 - 1072*Cos[5*(c + d*x)] + 600*Cos[6 *(c + d*x)] - 220*Cos[7*(c + d*x)] + 90*Cos[8*(c + d*x)] - 60*Cos[9*(c + d *x)] + 1890*Log[Cos[c + d*x]] + 675*Cos[6*(c + d*x)]*Log[Cos[c + d*x]] + 1 50*Cos[8*(c + d*x)]*Log[Cos[c + d*x]] + 15*Cos[10*(c + d*x)]*Log[Cos[c + d *x]] + 40*Cos[4*(c + d*x)]*(37 + 45*Log[Cos[c + d*x]])))*Sec[c + d*x]^10)/ (161280*d)
Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.69, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 4367, 27, 99, 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.
| \(\displaystyle \int \tan ^9(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^9 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^9 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^2dx\) |
| \(\Big \downarrow \) 4367 |
| \(\displaystyle -\frac {\int a^{10} (1-\cos (c+d x))^4 (\cos (c+d x)+1)^6 \sec ^{11}(c+d x)d\cos (c+d x)}{a^8 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \int (1-\cos (c+d x))^4 (\cos (c+d x)+1)^6 \sec ^{11}(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {a^2 \int \left (\sec ^{11}(c+d x)+2 \sec ^{10}(c+d x)-3 \sec ^9(c+d x)-8 \sec ^8(c+d x)+2 \sec ^7(c+d x)+12 \sec ^6(c+d x)+2 \sec ^5(c+d x)-8 \sec ^4(c+d x)-3 \sec ^3(c+d x)+2 \sec ^2(c+d x)+\sec (c+d x)\right )d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \left (-\frac {1}{10} \sec ^{10}(c+d x)-\frac {2}{9} \sec ^9(c+d x)+\frac {3}{8} \sec ^8(c+d x)+\frac {8}{7} \sec ^7(c+d x)-\frac {1}{3} \sec ^6(c+d x)-\frac {12}{5} \sec ^5(c+d x)-\frac {1}{2} \sec ^4(c+d x)+\frac {8}{3} \sec ^3(c+d x)+\frac {3}{2} \sec ^2(c+d x)-2 \sec (c+d x)+\log (\cos (c+d x))\right )}{d}\) |
-((a^2*(Log[Cos[c + d*x]] - 2*Sec[c + d*x] + (3*Sec[c + d*x]^2)/2 + (8*Sec [c + d*x]^3)/3 - Sec[c + d*x]^4/2 - (12*Sec[c + d*x]^5)/5 - Sec[c + d*x]^6 /3 + (8*Sec[c + d*x]^7)/7 + (3*Sec[c + d*x]^8)/8 - (2*Sec[c + d*x]^9)/9 - Sec[c + d*x]^10/10))/d)
3.1.19.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[1/(a^(m - n - 1)*b^n*d) Subst[Int[(a - b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x^(m + n)), x], x, Sin[c + d*x]], x] /; Fr eeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && Integer Q[n]
Time = 2.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.59
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right )^{10}}{10}+\frac {2 \sec \left (d x +c \right )^{9}}{9}-\frac {3 \sec \left (d x +c \right )^{8}}{8}-\frac {8 \sec \left (d x +c \right )^{7}}{7}+\frac {\sec \left (d x +c \right )^{6}}{3}+\frac {12 \sec \left (d x +c \right )^{5}}{5}+\frac {\sec \left (d x +c \right )^{4}}{2}-\frac {8 \sec \left (d x +c \right )^{3}}{3}-\frac {3 \sec \left (d x +c \right )^{2}}{2}+2 \sec \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(114\) |
| default | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right )^{10}}{10}+\frac {2 \sec \left (d x +c \right )^{9}}{9}-\frac {3 \sec \left (d x +c \right )^{8}}{8}-\frac {8 \sec \left (d x +c \right )^{7}}{7}+\frac {\sec \left (d x +c \right )^{6}}{3}+\frac {12 \sec \left (d x +c \right )^{5}}{5}+\frac {\sec \left (d x +c \right )^{4}}{2}-\frac {8 \sec \left (d x +c \right )^{3}}{3}-\frac {3 \sec \left (d x +c \right )^{2}}{2}+2 \sec \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(114\) |
| parts | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{8}}{8}-\frac {\tan \left (d x +c \right )^{6}}{6}+\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a^{2} \tan \left (d x +c \right )^{10}}{10 d}+\frac {2 a^{2} \left (\frac {\sec \left (d x +c \right )^{9}}{9}-\frac {4 \sec \left (d x +c \right )^{7}}{7}+\frac {6 \sec \left (d x +c \right )^{5}}{5}-\frac {4 \sec \left (d x +c \right )^{3}}{3}+\sec \left (d x +c \right )\right )}{d}\) | \(134\) |
| risch | \(i a^{2} x +\frac {2 i a^{2} c}{d}+\frac {2 a^{2} \left (630 \,{\mathrm e}^{19 i \left (d x +c \right )}-945 \,{\mathrm e}^{18 i \left (d x +c \right )}+2310 \,{\mathrm e}^{17 i \left (d x +c \right )}-6300 \,{\mathrm e}^{16 i \left (d x +c \right )}+11256 \,{\mathrm e}^{15 i \left (d x +c \right )}-15540 \,{\mathrm e}^{14 i \left (d x +c \right )}+19800 \,{\mathrm e}^{13 i \left (d x +c \right )}-35700 \,{\mathrm e}^{12 i \left (d x +c \right )}+31540 \,{\mathrm e}^{11 i \left (d x +c \right )}-34902 \,{\mathrm e}^{10 i \left (d x +c \right )}+31540 \,{\mathrm e}^{9 i \left (d x +c \right )}-35700 \,{\mathrm e}^{8 i \left (d x +c \right )}+19800 \,{\mathrm e}^{7 i \left (d x +c \right )}-15540 \,{\mathrm e}^{6 i \left (d x +c \right )}+11256 \,{\mathrm e}^{5 i \left (d x +c \right )}-6300 \,{\mathrm e}^{4 i \left (d x +c \right )}+2310 \,{\mathrm e}^{3 i \left (d x +c \right )}-945 \,{\mathrm e}^{2 i \left (d x +c \right )}+630 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{315 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(270\) |
a^2/d*(1/10*sec(d*x+c)^10+2/9*sec(d*x+c)^9-3/8*sec(d*x+c)^8-8/7*sec(d*x+c) ^7+1/3*sec(d*x+c)^6+12/5*sec(d*x+c)^5+1/2*sec(d*x+c)^4-8/3*sec(d*x+c)^3-3/ 2*sec(d*x+c)^2+2*sec(d*x+c)+ln(sec(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.81 \[ \int (a+a \sec (c+d x))^2 \tan ^9(c+d x) \, dx=-\frac {2520 \, a^{2} \cos \left (d x + c\right )^{10} \log \left (-\cos \left (d x + c\right )\right ) - 5040 \, a^{2} \cos \left (d x + c\right )^{9} + 3780 \, a^{2} \cos \left (d x + c\right )^{8} + 6720 \, a^{2} \cos \left (d x + c\right )^{7} - 1260 \, a^{2} \cos \left (d x + c\right )^{6} - 6048 \, a^{2} \cos \left (d x + c\right )^{5} - 840 \, a^{2} \cos \left (d x + c\right )^{4} + 2880 \, a^{2} \cos \left (d x + c\right )^{3} + 945 \, a^{2} \cos \left (d x + c\right )^{2} - 560 \, a^{2} \cos \left (d x + c\right ) - 252 \, a^{2}}{2520 \, d \cos \left (d x + c\right )^{10}} \]
-1/2520*(2520*a^2*cos(d*x + c)^10*log(-cos(d*x + c)) - 5040*a^2*cos(d*x + c)^9 + 3780*a^2*cos(d*x + c)^8 + 6720*a^2*cos(d*x + c)^7 - 1260*a^2*cos(d* x + c)^6 - 6048*a^2*cos(d*x + c)^5 - 840*a^2*cos(d*x + c)^4 + 2880*a^2*cos (d*x + c)^3 + 945*a^2*cos(d*x + c)^2 - 560*a^2*cos(d*x + c) - 252*a^2)/(d* cos(d*x + c)^10)
Time = 2.47 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.64 \[ \int (a+a \sec (c+d x))^2 \tan ^9(c+d x) \, dx=\begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{8}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{10 d} + \frac {2 a^{2} \tan ^{8}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{9 d} + \frac {a^{2} \tan ^{8}{\left (c + d x \right )}}{8 d} - \frac {a^{2} \tan ^{6}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{10 d} - \frac {16 a^{2} \tan ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{63 d} - \frac {a^{2} \tan ^{6}{\left (c + d x \right )}}{6 d} + \frac {a^{2} \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{10 d} + \frac {32 a^{2} \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{105 d} + \frac {a^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {a^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{10 d} - \frac {128 a^{2} \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{315 d} - \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {a^{2} \sec ^{2}{\left (c + d x \right )}}{10 d} + \frac {256 a^{2} \sec {\left (c + d x \right )}}{315 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\left (c \right )} + a\right )^{2} \tan ^{9}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**8*sec( c + d*x)**2/(10*d) + 2*a**2*tan(c + d*x)**8*sec(c + d*x)/(9*d) + a**2*tan( c + d*x)**8/(8*d) - a**2*tan(c + d*x)**6*sec(c + d*x)**2/(10*d) - 16*a**2* tan(c + d*x)**6*sec(c + d*x)/(63*d) - a**2*tan(c + d*x)**6/(6*d) + a**2*ta n(c + d*x)**4*sec(c + d*x)**2/(10*d) + 32*a**2*tan(c + d*x)**4*sec(c + d*x )/(105*d) + a**2*tan(c + d*x)**4/(4*d) - a**2*tan(c + d*x)**2*sec(c + d*x) **2/(10*d) - 128*a**2*tan(c + d*x)**2*sec(c + d*x)/(315*d) - a**2*tan(c + d*x)**2/(2*d) + a**2*sec(c + d*x)**2/(10*d) + 256*a**2*sec(c + d*x)/(315*d ), Ne(d, 0)), (x*(a*sec(c) + a)**2*tan(c)**9, True))
Time = 0.21 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.78 \[ \int (a+a \sec (c+d x))^2 \tan ^9(c+d x) \, dx=-\frac {2520 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {5040 \, a^{2} \cos \left (d x + c\right )^{9} - 3780 \, a^{2} \cos \left (d x + c\right )^{8} - 6720 \, a^{2} \cos \left (d x + c\right )^{7} + 1260 \, a^{2} \cos \left (d x + c\right )^{6} + 6048 \, a^{2} \cos \left (d x + c\right )^{5} + 840 \, a^{2} \cos \left (d x + c\right )^{4} - 2880 \, a^{2} \cos \left (d x + c\right )^{3} - 945 \, a^{2} \cos \left (d x + c\right )^{2} + 560 \, a^{2} \cos \left (d x + c\right ) + 252 \, a^{2}}{\cos \left (d x + c\right )^{10}}}{2520 \, d} \]
-1/2520*(2520*a^2*log(cos(d*x + c)) - (5040*a^2*cos(d*x + c)^9 - 3780*a^2* cos(d*x + c)^8 - 6720*a^2*cos(d*x + c)^7 + 1260*a^2*cos(d*x + c)^6 + 6048* a^2*cos(d*x + c)^5 + 840*a^2*cos(d*x + c)^4 - 2880*a^2*cos(d*x + c)^3 - 94 5*a^2*cos(d*x + c)^2 + 560*a^2*cos(d*x + c) + 252*a^2)/cos(d*x + c)^10)/d
Time = 6.02 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.78 \[ \int (a+a \sec (c+d x))^2 \tan ^9(c+d x) \, dx=\frac {2520 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 2520 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {11477 \, a^{2} + \frac {119810 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {566865 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1605720 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3031770 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {2995020 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2171610 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {1114200 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {382545 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {78850 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {7381 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{10}}}{2520 \, d} \]
1/2520*(2520*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 25 20*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) + (11477*a^2 + 119810*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 566865*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1605720*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 3031770*a^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 299 5020*a^2*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 2171610*a^2*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 1114200*a^2*(cos(d*x + c) - 1)^7/(cos(d *x + c) + 1)^7 + 382545*a^2*(cos(d*x + c) - 1)^8/(cos(d*x + c) + 1)^8 + 78 850*a^2*(cos(d*x + c) - 1)^9/(cos(d*x + c) + 1)^9 + 7381*a^2*(cos(d*x + c) - 1)^10/(cos(d*x + c) + 1)^10)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1 )^10)/d
Time = 18.11 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.60 \[ \int (a+a \sec (c+d x))^2 \tan ^9(c+d x) \, dx=\frac {2\,a^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+\frac {272\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{3}-\frac {740\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{3}+\frac {2252\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{5}-588\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {2000\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{7}-\frac {652\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}+\frac {1150\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{63}-\frac {512\,a^2}{315}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{20}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}+45\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-120\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+210\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-252\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+210\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-120\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+45\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(2*a^2*atanh(tan(c/2 + (d*x)/2)^2))/d - ((1150*a^2*tan(c/2 + (d*x)/2)^2)/6 3 - (652*a^2*tan(c/2 + (d*x)/2)^4)/7 + (2000*a^2*tan(c/2 + (d*x)/2)^6)/7 - 588*a^2*tan(c/2 + (d*x)/2)^8 + (2252*a^2*tan(c/2 + (d*x)/2)^10)/5 - (740* a^2*tan(c/2 + (d*x)/2)^12)/3 + (272*a^2*tan(c/2 + (d*x)/2)^14)/3 - 20*a^2* tan(c/2 + (d*x)/2)^16 + 2*a^2*tan(c/2 + (d*x)/2)^18 - (512*a^2)/315)/(d*(4 5*tan(c/2 + (d*x)/2)^4 - 10*tan(c/2 + (d*x)/2)^2 - 120*tan(c/2 + (d*x)/2)^ 6 + 210*tan(c/2 + (d*x)/2)^8 - 252*tan(c/2 + (d*x)/2)^10 + 210*tan(c/2 + ( d*x)/2)^12 - 120*tan(c/2 + (d*x)/2)^14 + 45*tan(c/2 + (d*x)/2)^16 - 10*tan (c/2 + (d*x)/2)^18 + tan(c/2 + (d*x)/2)^20 + 1))