Integrand size = 33, antiderivative size = 227 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A b (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]
1/8*b*(4*b^2*(A+2*C)+3*a^2*(3*A+4*C))*arctanh(sin(d*x+c))/d+1/15*a*(15*b^2 *(2*A+3*C)+2*a^2*(4*A+5*C))*tan(d*x+c)/d+3/40*b*(2*A*b^2+5*a^2*(3*A+4*C))* sec(d*x+c)*tan(d*x+c)/d+1/30*a*(3*A*b^2+2*a^2*(4*A+5*C))*sec(d*x+c)^2*tan( d*x+c)/d+3/20*A*b*(a+b*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c)/d+1/5*A*(a+b* cos(d*x+c))^3*sec(d*x+c)^4*tan(d*x+c)/d
Time = 2.51 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.66 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 b \left (4 A b^2+3 a^2 (3 A+4 C)\right ) \sec (c+d x)+90 a^2 A b \sec ^3(c+d x)+8 a \left (15 \left (a^2+3 b^2\right ) (A+C)+5 \left (3 A b^2+a^2 (2 A+C)\right ) \tan ^2(c+d x)+3 a^2 A \tan ^4(c+d x)\right )\right )}{120 d} \]
(15*b*(4*b^2*(A + 2*C) + 3*a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*b*(4*A*b^2 + 3*a^2*(3*A + 4*C))*Sec[c + d*x] + 90*a^2*A*b*Sec[c + d*x]^3 + 8*a*(15*(a^2 + 3*b^2)*(A + C) + 5*(3*A*b^2 + a^2*(2*A + C))*Ta n[c + d*x]^2 + 3*a^2*A*Tan[c + d*x]^4)))/(120*d)
Time = 1.61 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3527, 3042, 3526, 3042, 3510, 25, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.
\(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
\(\Big \downarrow \) 3527 |
\(\displaystyle \frac {1}{5} \int (a+b \cos (c+d x))^2 \left (b (A+5 C) \cos ^2(c+d x)+a (4 A+5 C) \cos (c+d x)+3 A b\right ) \sec ^5(c+d x)dx+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (b (A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int (a+b \cos (c+d x)) \left (b^2 (7 A+20 C) \cos ^2(c+d x)+a b (29 A+40 C) \cos (c+d x)+2 \left (2 (4 A+5 C) a^2+3 A b^2\right )\right ) \sec ^4(c+d x)dx+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b^2 (7 A+20 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a b (29 A+40 C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left (2 (4 A+5 C) a^2+3 A b^2\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 3510 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {1}{3} \int -\left (\left (3 (7 A+20 C) \cos ^2(c+d x) b^3+9 \left (5 (3 A+4 C) a^2+2 A b^2\right ) b+4 a \left (2 (4 A+5 C) a^2+15 b^2 (2 A+3 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x)\right )dx\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \left (3 (7 A+20 C) \cos ^2(c+d x) b^3+9 \left (5 (3 A+4 C) a^2+2 A b^2\right ) b+4 a \left (2 (4 A+5 C) a^2+15 b^2 (2 A+3 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \frac {3 (7 A+20 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^3+9 \left (5 (3 A+4 C) a^2+2 A b^2\right ) b+4 a \left (2 (4 A+5 C) a^2+15 b^2 (2 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 3500 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (8 a \left (2 (4 A+5 C) a^2+15 b^2 (2 A+3 C)\right )+15 b \left (3 (3 A+4 C) a^2+4 b^2 (A+2 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {9 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {8 a \left (2 (4 A+5 C) a^2+15 b^2 (2 A+3 C)\right )+15 b \left (3 (3 A+4 C) a^2+4 b^2 (A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {9 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (8 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \int \sec ^2(c+d x)dx+15 b \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int \sec (c+d x)dx\right )+\frac {9 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 b \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+8 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {9 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 b \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {8 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {9 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 b \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {8 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \tan (c+d x)}{d}\right )+\frac {9 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {15 b \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {8 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \tan (c+d x)}{d}\right )+\frac {9 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
(A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((3*A*b*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((2*a*(3*A*b^2 + 2*a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((9*b*(2*A*b^2 + 5 *a^2*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((15*b*(4*b^2*(A + 2* C) + 3*a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/d + (8*a*(15*b^2*(2*A + 3*C ) + 2*a^2*(4*A + 5*C))*Tan[c + d*x])/d)/2)/3)/4)/5
3.6.47.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 11.62 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.95
method | result | size |
parts | \(-\frac {A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{3}+3 C \,a^{2} b \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (3 A a \,b^{2}+C \,a^{3}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 A \,a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {3 C a \,b^{2} \tan \left (d x +c \right )}{d}\) | \(215\) |
derivativedivides | \(\frac {-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-C \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 A \,a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 C \,a^{2} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 A a \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 C \tan \left (d x +c \right ) a \,b^{2}+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(246\) |
default | \(\frac {-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-C \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 A \,a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 C \,a^{2} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 A a \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 C \tan \left (d x +c \right ) a \,b^{2}+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(246\) |
parallelrisch | \(\frac {-135 b \left (\frac {4 b^{2} \left (A +2 C \right )}{9}+\left (A +\frac {4 C}{3}\right ) a^{2}\right ) \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+135 b \left (\frac {4 b^{2} \left (A +2 C \right )}{9}+\left (A +\frac {4 C}{3}\right ) a^{2}\right ) \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+320 \left (\frac {3 \left (5 A +\frac {9 C}{2}\right ) b^{2}}{4}+a^{2} \left (A +\frac {5 C}{4}\right )\right ) a \sin \left (3 d x +3 c \right )+64 \left (\frac {15 b^{2} \left (A +\frac {3 C}{2}\right )}{4}+a^{2} \left (A +\frac {5 C}{4}\right )\right ) a \sin \left (5 d x +5 c \right )+1260 \left (\frac {4 A \,b^{2}}{21}+a^{2} \left (A +\frac {4 C}{7}\right )\right ) b \sin \left (2 d x +2 c \right )+270 b \left (\frac {4 A \,b^{2}}{9}+\left (A +\frac {4 C}{3}\right ) a^{2}\right ) \sin \left (4 d x +4 c \right )+640 \sin \left (d x +c \right ) \left (\frac {3 \left (A +\frac {3 C}{4}\right ) b^{2}}{2}+a^{2} \left (A +\frac {C}{2}\right )\right ) a}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(317\) |
risch | \(-\frac {i \left (-64 A \,a^{3}-360 C a \,b^{2}-240 A a \,b^{2}-60 A \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+120 A \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-320 A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-400 C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+60 A \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}-240 C \,a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-640 A \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-120 A \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-560 C \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+630 A \,a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-360 C \,a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-1440 C a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-630 A \,a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-720 A a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-1680 A a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-1200 A a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+180 C \,a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}-360 C a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+360 C \,a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-135 A \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-180 C \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-1440 C a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-2160 C a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+135 A \,a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}-80 C \,a^{3}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {9 A \,a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} C}{2 d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {9 A \,a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} C}{2 d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(602\) |
-A*a^3/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(A*b^3+3*C* a^2*b)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-(3*A*a* b^2+C*a^3)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+C*b^3/d*ln(sec(d*x+c)+tan( d*x+c))+3*A*a^2*b/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)))+3*C*a*b^2/d*tan(d*x+c)
Time = 0.29 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.99 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (90 \, A a^{2} b \cos \left (d x + c\right ) + 8 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, {\left (2 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{4} + 24 \, A a^{3} + 15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
1/240*(15*(3*(3*A + 4*C)*a^2*b + 4*(A + 2*C)*b^3)*cos(d*x + c)^5*log(sin(d *x + c) + 1) - 15*(3*(3*A + 4*C)*a^2*b + 4*(A + 2*C)*b^3)*cos(d*x + c)^5*l og(-sin(d*x + c) + 1) + 2*(90*A*a^2*b*cos(d*x + c) + 8*(2*(4*A + 5*C)*a^3 + 15*(2*A + 3*C)*a*b^2)*cos(d*x + c)^4 + 24*A*a^3 + 15*(3*(3*A + 4*C)*a^2* b + 4*A*b^3)*cos(d*x + c)^3 + 8*((4*A + 5*C)*a^3 + 15*A*a*b^2)*cos(d*x + c )^2)*sin(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.30 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} - 45 \, A a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, C a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 720 \, C a b^{2} \tan \left (d x + c\right )}{240 \, d} \]
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 + 240*(tan(d*x + c)^3 + 3*tan( d*x + c))*A*a*b^2 - 45*A*a^2*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)) - 180*C*a^2*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(si n(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*A*b^3*(2*sin(d*x + c)/(sin(d *x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*C*b^ 3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 720*C*a*b^2*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 656 vs. \(2 (215) = 430\).
Time = 0.34 (sec) , antiderivative size = 656, normalized size of antiderivative = 2.89 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (9 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 8 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (9 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 8 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 320 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 90 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 960 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1440 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2160 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 90 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 360 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 960 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1440 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{3} \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*(15*(9*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 8*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(9*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 8*C*b^3)*log(abs( tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 120*C*a ^3*tan(1/2*d*x + 1/2*c)^9 - 225*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 180*C*a^2 *b*tan(1/2*d*x + 1/2*c)^9 + 360*A*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 360*C*a*b ^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*b^3*tan(1/2*d*x + 1/2*c)^9 - 160*A*a^3*ta n(1/2*d*x + 1/2*c)^7 - 320*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 90*A*a^2*b*tan(1 /2*d*x + 1/2*c)^7 + 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 960*A*a*b^2*tan(1 /2*d*x + 1/2*c)^7 - 1440*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*A*b^3*tan(1/ 2*d*x + 1/2*c)^7 + 464*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 400*C*a^3*tan(1/2*d* x + 1/2*c)^5 + 1200*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 2160*C*a*b^2*tan(1/2* d*x + 1/2*c)^5 - 160*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 320*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 90*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 1440*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 120*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^3*tan(1/2*d*x + 1/ 2*c) + 120*C*a^3*tan(1/2*d*x + 1/2*c) + 225*A*a^2*b*tan(1/2*d*x + 1/2*c) + 180*C*a^2*b*tan(1/2*d*x + 1/2*c) + 360*A*a*b^2*tan(1/2*d*x + 1/2*c) + 360 *C*a*b^2*tan(1/2*d*x + 1/2*c) + 60*A*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d* x + 1/2*c)^2 - 1)^5)/d
Time = 5.55 (sec) , antiderivative size = 445, normalized size of antiderivative = 1.96 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {b\,\mathrm {atanh}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,A\,a^2+4\,A\,b^2+12\,C\,a^2+8\,C\,b^2\right )}{2\,\left (2\,A\,b^3+4\,C\,b^3+\frac {9\,A\,a^2\,b}{2}+6\,C\,a^2\,b\right )}\right )\,\left (9\,A\,a^2+4\,A\,b^2+12\,C\,a^2+8\,C\,b^2\right )}{4\,d}-\frac {\left (2\,A\,a^3-A\,b^3+2\,C\,a^3+6\,A\,a\,b^2-\frac {15\,A\,a^2\,b}{4}+6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,b^3-\frac {8\,A\,a^3}{3}-\frac {16\,C\,a^3}{3}-16\,A\,a\,b^2+\frac {3\,A\,a^2\,b}{2}-24\,C\,a\,b^2+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^3}{15}+\frac {20\,C\,a^3}{3}+20\,A\,a\,b^2+36\,C\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^3}{3}-2\,A\,b^3-\frac {16\,C\,a^3}{3}-16\,A\,a\,b^2-\frac {3\,A\,a^2\,b}{2}-24\,C\,a\,b^2-6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+A\,b^3+2\,C\,a^3+6\,A\,a\,b^2+\frac {15\,A\,a^2\,b}{4}+6\,C\,a\,b^2+3\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(b*atanh((b*tan(c/2 + (d*x)/2)*(9*A*a^2 + 4*A*b^2 + 12*C*a^2 + 8*C*b^2))/( 2*(2*A*b^3 + 4*C*b^3 + (9*A*a^2*b)/2 + 6*C*a^2*b)))*(9*A*a^2 + 4*A*b^2 + 1 2*C*a^2 + 8*C*b^2))/(4*d) - (tan(c/2 + (d*x)/2)^9*(2*A*a^3 - A*b^3 + 2*C*a ^3 + 6*A*a*b^2 - (15*A*a^2*b)/4 + 6*C*a*b^2 - 3*C*a^2*b) - tan(c/2 + (d*x) /2)^3*((8*A*a^3)/3 + 2*A*b^3 + (16*C*a^3)/3 + 16*A*a*b^2 + (3*A*a^2*b)/2 + 24*C*a*b^2 + 6*C*a^2*b) - tan(c/2 + (d*x)/2)^7*((8*A*a^3)/3 - 2*A*b^3 + ( 16*C*a^3)/3 + 16*A*a*b^2 - (3*A*a^2*b)/2 + 24*C*a*b^2 - 6*C*a^2*b) + tan(c /2 + (d*x)/2)^5*((116*A*a^3)/15 + (20*C*a^3)/3 + 20*A*a*b^2 + 36*C*a*b^2) + tan(c/2 + (d*x)/2)*(2*A*a^3 + A*b^3 + 2*C*a^3 + 6*A*a*b^2 + (15*A*a^2*b) /4 + 6*C*a*b^2 + 3*C*a^2*b))/(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))