Integrand size = 41, antiderivative size = 222 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (14 A+12 B+11 C) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {a^2 (10 A+9 B+8 C) \tan (c+d x)}{5 d}+\frac {a^2 (14 A+12 B+11 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^2 (10 A+12 B+9 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {a^2 (10 A+9 B+8 C) \tan ^3(c+d x)}{15 d} \]
1/16*a^2*(14*A+12*B+11*C)*arctanh(sin(d*x+c))/d+1/5*a^2*(10*A+9*B+8*C)*tan (d*x+c)/d+1/16*a^2*(14*A+12*B+11*C)*sec(d*x+c)*tan(d*x+c)/d+1/40*a^2*(10*A +12*B+9*C)*sec(d*x+c)^3*tan(d*x+c)/d+1/6*C*sec(d*x+c)^3*(a+a*sec(d*x+c))^2 *tan(d*x+c)/d+1/15*(3*B+C)*sec(d*x+c)^3*(a^2+a^2*sec(d*x+c))*tan(d*x+c)/d+ 1/15*a^2*(10*A+9*B+8*C)*tan(d*x+c)^3/d
Time = 7.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.59 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \left (15 (14 A+12 B+11 C) \text {arctanh}(\sin (c+d x))+\sec (c+d x) \left (15 (14 A+12 B+11 C)+16 (10 A+9 B+8 C) (2+\cos (2 (c+d x))) \sec (c+d x)+10 (6 A+12 B+11 C) \sec ^2(c+d x)+48 (B+2 C) \sec ^3(c+d x)+40 C \sec ^4(c+d x)\right ) \tan (c+d x)\right )}{240 d} \]
(a^2*(15*(14*A + 12*B + 11*C)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(15*(14 *A + 12*B + 11*C) + 16*(10*A + 9*B + 8*C)*(2 + Cos[2*(c + d*x)])*Sec[c + d *x] + 10*(6*A + 12*B + 11*C)*Sec[c + d*x]^2 + 48*(B + 2*C)*Sec[c + d*x]^3 + 40*C*Sec[c + d*x]^4)*Tan[c + d*x]))/(240*d)
Time = 1.33 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.97, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 4576, 3042, 4506, 27, 3042, 4485, 3042, 4274, 3042, 4254, 2009, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \sec (c+d x)+a)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \frac {\int \sec ^3(c+d x) (\sec (c+d x) a+a)^2 (3 a (2 A+C)+2 a (3 B+C) \sec (c+d x))dx}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 a (2 A+C)+2 a (3 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{5} \int 3 \sec ^3(c+d x) (\sec (c+d x) a+a) \left ((10 A+6 B+7 C) a^2+(10 A+12 B+9 C) \sec (c+d x) a^2\right )dx+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3}{5} \int \sec ^3(c+d x) (\sec (c+d x) a+a) \left ((10 A+6 B+7 C) a^2+(10 A+12 B+9 C) \sec (c+d x) a^2\right )dx+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((10 A+6 B+7 C) a^2+(10 A+12 B+9 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \int \sec ^3(c+d x) \left (5 (14 A+12 B+11 C) a^3+8 (10 A+9 B+8 C) \sec (c+d x) a^3\right )dx+\frac {a^3 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (5 (14 A+12 B+11 C) a^3+8 (10 A+9 B+8 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {a^3 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (8 a^3 (10 A+9 B+8 C) \int \sec ^4(c+d x)dx+5 a^3 (14 A+12 B+11 C) \int \sec ^3(c+d x)dx\right )+\frac {a^3 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (14 A+12 B+11 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+8 a^3 (10 A+9 B+8 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {a^3 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (14 A+12 B+11 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {8 a^3 (10 A+9 B+8 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {a^3 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (14 A+12 B+11 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {8 a^3 (10 A+9 B+8 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^3 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (14 A+12 B+11 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (10 A+9 B+8 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^3 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (14 A+12 B+11 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (10 A+9 B+8 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^3 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (14 A+12 B+11 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (10 A+9 B+8 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^3 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {2 (3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
(C*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(6*d) + ((2*(3*B + C)*Sec[c + d*x]^3*(a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(5*d) + (3*((a^3* (10*A + 12*B + 9*C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a^3*(14*A + 12 *B + 11*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d )) - (8*a^3*(10*A + 9*B + 8*C)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/4))/ 5)/(6*a)
3.5.17.3.1 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[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_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Time = 0.86 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.12
| method | result | size |
| parts | \(-\frac {\left (2 a^{2} A +B \,a^{2}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {\left (B \,a^{2}+2 C \,a^{2}\right ) \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (a^{2} A +2 B \,a^{2}+C \,a^{2}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}+\frac {a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(248\) |
| norman | \(\frac {\frac {17 a^{2} \left (14 A +12 B +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}-\frac {a^{2} \left (14 A +12 B +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}+\frac {a^{2} \left (50 A +52 B +53 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {a^{2} \left (430 A +428 B +331 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 d}-\frac {a^{2} \left (466 A +372 B +261 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}+\frac {a^{2} \left (530 A +468 B +501 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6}}-\frac {a^{2} \left (14 A +12 B +11 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {a^{2} \left (14 A +12 B +11 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\) | \(251\) |
| parallelrisch | \(-\frac {7 a^{2} \left (\left (A +\frac {6 B}{7}+\frac {11 C}{14}\right ) \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A +\frac {6 B}{7}+\frac {11 C}{14}\right ) \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-\frac {96 A}{7}-16 B -\frac {128 C}{7}\right ) \sin \left (2 d x +2 c \right )+\left (-\frac {187 C}{21}-\frac {58 A}{7}-\frac {68 B}{7}\right ) \sin \left (3 d x +3 c \right )+\left (-\frac {288 B}{35}-\frac {256 C}{35}-\frac {64 A}{7}\right ) \sin \left (4 d x +4 c \right )+\left (-\frac {11 C}{7}-2 A -\frac {12 B}{7}\right ) \sin \left (5 d x +5 c \right )+\left (-\frac {48 B}{35}-\frac {32 A}{21}-\frac {128 C}{105}\right ) \sin \left (6 d x +6 c \right )-\frac {44 \left (A +\frac {14 B}{11}+\frac {47 C}{22}\right ) \sin \left (d x +c \right )}{7}\right )}{8 d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\) | \(270\) |
| derivativedivides | \(\frac {a^{2} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )-2 a^{2} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(365\) |
| default | \(\frac {a^{2} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )-2 a^{2} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(365\) |
| risch | \(-\frac {i a^{2} \left (-256 C -320 A -288 B -960 A \,{\mathrm e}^{8 i \left (d x +c \right )}+660 A \,{\mathrm e}^{7 i \left (d x +c \right )}-660 A \,{\mathrm e}^{5 i \left (d x +c \right )}-870 A \,{\mathrm e}^{3 i \left (d x +c \right )}-210 A \,{\mathrm e}^{i \left (d x +c \right )}-3840 B \,{\mathrm e}^{4 i \left (d x +c \right )}-3200 A \,{\mathrm e}^{6 i \left (d x +c \right )}-2560 C \,{\mathrm e}^{6 i \left (d x +c \right )}-3840 A \,{\mathrm e}^{4 i \left (d x +c \right )}-3840 C \,{\mathrm e}^{4 i \left (d x +c \right )}-1920 A \,{\mathrm e}^{2 i \left (d x +c \right )}-1536 C \,{\mathrm e}^{2 i \left (d x +c \right )}+1410 C \,{\mathrm e}^{7 i \left (d x +c \right )}-1410 C \,{\mathrm e}^{5 i \left (d x +c \right )}-935 C \,{\mathrm e}^{3 i \left (d x +c \right )}-1728 B \,{\mathrm e}^{2 i \left (d x +c \right )}-165 C \,{\mathrm e}^{i \left (d x +c \right )}+210 A \,{\mathrm e}^{11 i \left (d x +c \right )}+165 C \,{\mathrm e}^{11 i \left (d x +c \right )}+870 A \,{\mathrm e}^{9 i \left (d x +c \right )}+935 C \,{\mathrm e}^{9 i \left (d x +c \right )}-180 B \,{\mathrm e}^{i \left (d x +c \right )}-840 B \,{\mathrm e}^{5 i \left (d x +c \right )}-480 B \,{\mathrm e}^{8 i \left (d x +c \right )}+840 B \,{\mathrm e}^{7 i \left (d x +c \right )}+180 B \,{\mathrm e}^{11 i \left (d x +c \right )}-1020 B \,{\mathrm e}^{3 i \left (d x +c \right )}+1020 B \,{\mathrm e}^{9 i \left (d x +c \right )}-2880 B \,{\mathrm e}^{6 i \left (d x +c \right )}\right )}{120 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{4 d}-\frac {11 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{16 d}+\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{4 d}+\frac {11 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{16 d}\) | \(514\) |
-(2*A*a^2+B*a^2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)-(B*a^2+2*C*a^2)/d*(- 8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(A*a^2+2*B*a^2+C*a^2)/ d*(-(-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*a^2/d*(-(-1/6*sec(d*x+c)^5-5/24*sec(d*x+c)^3-5/16*sec(d*x+c))*tan (d*x+c)+5/16*ln(sec(d*x+c)+tan(d*x+c)))+a^2*A/d*(1/2*sec(d*x+c)*tan(d*x+c) +1/2*ln(sec(d*x+c)+tan(d*x+c)))
Time = 0.26 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.91 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, {\left (10 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 15 \, {\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 16 \, {\left (10 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (6 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 40 \, C a^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]
integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
1/480*(15*(14*A + 12*B + 11*C)*a^2*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(14*A + 12*B + 11*C)*a^2*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(32* (10*A + 9*B + 8*C)*a^2*cos(d*x + c)^5 + 15*(14*A + 12*B + 11*C)*a^2*cos(d* x + c)^4 + 16*(10*A + 9*B + 8*C)*a^2*cos(d*x + c)^3 + 10*(6*A + 12*B + 11* C)*a^2*cos(d*x + c)^2 + 48*(B + 2*C)*a^2*cos(d*x + c) + 40*C*a^2)*sin(d*x + c))/(d*cos(d*x + c)^6)
\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{7}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(A*sec(c + d*x)**3, x) + Integral(2*A*sec(c + d*x)**4, x) + Integral(A*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**4, x) + Integral (2*B*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**6, x) + Integral(C*sec (c + d*x)**5, x) + Integral(2*C*sec(c + d*x)**6, x) + Integral(C*sec(c + d *x)**7, x))
Leaf count of result is larger than twice the leaf count of optimal. 477 vs. \(2 (208) = 416\).
Time = 0.27 (sec) , antiderivative size = 477, normalized size of antiderivative = 2.15 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 32 \, {\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 )} B a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 64 \, {\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 a^{2} - 5 \, C a^{2} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, A a^{2} {\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 )} - 60 \, B a^{2} {\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 )} - 30 \, C a^{2} {\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 )} - 120 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \]
integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
1/480*(320*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^2 + 160*(tan(d*x + c)^3 + 3*tan (d*x + c))*B*a^2 + 64*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^2 - 5*C*a^2*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d* x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*lo g(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 30*A*a^2*(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(si n(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*B*a^2*(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)) - 30*C*a^2*(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)) - 120*A*a^2*(2*sin(d*x + c)/(sin(d*x + c) ^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
Time = 0.38 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.77 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (14 \, A a^{2} + 12 \, B a^{2} + 11 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (14 \, A a^{2} + 12 \, B a^{2} + 11 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (210 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 180 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 165 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1190 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1020 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 935 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2580 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2568 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1986 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3180 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2808 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3006 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2330 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1860 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1305 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 750 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 780 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 795 \, C a^{2} \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} \]
integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
1/240*(15*(14*A*a^2 + 12*B*a^2 + 11*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(14*A*a^2 + 12*B*a^2 + 11*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1 )) - 2*(210*A*a^2*tan(1/2*d*x + 1/2*c)^11 + 180*B*a^2*tan(1/2*d*x + 1/2*c) ^11 + 165*C*a^2*tan(1/2*d*x + 1/2*c)^11 - 1190*A*a^2*tan(1/2*d*x + 1/2*c)^ 9 - 1020*B*a^2*tan(1/2*d*x + 1/2*c)^9 - 935*C*a^2*tan(1/2*d*x + 1/2*c)^9 + 2580*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 2568*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 1 986*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 3180*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 280 8*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 3006*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 2330* A*a^2*tan(1/2*d*x + 1/2*c)^3 + 1860*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 1305*C* a^2*tan(1/2*d*x + 1/2*c)^3 - 750*A*a^2*tan(1/2*d*x + 1/2*c) - 780*B*a^2*ta n(1/2*d*x + 1/2*c) - 795*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c) ^2 - 1)^6)/d
Time = 19.83 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.52 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2\,\mathrm {atanh}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (14\,A+12\,B+11\,C\right )}{4\,\left (\frac {7\,A\,a^2}{2}+3\,B\,a^2+\frac {11\,C\,a^2}{4}\right )}\right )\,\left (14\,A+12\,B+11\,C\right )}{8\,d}-\frac {\left (\frac {7\,A\,a^2}{4}+\frac {3\,B\,a^2}{2}+\frac {11\,C\,a^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (-\frac {119\,A\,a^2}{12}-\frac {17\,B\,a^2}{2}-\frac {187\,C\,a^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {43\,A\,a^2}{2}+\frac {107\,B\,a^2}{5}+\frac {331\,C\,a^2}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-\frac {53\,A\,a^2}{2}-\frac {117\,B\,a^2}{5}-\frac {501\,C\,a^2}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {233\,A\,a^2}{12}+\frac {31\,B\,a^2}{2}+\frac {87\,C\,a^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-\frac {25\,A\,a^2}{4}-\frac {13\,B\,a^2}{2}-\frac {53\,C\,a^2}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(a^2*atanh((a^2*tan(c/2 + (d*x)/2)*(14*A + 12*B + 11*C))/(4*((7*A*a^2)/2 + 3*B*a^2 + (11*C*a^2)/4)))*(14*A + 12*B + 11*C))/(8*d) - (tan(c/2 + (d*x)/ 2)^11*((7*A*a^2)/4 + (3*B*a^2)/2 + (11*C*a^2)/8) - tan(c/2 + (d*x)/2)^9*(( 119*A*a^2)/12 + (17*B*a^2)/2 + (187*C*a^2)/24) + tan(c/2 + (d*x)/2)^3*((23 3*A*a^2)/12 + (31*B*a^2)/2 + (87*C*a^2)/8) + tan(c/2 + (d*x)/2)^7*((43*A*a ^2)/2 + (107*B*a^2)/5 + (331*C*a^2)/20) - tan(c/2 + (d*x)/2)^5*((53*A*a^2) /2 + (117*B*a^2)/5 + (501*C*a^2)/20) - tan(c/2 + (d*x)/2)*((25*A*a^2)/4 + (13*B*a^2)/2 + (53*C*a^2)/8))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d *x)/2)^2 - 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))