Integrand size = 33, antiderivative size = 335 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 b^2 \left (5 a^2 A b^2-4 A b^4+3 a^4 C-2 a^2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {b \left (4 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{a^5 d}-\frac {\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac {b \left (2 A b^2-a^2 (A-C)\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
2*b^2*(5*A*a^2*b^2-4*A*b^4+3*C*a^4-2*C*a^2*b^2)*arctan((a-b)^(1/2)*tan(1/2 *d*x+1/2*c)/(a+b)^(1/2))/a^5/(a-b)^(3/2)/(a+b)^(3/2)/d-b*(4*A*b^2+a^2*(A+2 *C))*arctanh(sin(d*x+c))/a^5/d-1/3*(12*A*b^4-a^2*b^2*(7*A-6*C)-a^4*(2*A+3* C))*tan(d*x+c)/a^4/(a^2-b^2)/d+b*(2*A*b^2-a^2*(A-C))*sec(d*x+c)*tan(d*x+c) /a^3/(a^2-b^2)/d-1/3*(4*A*b^2-a^2*(A-3*C))*sec(d*x+c)^2*tan(d*x+c)/a^2/(a^ 2-b^2)/d+(A*b^2+C*a^2)*sec(d*x+c)^2*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+ c))
Time = 7.51 (sec) , antiderivative size = 593, normalized size of antiderivative = 1.77 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b^2 \left (5 a^2 A b^2-4 A b^4+3 a^4 C-2 a^2 b^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{a^5 \left (a^2-b^2\right ) \sqrt {-a^2+b^2} d}+\frac {\left (a^2 A b+4 A b^3+2 a^2 b C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {\left (-a^2 A b-4 A b^3-2 a^2 b C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {A (a-6 b)}{12 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {A (a-6 b)}{12 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 a^2 A \sin \left (\frac {1}{2} (c+d x)\right )+9 A b^2 \sin \left (\frac {1}{2} (c+d x)\right )+3 a^2 C \sin \left (\frac {1}{2} (c+d x)\right )}{3 a^4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 a^2 A \sin \left (\frac {1}{2} (c+d x)\right )+9 A b^2 \sin \left (\frac {1}{2} (c+d x)\right )+3 a^2 C \sin \left (\frac {1}{2} (c+d x)\right )}{3 a^4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {-A b^5 \sin (c+d x)-a^2 b^3 C \sin (c+d x)}{a^4 (a-b) (a+b) d (a+b \cos (c+d x))} \]
(-2*b^2*(5*a^2*A*b^2 - 4*A*b^4 + 3*a^4*C - 2*a^2*b^2*C)*ArcTanh[((a - b)*T an[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(a^5*(a^2 - b^2)*Sqrt[-a^2 + b^2]*d) + ((a^2*A*b + 4*A*b^3 + 2*a^2*b*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] )/(a^5*d) + ((-(a^2*A*b) - 4*A*b^3 - 2*a^2*b*C)*Log[Cos[(c + d*x)/2] + Sin [(c + d*x)/2]])/(a^5*d) + (A*(a - 6*b))/(12*a^3*d*(Cos[(c + d*x)/2] - Sin[ (c + d*x)/2])^2) + (A*Sin[(c + d*x)/2])/(6*a^2*d*(Cos[(c + d*x)/2] - Sin[( c + d*x)/2])^3) + (A*Sin[(c + d*x)/2])/(6*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - (A*(a - 6*b))/(12*a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x) /2])^2) + (2*a^2*A*Sin[(c + d*x)/2] + 9*A*b^2*Sin[(c + d*x)/2] + 3*a^2*C*S in[(c + d*x)/2])/(3*a^4*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*a^2* A*Sin[(c + d*x)/2] + 9*A*b^2*Sin[(c + d*x)/2] + 3*a^2*C*Sin[(c + d*x)/2])/ (3*a^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (-(A*b^5*Sin[c + d*x]) - a^2*b^3*C*Sin[c + d*x])/(a^4*(a - b)*(a + b)*d*(a + b*Cos[c + d*x]))
Time = 2.43 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 3535, 25, 3042, 3534, 25, 3042, 3534, 27, 3042, 3534, 27, 3042, 3480, 3042, 3138, 218, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {\left (-\left ((A-3 C) a^2\right )+b (A+C) \cos (c+d x) a+4 A b^2-3 \left (C a^2+A b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {\left (-\left ((A-3 C) a^2\right )+b (A+C) \cos (c+d x) a+4 A b^2-3 \left (C a^2+A b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((A-3 C) a^2\right )+b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+4 A b^2-3 \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\int -\frac {\left (-2 b \left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x)+a \left ((2 A+3 C) a^2+A b^2\right ) \cos (c+d x)+6 b \left (2 A b^2-a^2 (A-C)\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}+\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {\left (-2 b \left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x)+a \left ((2 A+3 C) a^2+A b^2\right ) \cos (c+d x)+6 b \left (2 A b^2-a^2 (A-C)\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {-2 b \left (4 A b^2-a^2 (A-3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left ((2 A+3 C) a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+6 b \left (2 A b^2-a^2 (A-C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\int -\frac {2 \left (-\frac {1}{2} (4 A+6 C) a^4-b^2 (7 A-6 C) a^2+b \left ((A+3 C) a^2+2 A b^2\right ) \cos (c+d x) a+12 A b^4-3 b^2 \left (2 A b^2-a^2 (A-C)\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\int \frac {\left (-\left ((2 A+3 C) a^4\right )-b^2 (7 A-6 C) a^2+b \left ((A+3 C) a^2+2 A b^2\right ) \cos (c+d x) a+12 A b^4-3 b^2 \left (2 A b^2-a^2 (A-C)\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\int \frac {-\left ((2 A+3 C) a^4\right )-b^2 (7 A-6 C) a^2+b \left ((A+3 C) a^2+2 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+12 A b^4-3 b^2 \left (2 A b^2-a^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {\int \frac {3 \left (b \left (a^2-b^2\right ) \left ((A+2 C) a^2+4 A b^2\right )-a b^2 \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\left (-\left (a^4 (2 A+3 C)\right )-a^2 b^2 (7 A-6 C)+12 A b^4\right ) \tan (c+d x)}{a d}}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 \int \frac {\left (b \left (a^2-b^2\right ) \left ((A+2 C) a^2+4 A b^2\right )-a b^2 \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\left (-\left (a^4 (2 A+3 C)\right )-a^2 b^2 (7 A-6 C)+12 A b^4\right ) \tan (c+d x)}{a d}}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 \int \frac {b \left (a^2-b^2\right ) \left ((A+2 C) a^2+4 A b^2\right )-a b^2 \left (2 A b^2-a^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {\left (-\left (a^4 (2 A+3 C)\right )-a^2 b^2 (7 A-6 C)+12 A b^4\right ) \tan (c+d x)}{a d}}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 \left (\frac {b \left (a^2-b^2\right ) \left (a^2 (A+2 C)+4 A b^2\right ) \int \sec (c+d x)dx}{a}+\frac {b^2 \left (-3 a^4 C-a^2 b^2 (5 A-2 C)+4 A b^4\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}+\frac {\left (-\left (a^4 (2 A+3 C)\right )-a^2 b^2 (7 A-6 C)+12 A b^4\right ) \tan (c+d x)}{a d}}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 \left (\frac {b \left (a^2-b^2\right ) \left (a^2 (A+2 C)+4 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {b^2 \left (-3 a^4 C-a^2 b^2 (5 A-2 C)+4 A b^4\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}+\frac {\left (-\left (a^4 (2 A+3 C)\right )-a^2 b^2 (7 A-6 C)+12 A b^4\right ) \tan (c+d x)}{a d}}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 \left (\frac {b \left (a^2-b^2\right ) \left (a^2 (A+2 C)+4 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 b^2 \left (-3 a^4 C-a^2 b^2 (5 A-2 C)+4 A b^4\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}+\frac {\left (-\left (a^4 (2 A+3 C)\right )-a^2 b^2 (7 A-6 C)+12 A b^4\right ) \tan (c+d x)}{a d}}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 \left (\frac {b \left (a^2-b^2\right ) \left (a^2 (A+2 C)+4 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 b^2 \left (-3 a^4 C-a^2 b^2 (5 A-2 C)+4 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}+\frac {\left (-\left (a^4 (2 A+3 C)\right )-a^2 b^2 (7 A-6 C)+12 A b^4\right ) \tan (c+d x)}{a d}}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a d}-\frac {\frac {3 \left (\frac {b \left (a^2-b^2\right ) \left (a^2 (A+2 C)+4 A b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}+\frac {2 b^2 \left (-3 a^4 C-a^2 b^2 (5 A-2 C)+4 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}+\frac {\left (-\left (a^4 (2 A+3 C)\right )-a^2 b^2 (7 A-6 C)+12 A b^4\right ) \tan (c+d x)}{a d}}{a}}{3 a}}{a \left (a^2-b^2\right )}\) |
((A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[ c + d*x])) - (((4*A*b^2 - a^2*(A - 3*C))*Sec[c + d*x]^2*Tan[c + d*x])/(3*a *d) - ((3*b*(2*A*b^2 - a^2*(A - C))*Sec[c + d*x]*Tan[c + d*x])/(a*d) - ((3 *((2*b^2*(4*A*b^4 - a^2*b^2*(5*A - 2*C) - 3*a^4*C)*ArcTan[(Sqrt[a - b]*Tan [(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + (b*(a^2 - b^2 )*(4*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(a*d)))/a + ((12*A*b^4 - a^2*b^2*(7*A - 6*C) - a^4*(2*A + 3*C))*Tan[c + d*x])/(a*d))/a)/(3*a))/(a *(a^2 - b^2))
3.6.77.3.1 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_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
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[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin [e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 3.04 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.19
| method | result | size |
| derivativedivides | \(\frac {-\frac {A}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {A \,a^{2}+A a b +3 A \,b^{2}+a^{2} C}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A \left (a +2 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {b \left (A \,a^{2}+4 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{5}}-\frac {A}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {A \,a^{2}+A a b +3 A \,b^{2}+a^{2} C}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A \left (a +2 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {b \left (A \,a^{2}+4 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{5}}+\frac {2 b^{2} \left (-\frac {a \left (A \,b^{2}+a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (5 A \,a^{2} b^{2}-4 A \,b^{4}+3 C \,a^{4}-2 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5}}}{d}\) | \(399\) |
| default | \(\frac {-\frac {A}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {A \,a^{2}+A a b +3 A \,b^{2}+a^{2} C}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A \left (a +2 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {b \left (A \,a^{2}+4 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{5}}-\frac {A}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {A \,a^{2}+A a b +3 A \,b^{2}+a^{2} C}{a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A \left (a +2 b \right )}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {b \left (A \,a^{2}+4 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{5}}+\frac {2 b^{2} \left (-\frac {a \left (A \,b^{2}+a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (5 A \,a^{2} b^{2}-4 A \,b^{4}+3 C \,a^{4}-2 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5}}}{d}\) | \(399\) |
| risch | \(\text {Expression too large to display}\) | \(1478\) |
1/d*(-1/3*A/a^2/(tan(1/2*d*x+1/2*c)-1)^3-(A*a^2+A*a*b+3*A*b^2+C*a^2)/a^4/( tan(1/2*d*x+1/2*c)-1)-1/2*A*(a+2*b)/a^3/(tan(1/2*d*x+1/2*c)-1)^2+b*(A*a^2+ 4*A*b^2+2*C*a^2)/a^5*ln(tan(1/2*d*x+1/2*c)-1)-1/3*A/a^2/(tan(1/2*d*x+1/2*c )+1)^3-(A*a^2+A*a*b+3*A*b^2+C*a^2)/a^4/(tan(1/2*d*x+1/2*c)+1)+1/2*A*(a+2*b )/a^3/(tan(1/2*d*x+1/2*c)+1)^2-b*(A*a^2+4*A*b^2+2*C*a^2)/a^5*ln(tan(1/2*d* x+1/2*c)+1)+2*b^2/a^5*(-a*(A*b^2+C*a^2)*b/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(ta n(1/2*d*x+1/2*c)^2*a-b*tan(1/2*d*x+1/2*c)^2+a+b)+(5*A*a^2*b^2-4*A*b^4+3*C* a^4-2*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))))
Time = 6.14 (sec) , antiderivative size = 1305, normalized size of antiderivative = 3.90 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
[-1/6*(3*((3*C*a^4*b^3 + (5*A - 2*C)*a^2*b^5 - 4*A*b^7)*cos(d*x + c)^4 + ( 3*C*a^5*b^2 + (5*A - 2*C)*a^3*b^4 - 4*A*a*b^6)*cos(d*x + c)^3)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(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)) + 3*(((A + 2*C)*a^6*b^2 + 2*(A - 2*C)*a^4* b^4 - (7*A - 2*C)*a^2*b^6 + 4*A*b^8)*cos(d*x + c)^4 + ((A + 2*C)*a^7*b + 2 *(A - 2*C)*a^5*b^3 - (7*A - 2*C)*a^3*b^5 + 4*A*a*b^7)*cos(d*x + c)^3)*log( sin(d*x + c) + 1) - 3*(((A + 2*C)*a^6*b^2 + 2*(A - 2*C)*a^4*b^4 - (7*A - 2 *C)*a^2*b^6 + 4*A*b^8)*cos(d*x + c)^4 + ((A + 2*C)*a^7*b + 2*(A - 2*C)*a^5 *b^3 - (7*A - 2*C)*a^3*b^5 + 4*A*a*b^7)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(A*a^8 - 2*A*a^6*b^2 + A*a^4*b^4 + ((2*A + 3*C)*a^7*b + (5*A - 9* C)*a^5*b^3 - (19*A - 6*C)*a^3*b^5 + 12*A*a*b^7)*cos(d*x + c)^3 + ((2*A + 3 *C)*a^8 + 2*(A - 3*C)*a^6*b^2 - (10*A - 3*C)*a^4*b^4 + 6*A*a^2*b^6)*cos(d* x + c)^2 - 2*(A*a^7*b - 2*A*a^5*b^3 + A*a^3*b^5)*cos(d*x + c))*sin(d*x + c ))/((a^9*b - 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^4 + (a^10 - 2*a^8*b^2 + a ^6*b^4)*d*cos(d*x + c)^3), 1/6*(6*((3*C*a^4*b^3 + (5*A - 2*C)*a^2*b^5 - 4* A*b^7)*cos(d*x + c)^4 + (3*C*a^5*b^2 + (5*A - 2*C)*a^3*b^4 - 4*A*a*b^6)*co s(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2 )*sin(d*x + c))) - 3*(((A + 2*C)*a^6*b^2 + 2*(A - 2*C)*a^4*b^4 - (7*A - 2* C)*a^2*b^6 + 4*A*b^8)*cos(d*x + c)^4 + ((A + 2*C)*a^7*b + 2*(A - 2*C)*a...
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.38 (sec) , antiderivative size = 483, normalized size of antiderivative = 1.44 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {6 \, {\left (3 \, C a^{4} b^{2} + 5 \, A a^{2} b^{4} - 2 \, C a^{2} b^{4} - 4 \, A b^{6}\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^{7} - a^{5} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {6 \, {\left (C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\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 (A a^{2} b + 2 \, C a^{2} b + 4 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {3 \, {\left (A a^{2} b + 2 \, C a^{2} b + 4 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} + \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, A 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} a^{4}}}{3 \, d} \]
-1/3*(6*(3*C*a^4*b^2 + 5*A*a^2*b^4 - 2*C*a^2*b^4 - 4*A*b^6)*(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^7 - a^5*b^2)*sqrt(a^2 - b^2)) + 6*(C*a^2*b^3*tan(1/2*d*x + 1/2*c) + A*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - a^4*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) + 3*(A*a^2*b + 2*C*a^2*b + 4*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - 3*(A*a^2*b + 2*C*a^2*b + 4*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^5 + 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*A *a*b*tan(1/2*d*x + 1/2*c)^5 + 9*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 2*A*a^2*tan (1/2*d*x + 1/2*c)^3 - 6*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 18*A*b^2*tan(1/2*d* x + 1/2*c)^3 + 3*A*a^2*tan(1/2*d*x + 1/2*c) + 3*C*a^2*tan(1/2*d*x + 1/2*c) - 3*A*a*b*tan(1/2*d*x + 1/2*c) + 9*A*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2* d*x + 1/2*c)^2 - 1)^3*a^4))/d
Time = 11.50 (sec) , antiderivative size = 6976, normalized size of antiderivative = 20.82 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
((2*tan(c/2 + (d*x)/2)^3*(A*a^5 + 36*A*b^5 - 3*C*a^5 - 19*A*a^2*b^3 - 7*A* a^3*b^2 + 18*C*a^2*b^3 + 3*C*a^3*b^2 + 6*A*a*b^4 - 8*A*a^4*b - 9*C*a^4*b)) /(3*a^4*(a + b)*(a - b)) + (2*tan(c/2 + (d*x)/2)^5*(A*a^5 - 36*A*b^5 - 3*C *a^5 + 19*A*a^2*b^3 - 7*A*a^3*b^2 - 18*C*a^2*b^3 + 3*C*a^3*b^2 + 6*A*a*b^4 + 8*A*a^4*b + 9*C*a^4*b))/(3*a^4*(a + b)*(a - b)) + (2*tan(c/2 + (d*x)/2) ^7*(A*a^5 + 4*A*b^5 + C*a^5 - 3*A*a^2*b^3 + A*a^3*b^2 + 2*C*a^2*b^3 - C*a^ 3*b^2 - 2*A*a*b^4 - C*a^4*b))/(a^4*(a + b)*(a - b)) + (2*tan(c/2 + (d*x)/2 )*(A*a^5 - 4*A*b^5 + C*a^5 + 3*A*a^2*b^3 + A*a^3*b^2 - 2*C*a^2*b^3 - C*a^3 *b^2 - 2*A*a*b^4 + C*a^4*b))/(a^4*(a + b)*(a - b)))/(d*(a + b - tan(c/2 + (d*x)/2)^8*(a - b) - tan(c/2 + (d*x)/2)^2*(2*a + 4*b) + tan(c/2 + (d*x)/2) ^6*(2*a - 4*b) + 6*b*tan(c/2 + (d*x)/2)^4)) - (atan((((4*A*b^3 + a^2*(A*b + 2*C*b))*((((32*(2*A*a^11*b^7 - 4*A*a^10*b^8 + 9*A*a^12*b^6 - 4*A*a^13*b^ 5 - 5*A*a^14*b^4 + A*a^15*b^3 - 2*C*a^12*b^6 + C*a^13*b^5 + 5*C*a^14*b^4 - 3*C*a^15*b^3 - 3*C*a^16*b^2 + A*a^17*b + 2*C*a^17*b))/(a^14*b + a^15 - a^ 12*b^3 - a^13*b^2) - (32*tan(c/2 + (d*x)/2)*(4*A*b^3 + a^2*(A*b + 2*C*b))* (2*a^15*b - 2*a^10*b^6 + 2*a^11*b^5 + 4*a^12*b^4 - 4*a^13*b^3 - 2*a^14*b^2 ))/(a^5*(a^10*b + a^11 - a^8*b^3 - a^9*b^2)))*(4*A*b^3 + a^2*(A*b + 2*C*b) ))/a^5 - (32*tan(c/2 + (d*x)/2)*(32*A^2*b^12 - 32*A^2*a*b^11 - 48*A^2*a^2* b^10 + 48*A^2*a^3*b^9 + 2*A^2*a^4*b^8 - 2*A^2*a^5*b^7 + 7*A^2*a^6*b^6 - 12 *A^2*a^7*b^5 + 7*A^2*a^8*b^4 - 2*A^2*a^9*b^3 + A^2*a^10*b^2 + 8*C^2*a^4...