Integrand size = 41, antiderivative size = 398 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) x}{2 b^5}+\frac {2 a^2 \left (2 a^2 A b^2-3 A b^4-3 a^3 b B+4 a b^3 B+4 a^4 C-5 a^2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^5 (a+b)^{3/2} d}-\frac {\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac {\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
1/2*(6*B*a^2*b+B*b^3-8*a^3*C-2*a*b^2*(2*A+C))*x/b^5+2*a^2*(2*A*a^2*b^2-3*A *b^4-3*B*a^3*b+4*B*a*b^3+4*C*a^4-5*C*a^2*b^2)*arctan((a-b)^(1/2)*tan(1/2*d *x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^5/(a+b)^(3/2)/d-1/3*(9*B*a^3*b-6*B*a* b^3-a^2*b^2*(6*A-7*C)-12*a^4*C+b^4*(3*A+2*C))*sin(d*x+c)/b^4/(a^2-b^2)/d+1 /2*(3*B*a^2*b-B*b^3-2*a*b^2*(A-C)-4*a^3*C)*cos(d*x+c)*sin(d*x+c)/b^3/(a^2- b^2)/d+1/3*(3*A*b^2-3*B*a*b+4*C*a^2-C*b^2)*cos(d*x+c)^2*sin(d*x+c)/b^2/(a^ 2-b^2)/d-(A*b^2-a*(B*b-C*a))*cos(d*x+c)^3*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*co s(d*x+c))
Time = 4.02 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {6 \left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) (c+d x)+\frac {24 a^2 \left (-3 A b^4-3 a^3 b B+4 a b^3 B+a^2 b^2 (2 A-5 C)+4 a^4 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+3 b \left (4 A b^2-8 a b B+12 a^2 C+3 b^2 C\right ) \sin (c+d x)+\frac {12 a^3 b \left (A b^2+a (-b B+a C)\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+3 b^2 (b B-2 a C) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^5 d} \]
(6*(6*a^2*b*B + b^3*B - 8*a^3*C - 2*a*b^2*(2*A + C))*(c + d*x) + (24*a^2*( -3*A*b^4 - 3*a^3*b*B + 4*a*b^3*B + a^2*b^2*(2*A - 5*C) + 4*a^4*C)*ArcTanh[ ((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 3*b*(4* A*b^2 - 8*a*b*B + 12*a^2*C + 3*b^2*C)*Sin[c + d*x] + (12*a^3*b*(A*b^2 + a* (-(b*B) + a*C))*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])) + 3*b ^2*(b*B - 2*a*C)*Sin[2*(c + d*x)] + b^3*C*Sin[3*(c + d*x)])/(12*b^5*d)
Time = 2.18 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.02, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 3526, 3042, 3528, 25, 3042, 3528, 3042, 3502, 27, 3042, 3214, 3042, 3138, 218}
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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle -\frac {\int \frac {\cos ^2(c+d x) \left (-\left (\left (4 C a^2-3 b B a+3 A b^2-b^2 C\right ) \cos ^2(c+d x)\right )+b (b B-a (A+C)) \cos (c+d x)+3 \left (A b^2-a (b B-a C)\right )\right )}{a+b \cos (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (-4 C a^2+3 b B a-3 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (b B-a (A+C)) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (A b^2-a (b B-a C)\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle -\frac {\frac {\int -\frac {\cos (c+d x) \left (3 \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right ) \cos ^2(c+d x)-b \left (C a^2-3 b B a+3 A b^2+2 b^2 C\right ) \cos (c+d x)+2 a \left (4 C a^2-3 b B a+3 A b^2-b^2 C\right )\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {\int \frac {\cos (c+d x) \left (3 \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right ) \cos ^2(c+d x)-b \left (C a^2-3 b B a+3 A b^2+2 b^2 C\right ) \cos (c+d x)+2 a \left (4 C a^2-3 b B a+3 A b^2-b^2 C\right )\right )}{a+b \cos (c+d x)}dx}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (3 \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (C a^2-3 b B a+3 A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a \left (4 C a^2-3 b B a+3 A b^2-b^2 C\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {-2 \left (-12 C a^4+9 b B a^3-b^2 (6 A-7 C) a^2-6 b^3 B a+b^4 (3 A+2 C)\right ) \cos ^2(c+d x)-b \left (-4 C a^3+3 b B a^2-2 b^2 (3 A+C) a+3 b^3 B\right ) \cos (c+d x)+3 a \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )}{a+b \cos (c+d x)}dx}{2 b}+\frac {3 \sin (c+d x) \cos (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {-2 \left (-12 C a^4+9 b B a^3-b^2 (6 A-7 C) a^2-6 b^3 B a+b^4 (3 A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (-4 C a^3+3 b B a^2-2 b^2 (3 A+C) a+3 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}+\frac {3 \sin (c+d x) \cos (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {\int \frac {3 \left (a b \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )+\left (a^2-b^2\right ) \left (-8 C a^3+6 b B a^2-2 b^2 (2 A+C) a+b^3 B\right ) \cos (c+d x)\right )}{a+b \cos (c+d x)}dx}{b}-\frac {2 \sin (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \sin (c+d x) \cos (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {3 \int \frac {a b \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )+\left (a^2-b^2\right ) \left (-8 C a^3+6 b B a^2-2 b^2 (2 A+C) a+b^3 B\right ) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b}-\frac {2 \sin (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \sin (c+d x) \cos (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {3 \int \frac {a b \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )+\left (a^2-b^2\right ) \left (-8 C a^3+6 b B a^2-2 b^2 (2 A+C) a+b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \sin (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \sin (c+d x) \cos (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {3 \left (\frac {x \left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right )}{b}-\frac {2 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}\right )}{b}-\frac {2 \sin (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \sin (c+d x) \cos (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {3 \left (\frac {x \left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right )}{b}-\frac {2 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {2 \sin (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \sin (c+d x) \cos (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {3 \left (\frac {x \left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right )}{b}-\frac {4 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {2 \sin (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \sin (c+d x) \cos (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {-\frac {\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}-\frac {\frac {3 \sin (c+d x) \cos (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}+\frac {\frac {3 \left (\frac {x \left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right )}{b}-\frac {4 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {2 \sin (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}}{3 b}}{b \left (a^2-b^2\right )}\) |
-(((A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^3*Sin[c + d*x])/(b*(a^2 - b^2)*d*( a + b*Cos[c + d*x]))) - (-1/3*((3*A*b^2 - 3*a*b*B + 4*a^2*C - b^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(b*d) - ((3*(3*a^2*b*B - b^3*B - 2*a*b^2*(A - C) - 4*a^3*C)*Cos[c + d*x]*Sin[c + d*x])/(2*b*d) + ((3*(((a^2 - b^2)*(6*a^2*b* B + b^3*B - 8*a^3*C - 2*a*b^2*(2*A + C))*x)/b - (4*a^2*(3*A*b^4 + 3*a^3*b* B - 4*a*b^3*B - a^2*b^2*(2*A - 5*C) - 4*a^4*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)))/b - (2*(9*a^3*b*B - 6*a*b^3*B - a^2*b^2*(6*A - 7*C) - 12*a^4*C + b^4*(3*A + 2*C))*Sin[c + d *x])/(b*d))/(2*b))/(3*b))/(b*(a^2 - b^2))
3.10.86.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
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_)])^(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_.) + (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])))
Time = 0.74 (sec) , antiderivative size = 391, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {\frac {2 a^{2} \left (\frac {a \left (A \,b^{2}-B a b +a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {\left (2 A \,a^{2} b^{2}-3 A \,b^{4}-3 B \,a^{3} b +4 B a \,b^{3}+4 a^{4} C -5 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5}}-\frac {2 \left (\frac {\left (-A \,b^{3}+2 B a \,b^{2}+\frac {1}{2} B \,b^{3}-3 a^{2} b C -C a \,b^{2}-C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A \,b^{3}+4 B a \,b^{2}-6 a^{2} b C -\frac {2}{3} C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A \,b^{3}+2 B a \,b^{2}-3 a^{2} b C -C \,b^{3}-\frac {1}{2} B \,b^{3}+C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\left (4 a A \,b^{2}-6 B \,a^{2} b -B \,b^{3}+8 a^{3} C +2 C a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{5}}}{d}\) | \(391\) |
| default | \(\frac {\frac {2 a^{2} \left (\frac {a \left (A \,b^{2}-B a b +a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {\left (2 A \,a^{2} b^{2}-3 A \,b^{4}-3 B \,a^{3} b +4 B a \,b^{3}+4 a^{4} C -5 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5}}-\frac {2 \left (\frac {\left (-A \,b^{3}+2 B a \,b^{2}+\frac {1}{2} B \,b^{3}-3 a^{2} b C -C a \,b^{2}-C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A \,b^{3}+4 B a \,b^{2}-6 a^{2} b C -\frac {2}{3} C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A \,b^{3}+2 B a \,b^{2}-3 a^{2} b C -C \,b^{3}-\frac {1}{2} B \,b^{3}+C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\left (4 a A \,b^{2}-6 B \,a^{2} b -B \,b^{3}+8 a^{3} C +2 C a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{5}}}{d}\) | \(391\) |
| risch | \(\text {Expression too large to display}\) | \(1436\) |
1/d*(2*a^2/b^5*(a*(A*b^2-B*a*b+C*a^2)*b/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan( 1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)+(2*A*a^2*b^2-3*A*b^4-3*B*a^ 3*b+4*B*a*b^3+4*C*a^4-5*C*a^2*b^2)/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan( (a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))-2/b^5*(((-A*b^3+2*B*a*b^2+1 /2*B*b^3-3*a^2*b*C-C*a*b^2-C*b^3)*tan(1/2*d*x+1/2*c)^5+(-2*A*b^3+4*B*a*b^2 -6*a^2*b*C-2/3*C*b^3)*tan(1/2*d*x+1/2*c)^3+(-A*b^3+2*B*a*b^2-3*a^2*b*C-C*b ^3-1/2*B*b^3+C*a*b^2)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*( 4*A*a*b^2-6*B*a^2*b-B*b^3+8*C*a^3+2*C*a*b^2)*arctan(tan(1/2*d*x+1/2*c))))
Time = 0.40 (sec) , antiderivative size = 1357, normalized size of antiderivative = 3.41 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2, x, algorithm="fricas")
[-1/6*(3*(8*C*a^7*b - 6*B*a^6*b^2 + 2*(2*A - 7*C)*a^5*b^3 + 11*B*a^4*b^4 - 4*(2*A - C)*a^3*b^5 - 4*B*a^2*b^6 + 2*(2*A + C)*a*b^7 - B*b^8)*d*x*cos(d* x + c) + 3*(8*C*a^8 - 6*B*a^7*b + 2*(2*A - 7*C)*a^6*b^2 + 11*B*a^5*b^3 - 4 *(2*A - C)*a^4*b^4 - 4*B*a^3*b^5 + 2*(2*A + C)*a^2*b^6 - B*a*b^7)*d*x + 3* (4*C*a^7 - 3*B*a^6*b + (2*A - 5*C)*a^5*b^2 + 4*B*a^4*b^3 - 3*A*a^3*b^4 + ( 4*C*a^6*b - 3*B*a^5*b^2 + (2*A - 5*C)*a^4*b^3 + 4*B*a^3*b^4 - 3*A*a^2*b^5) *cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*co s(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - (24*C*a^7*b - 18*B*a^6*b^2 + 2*(6*A - 19*C)*a^5*b^3 + 30*B*a^4*b^4 - 2*(9*A - 5*C)*a^3*b ^5 - 12*B*a^2*b^6 + 2*(3*A + 2*C)*a*b^7 + 2*(C*a^4*b^4 - 2*C*a^2*b^6 + C*b ^8)*cos(d*x + c)^3 - (4*C*a^5*b^3 - 3*B*a^4*b^4 - 8*C*a^3*b^5 + 6*B*a^2*b^ 6 + 4*C*a*b^7 - 3*B*b^8)*cos(d*x + c)^2 + (12*C*a^6*b^2 - 9*B*a^5*b^3 + 2* (3*A - 10*C)*a^4*b^4 + 18*B*a^3*b^5 - 4*(3*A - C)*a^2*b^6 - 9*B*a*b^7 + 2* (3*A + 2*C)*b^8)*cos(d*x + c))*sin(d*x + c))/((a^4*b^6 - 2*a^2*b^8 + b^10) *d*cos(d*x + c) + (a^5*b^5 - 2*a^3*b^7 + a*b^9)*d), -1/6*(3*(8*C*a^7*b - 6 *B*a^6*b^2 + 2*(2*A - 7*C)*a^5*b^3 + 11*B*a^4*b^4 - 4*(2*A - C)*a^3*b^5 - 4*B*a^2*b^6 + 2*(2*A + C)*a*b^7 - B*b^8)*d*x*cos(d*x + c) + 3*(8*C*a^8 - 6 *B*a^7*b + 2*(2*A - 7*C)*a^6*b^2 + 11*B*a^5*b^3 - 4*(2*A - C)*a^4*b^4 - 4* B*a^3*b^5 + 2*(2*A + C)*a^2*b^6 - B*a*b^7)*d*x - 6*(4*C*a^7 - 3*B*a^6*b...
Timed out. \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2, x, algorithm="maxima")
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.35 (sec) , antiderivative size = 563, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {12 \, {\left (4 \, C a^{6} - 3 \, B a^{5} b + 2 \, A a^{4} b^{2} - 5 \, C a^{4} b^{2} + 4 \, B a^{3} b^{3} - 3 \, A a^{2} b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{5} - b^{7}\right )} \sqrt {a^{2} - b^{2}}} - \frac {12 \, {\left (C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} + \frac {3 \, {\left (8 \, C a^{3} - 6 \, B a^{2} b + 4 \, A a b^{2} + 2 \, C a b^{2} - B b^{3}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {2 \, {\left (18 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{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 )}^{3} b^{4}}}{6 \, d} \]
integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2, x, algorithm="giac")
-1/6*(12*(4*C*a^6 - 3*B*a^5*b + 2*A*a^4*b^2 - 5*C*a^4*b^2 + 4*B*a^3*b^3 - 3*A*a^2*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-( a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^2*b ^5 - b^7)*sqrt(a^2 - b^2)) - 12*(C*a^5*tan(1/2*d*x + 1/2*c) - B*a^4*b*tan( 1/2*d*x + 1/2*c) + A*a^3*b^2*tan(1/2*d*x + 1/2*c))/((a^2*b^4 - b^6)*(a*tan (1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) + 3*(8*C*a^3 - 6* B*a^2*b + 4*A*a*b^2 + 2*C*a*b^2 - B*b^3)*(d*x + c)/b^5 - 2*(18*C*a^2*tan(1 /2*d*x + 1/2*c)^5 - 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^2*tan(1/2*d*x + 1/2*c) ^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 36*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 24 *B*a*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*C*b^2* tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*tan(1/2*d*x + 1/2*c) - 12*B*a*b*tan(1/2* d*x + 1/2*c) - 6*C*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x + 1/2*c) + 6*C*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2* d*x + 1/2*c)^2 + 1)^3*b^4))/d
Time = 14.76 (sec) , antiderivative size = 11768, normalized size of antiderivative = 29.57 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
- ((tan(c/2 + (d*x)/2)*(2*A*b^5 + B*b^5 - 8*C*a^5 + 2*C*b^5 - 2*A*a^2*b^3 - 4*A*a^3*b^2 - 5*B*a^2*b^3 + 3*B*a^3*b^2 + 2*C*a^2*b^3 + 6*C*a^3*b^2 + 2* A*a*b^4 - 3*B*a*b^4 + 6*B*a^4*b - 4*C*a^4*b))/(b^4*(a + b)*(a - b)) - (tan (c/2 + (d*x)/2)^3*(3*B*b^5 - 6*A*b^5 + 72*C*a^5 + 2*C*b^5 + 6*A*a^2*b^3 + 36*A*a^3*b^2 + 33*B*a^2*b^3 - 9*B*a^3*b^2 - 14*C*a^2*b^3 - 38*C*a^3*b^2 - 18*A*a*b^4 + 9*B*a*b^4 - 54*B*a^4*b - 16*C*a*b^4 + 12*C*a^4*b))/(3*b^4*(a + b)*(a - b)) + (tan(c/2 + (d*x)/2)^5*(2*C*b^5 - 3*B*b^5 - 72*C*a^5 - 6*A* b^5 + 6*A*a^2*b^3 - 36*A*a^3*b^2 - 33*B*a^2*b^3 - 9*B*a^3*b^2 - 14*C*a^2*b ^3 + 38*C*a^3*b^2 + 18*A*a*b^4 + 9*B*a*b^4 + 54*B*a^4*b + 16*C*a*b^4 + 12* C*a^4*b))/(3*b^4*(a + b)*(a - b)) - (tan(c/2 + (d*x)/2)^7*(2*A*b^5 - B*b^5 + 8*C*a^5 + 2*C*b^5 - 2*A*a^2*b^3 + 4*A*a^3*b^2 + 5*B*a^2*b^3 + 3*B*a^3*b ^2 + 2*C*a^2*b^3 - 6*C*a^3*b^2 - 2*A*a*b^4 - 3*B*a*b^4 - 6*B*a^4*b - 4*C*a ^4*b))/(b^4*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d*x)/2)^8*(a - b) + t an(c/2 + (d*x)/2)^2*(4*a + 2*b) + tan(c/2 + (d*x)/2)^6*(4*a - 2*b) + 6*a*t an(c/2 + (d*x)/2)^4)) - (atan(((((((8*(2*B*b^18 + 12*A*a^2*b^16 + 12*A*a^3 *b^15 - 20*A*a^4*b^14 - 4*A*a^5*b^13 + 8*A*a^6*b^12 + 6*B*a^2*b^16 - 16*B* a^3*b^15 - 14*B*a^4*b^14 + 28*B*a^5*b^13 + 6*B*a^6*b^12 - 12*B*a^7*b^11 - 4*C*a^3*b^15 + 20*C*a^4*b^14 + 16*C*a^5*b^13 - 36*C*a^6*b^12 - 8*C*a^7*b^1 1 + 16*C*a^8*b^10 - 8*A*a*b^17 - 4*C*a*b^17))/(a*b^14 + b^15 - a^2*b^13 - a^3*b^12) - (8*tan(c/2 + (d*x)/2)*((B*b^3*1i)/2 - C*a^3*4i - b^2*(A*a*2...