Integrand size = 33, antiderivative size = 219 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x+\frac {a^2 \left (12 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac {a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac {b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac {2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d} \]
2*a*b*(2*A*b^2+(2*a^2+b^2)*C)*x+1/2*a^2*(12*A*b^2+a^2*(A+2*C))*arctanh(sin (d*x+c))/d-1/6*b^2*(a^2*(39*A-34*C)-2*b^2*(3*A+2*C))*sin(d*x+c)/d-1/3*a*b^ 3*(9*A-4*C)*cos(d*x+c)*sin(d*x+c)/d-1/6*b^2*(15*A-2*C)*(a+b*cos(d*x+c))^2* sin(d*x+c)/d+2*A*b*(a+b*cos(d*x+c))^3*tan(d*x+c)/d+1/2*A*(a+b*cos(d*x+c))^ 4*sec(d*x+c)*tan(d*x+c)/d
Time = 7.84 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.47 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {24 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) (c+d x)-6 a^2 \left (12 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 a^2 \left (12 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {3 a^4 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {48 a^3 A b \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {3 a^4 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {48 a^3 A b \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+3 b^2 \left (4 A b^2+3 \left (8 a^2+b^2\right ) C\right ) \sin (c+d x)+12 a b^3 C \sin (2 (c+d x))+b^4 C \sin (3 (c+d x))}{12 d} \]
(24*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*(c + d*x) - 6*a^2*(12*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*a^2*(12*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (3*a^4*A)/(Cos[(c + d* x)/2] - Sin[(c + d*x)/2])^2 + (48*a^3*A*b*Sin[(c + d*x)/2])/(Cos[(c + d*x) /2] - Sin[(c + d*x)/2]) - (3*a^4*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^ 2 + (48*a^3*A*b*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3*b^2*(4*A*b^2 + 3*(8*a^2 + b^2)*C)*Sin[c + d*x] + 12*a*b^3*C*Sin[2*(c + d *x)] + b^4*C*Sin[3*(c + d*x)])/(12*d)
Time = 1.84 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 3527, 3042, 3526, 3042, 3528, 3042, 3512, 27, 3042, 3502, 27, 3042, 3214, 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.
\(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 3527 |
\(\displaystyle \frac {1}{2} \int (a+b \cos (c+d x))^3 \left (-b (3 A-2 C) \cos ^2(c+d x)+a (A+2 C) \cos (c+d x)+4 A b\right ) \sec ^2(c+d x)dx+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (-b (3 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )+4 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \frac {1}{2} \left (\int (a+b \cos (c+d x))^2 \left ((A+2 C) a^2-2 b (A-2 C) \cos (c+d x) a+12 A b^2-b^2 (15 A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x)dx+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left ((A+2 C) a^2-2 b (A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+12 A b^2-b^2 (15 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int (a+b \cos (c+d x)) \left (-4 a b^2 (9 A-4 C) \cos ^2(c+d x)-b \left (3 a^2 (A-6 C)-2 b^2 (3 A+2 C)\right ) \cos (c+d x)+3 a \left ((A+2 C) a^2+12 A b^2\right )\right ) \sec (c+d x)dx-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-4 a b^2 (9 A-4 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (3 a^2 (A-6 C)-2 b^2 (3 A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a \left ((A+2 C) a^2+12 A b^2\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3512 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{2} \int 2 \left (3 \left ((A+2 C) a^2+12 A b^2\right ) a^2+12 b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x) a-b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx-\frac {2 a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\int \left (3 \left ((A+2 C) a^2+12 A b^2\right ) a^2+12 b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x) a-b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx-\frac {2 a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\int \frac {3 \left ((A+2 C) a^2+12 A b^2\right ) a^2+12 b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\int 3 \left (\left ((A+2 C) a^2+12 A b^2\right ) a^2+4 b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x) a\right ) \sec (c+d x)dx-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}-\frac {2 a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (3 \int \left (\left ((A+2 C) a^2+12 A b^2\right ) a^2+4 b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x) a\right ) \sec (c+d x)dx-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}-\frac {2 a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (3 \int \frac {\left ((A+2 C) a^2+12 A b^2\right ) a^2+4 b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}-\frac {2 a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (3 \left (a^2 \left (a^2 (A+2 C)+12 A b^2\right ) \int \sec (c+d x)dx+4 a b x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )\right )-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}-\frac {2 a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (3 \left (a^2 \left (a^2 (A+2 C)+12 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+4 a b x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )\right )-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}-\frac {2 a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (3 \left (\frac {a^2 \left (a^2 (A+2 C)+12 A b^2\right ) \text {arctanh}(\sin (c+d x))}{d}+4 a b x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )\right )-\frac {b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}-\frac {2 a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{d}\right )-\frac {b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {4 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d}\) |
(A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (-1/3*(b^2*(1 5*A - 2*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/d + (3*(4*a*b*(2*A*b^2 + ( 2*a^2 + b^2)*C)*x + (a^2*(12*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]]) /d) - (b^2*(a^2*(39*A - 34*C) - 2*b^2*(3*A + 2*C))*Sin[c + d*x])/d - (2*a* b^3*(9*A - 4*C)*Cos[c + d*x]*Sin[c + d*x])/d)/3 + (4*A*b*(a + b*Cos[c + d* x])^3*Tan[c + d*x])/d)/2
3.6.53.3.1 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_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 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[((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)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 9.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.89
method | result | size |
parts | \(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (A \,b^{4}+6 C \,a^{2} b^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (4 A a \,b^{3}+4 C \,a^{3} b \right ) \left (d x +c \right )}{d}+\frac {\left (6 A \,a^{2} b^{2}+C \,a^{4}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {4 A \,a^{3} b \tan \left (d x +c \right )}{d}+\frac {4 C a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(196\) |
derivativedivides | \(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 A \,a^{3} b \tan \left (d x +c \right )+4 C \,a^{3} b \left (d x +c \right )+6 A \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \sin \left (d x +c \right ) a^{2} b^{2}+4 A a \,b^{3} \left (d x +c \right )+4 C a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b^{4}+\frac {C \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(198\) |
default | \(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 A \,a^{3} b \tan \left (d x +c \right )+4 C \,a^{3} b \left (d x +c \right )+6 A \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 C \sin \left (d x +c \right ) a^{2} b^{2}+4 A a \,b^{3} \left (d x +c \right )+4 C a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b^{4}+\frac {C \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(198\) |
parallelrisch | \(\frac {-12 \left (1+\cos \left (2 d x +2 c \right )\right ) a^{2} \left (12 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \left (1+\cos \left (2 d x +2 c \right )\right ) a^{2} \left (12 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+96 x b \left (a^{2} C +\left (A +\frac {C}{2}\right ) b^{2}\right ) d a \cos \left (2 d x +2 c \right )+12 \left (6 C \,a^{2} b^{2}+b^{4} \left (A +\frac {11 C}{12}\right )\right ) \sin \left (3 d x +3 c \right )+24 \left (4 A \,a^{3} b +C a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+12 C \sin \left (4 d x +4 c \right ) a \,b^{3}+C \sin \left (5 d x +5 c \right ) b^{4}+12 \left (2 a^{4} A +6 C \,a^{2} b^{2}+\left (A +\frac {5 C}{6}\right ) b^{4}\right ) \sin \left (d x +c \right )+96 x b \left (a^{2} C +\left (A +\frac {C}{2}\right ) b^{2}\right ) d a}{24 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(277\) |
risch | \(4 x A a \,b^{3}+4 C \,a^{3} b x +2 a \,b^{3} C x +\frac {i C \,b^{4} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{4}}{2 d}-\frac {i C a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {i C \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{2} b^{2}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{4}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,b^{4}}{8 d}-\frac {i A \,a^{3} \left ({\mathrm e}^{3 i \left (d x +c \right )} a -8 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}-8 b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {i C a \,b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,b^{4}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{2} b^{2}}{d}+\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(423\) |
norman | \(\frac {\left (4 A a \,b^{3}+4 C \,a^{3} b +2 C a \,b^{3}\right ) x +\left (-40 A a \,b^{3}-40 C \,a^{3} b -20 C a \,b^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-16 A a \,b^{3}-16 C \,a^{3} b -8 C a \,b^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-16 A a \,b^{3}-16 C \,a^{3} b -8 C a \,b^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 A a \,b^{3}+4 C \,a^{3} b +2 C a \,b^{3}\right ) x \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (16 A a \,b^{3}+16 C \,a^{3} b +8 C a \,b^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (16 A a \,b^{3}+16 C \,a^{3} b +8 C a \,b^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (16 A a \,b^{3}+16 C \,a^{3} b +8 C a \,b^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (16 A a \,b^{3}+16 C \,a^{3} b +8 C a \,b^{3}\right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (a^{4} A -8 A \,a^{3} b +2 A \,b^{4}+12 C \,a^{2} b^{2}-4 C a \,b^{3}+2 C \,b^{4}\right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (a^{4} A +8 A \,a^{3} b +2 A \,b^{4}+12 C \,a^{2} b^{2}+4 C a \,b^{3}+2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (21 a^{4} A -120 A \,a^{3} b +18 A \,b^{4}+108 C \,a^{2} b^{2}-12 C a \,b^{3}+10 C \,b^{4}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (21 a^{4} A +120 A \,a^{3} b +18 A \,b^{4}+108 C \,a^{2} b^{2}+12 C a \,b^{3}+10 C \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (63 a^{4} A -216 A \,a^{3} b +6 A \,b^{4}+36 C \,a^{2} b^{2}+36 C a \,b^{3}-2 C \,b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (63 a^{4} A +216 A \,a^{3} b +6 A \,b^{4}+36 C \,a^{2} b^{2}-36 C a \,b^{3}-2 C \,b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (105 a^{4} A -120 A \,a^{3} b -30 A \,b^{4}-180 C \,a^{2} b^{2}+36 C a \,b^{3}-14 C \,b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (105 a^{4} A +120 A \,a^{3} b -30 A \,b^{4}-180 C \,a^{2} b^{2}-36 C a \,b^{3}-14 C \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{2} \left (A \,a^{2}+12 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (A \,a^{2}+12 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(872\) |
a^4*A/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(A*b^4+6 *C*a^2*b^2)/d*sin(d*x+c)+(4*A*a*b^3+4*C*a^3*b)/d*(d*x+c)+(6*A*a^2*b^2+C*a^ 4)/d*ln(sec(d*x+c)+tan(d*x+c))+1/3*C*b^4/d*(2+cos(d*x+c)^2)*sin(d*x+c)+4*A *a^3*b/d*tan(d*x+c)+4*C*a*b^3/d*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)
Time = 0.31 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.96 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {24 \, {\left (2 \, C a^{3} b + {\left (2 \, A + C\right )} a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (A + 2 \, C\right )} a^{4} + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (A + 2 \, C\right )} a^{4} + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C b^{4} \cos \left (d x + c\right )^{4} + 12 \, C a b^{3} \cos \left (d x + c\right )^{3} + 24 \, A a^{3} b \cos \left (d x + c\right ) + 3 \, A a^{4} + 2 \, {\left (18 \, C a^{2} b^{2} + {\left (3 \, A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \]
1/12*(24*(2*C*a^3*b + (2*A + C)*a*b^3)*d*x*cos(d*x + c)^2 + 3*((A + 2*C)*a ^4 + 12*A*a^2*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*((A + 2*C)*a^4 + 12*A*a^2*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*C*b^4*cos(d* x + c)^4 + 12*C*a*b^3*cos(d*x + c)^3 + 24*A*a^3*b*cos(d*x + c) + 3*A*a^4 + 2*(18*C*a^2*b^2 + (3*A + 2*C)*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d *x + c)^2)
Timed out. \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {48 \, {\left (d x + c\right )} C a^{3} b + 48 \, {\left (d x + c\right )} A a b^{3} + 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{4} - 3 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 72 \, C a^{2} b^{2} \sin \left (d x + c\right ) + 12 \, A b^{4} \sin \left (d x + c\right ) + 48 \, A a^{3} b \tan \left (d x + c\right )}{12 \, d} \]
1/12*(48*(d*x + c)*C*a^3*b + 48*(d*x + c)*A*a*b^3 + 12*(2*d*x + 2*c + sin( 2*d*x + 2*c))*C*a*b^3 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*b^4 - 3*A*a^ 4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d *x + c) - 1)) + 6*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*A*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 72*C*a^2*b^ 2*sin(d*x + c) + 12*A*b^4*sin(d*x + c) + 48*A*a^3*b*tan(d*x + c))/d
Time = 0.37 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.81 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {12 \, {\left (2 \, C a^{3} b + 2 \, A a b^{3} + C a b^{3}\right )} {\left (d x + c\right )} + 3 \, {\left (A a^{4} + 2 \, C a^{4} + 12 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A a^{4} + 2 \, C a^{4} + 12 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {6 \, {\left (A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {4 \, {\left (18 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
1/6*(12*(2*C*a^3*b + 2*A*a*b^3 + C*a*b^3)*(d*x + c) + 3*(A*a^4 + 2*C*a^4 + 12*A*a^2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(A*a^4 + 2*C*a^4 + 1 2*A*a^2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(A*a^4*tan(1/2*d*x + 1 /2*c)^3 - 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 + A*a^4*tan(1/2*d*x + 1/2*c) + 8*A*a^3*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 4*(18*C*a ^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*b^4 *tan(1/2*d*x + 1/2*c)^5 + 3*C*b^4*tan(1/2*d*x + 1/2*c)^5 + 36*C*a^2*b^2*ta n(1/2*d*x + 1/2*c)^3 + 6*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 2*C*b^4*tan(1/2*d* x + 1/2*c)^3 + 18*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 6*C*a*b^3*tan(1/2*d*x + 1/2*c) + 3*A*b^4*tan(1/2*d*x + 1/2*c) + 3*C*b^4*tan(1/2*d*x + 1/2*c))/(ta n(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 3.53 (sec) , antiderivative size = 2658, normalized size of antiderivative = 12.14 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Too large to display} \]
(tan(c/2 + (d*x)/2)^5*(6*A*a^4 - 4*A*b^4 + (4*C*b^4)/3 - 24*C*a^2*b^2) + t an(c/2 + (d*x)/2)^3*(4*A*a^4 - (8*C*b^4)/3 + 16*A*a^3*b - 8*C*a*b^3) + tan (c/2 + (d*x)/2)^7*(4*A*a^4 - (8*C*b^4)/3 - 16*A*a^3*b + 8*C*a*b^3) + tan(c /2 + (d*x)/2)*(A*a^4 + 2*A*b^4 + 2*C*b^4 + 12*C*a^2*b^2 + 8*A*a^3*b + 4*C* a*b^3) + tan(c/2 + (d*x)/2)^9*(A*a^4 + 2*A*b^4 + 2*C*b^4 + 12*C*a^2*b^2 - 8*A*a^3*b - 4*C*a*b^3))/(d*(tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (atan(((((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2)*(16*A*a^4 + 32*C*a^4 + 192 *A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b) + tan(c/2 + (d*x)/2)* (8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192*A^2*a^6 *b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2))*((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2)*1i - (((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2)*(16*A*a^4 + 32*C*a^ 4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b) - tan(c/2 + (d *x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*b^4 + 192* A^2*a^6*b^2 + 128*C^2*a^2*b^6 + 512*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C *a^8 + 512*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 384*A*C*a^6*b^2))*((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2)*1i)/((((A*a^4)/2 + C*a^4 + 6*A*a^2*b^2)*(16*A*a^4 + 32*C*a^4 + 192*A*a^2*b^2 + 128*A*a*b^3 + 64*C*a*b^3 + 128*C*a^3*b) + tan(c /2 + (d*x)/2)*(8*A^2*a^8 + 32*C^2*a^8 + 512*A^2*a^2*b^6 + 1152*A^2*a^4*...