Integrand size = 33, antiderivative size = 313 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{b^4 d}+\frac {a \left (a^2 b^4 (A-8 C)-2 a^6 C+7 a^4 b^2 C+4 b^6 (A+2 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} b^4 (a+b)^{7/2} d}-\frac {\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a \left (2 A b^4-3 a^4 C+a^2 b^2 (3 A+8 C)\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac {\left (4 A b^6+9 a^6 C+2 a^2 b^4 (7 A+17 C)-a^4 b^2 (3 A+28 C)\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \]
C*arctanh(sin(d*x+c))/b^4/d+a*(a^2*b^4*(A-8*C)-2*a^6*C+7*a^4*b^2*C+4*b^6*( A+2*C))*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/b^ 4/(a+b)^(7/2)/d-1/3*(A*b^2+C*a^2)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a +b*sec(d*x+c))^3-1/6*a*(2*A*b^4-3*a^4*C+a^2*b^2*(3*A+8*C))*tan(d*x+c)/b^3/ (a^2-b^2)^2/d/(a+b*sec(d*x+c))^2-1/6*(4*A*b^6+9*a^6*C+2*a^2*b^4*(7*A+17*C) -a^4*b^2*(3*A+28*C))*tan(d*x+c)/b^3/(a^2-b^2)^3/d/(a+b*sec(d*x+c))
Result contains complex when optimal does not.
Time = 7.63 (sec) , antiderivative size = 531, normalized size of antiderivative = 1.70 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {(b+a \cos (c+d x)) \sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {6 C (b+a \cos (c+d x))^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^4}+\frac {6 C (b+a \cos (c+d x))^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^4}-\frac {6 a \left (a^2 b^4 (A-8 C)-2 a^6 C+7 a^4 b^2 C+4 b^6 (A+2 C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x))^3 (i \cos (c)+\sin (c))}{b^4 \left (a^2-b^2\right )^{7/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {2 \left (A b^2+a^2 C\right ) \sec (c) (b \sin (c)-a \sin (d x))}{a b \left (a^2-b^2\right )}+\frac {(b+a \cos (c+d x))^2 \sec (c) \left (-3 a b \left (a^2 b^2 (A-2 C)+a^4 C+2 b^4 (2 A+3 C)\right ) \sin (c)+\left (2 A b^6+6 a^6 C-17 a^4 b^2 C+13 a^2 b^4 (A+2 C)\right ) \sin (d x)\right )}{\left (-a^2 b+b^3\right )^3}+\frac {(b+a \cos (c+d x)) \sec (c) \left (a b \left (-5 A b^2+\left (a^2-6 b^2\right ) C\right ) \sin (c)+\left (2 A b^4-3 a^4 C+a^2 b^2 (3 A+8 C)\right ) \sin (d x)\right )}{\left (-a^2 b+b^3\right )^2}\right )}{3 d (A+2 C+A \cos (2 (c+d x))) (a+b \sec (c+d x))^4} \]
((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2)*((-6*C*(b + a* Cos[c + d*x])^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/b^4 + (6*C*(b + a*Cos[c + d*x])^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/b^4 - (6*a*(a^ 2*b^4*(A - 8*C) - 2*a^6*C + 7*a^4*b^2*C + 4*b^6*(A + 2*C))*ArcTan[((I*Cos[ c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*S qrt[(Cos[c] - I*Sin[c])^2])]*(b + a*Cos[c + d*x])^3*(I*Cos[c] + Sin[c]))/( b^4*(a^2 - b^2)^(7/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (2*(A*b^2 + a^2*C)*Se c[c]*(b*Sin[c] - a*Sin[d*x]))/(a*b*(a^2 - b^2)) + ((b + a*Cos[c + d*x])^2* Sec[c]*(-3*a*b*(a^2*b^2*(A - 2*C) + a^4*C + 2*b^4*(2*A + 3*C))*Sin[c] + (2 *A*b^6 + 6*a^6*C - 17*a^4*b^2*C + 13*a^2*b^4*(A + 2*C))*Sin[d*x]))/(-(a^2* b) + b^3)^3 + ((b + a*Cos[c + d*x])*Sec[c]*(a*b*(-5*A*b^2 + (a^2 - 6*b^2)* C)*Sin[c] + (2*A*b^4 - 3*a^4*C + a^2*b^2*(3*A + 8*C))*Sin[d*x]))/(-(a^2*b) + b^3)^2))/(3*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^4)
Time = 2.19 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.18, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4587, 3042, 4578, 25, 3042, 4568, 27, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^4}dx\) |
| \(\Big \downarrow \) 4587 |
| \(\displaystyle -\frac {\int \frac {\sec ^2(c+d x) \left (-3 \left (a^2-b^2\right ) C \sec ^2(c+d x)-3 a b (A+C) \sec (c+d x)+2 \left (C a^2+A b^2\right )\right )}{(a+b \sec (c+d x))^3}dx}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (-3 \left (a^2-b^2\right ) C \csc \left (c+d x+\frac {\pi }{2}\right )^2-3 a b (A+C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 \left (C a^2+A b^2\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 4578 |
| \(\displaystyle -\frac {\frac {\int -\frac {\sec (c+d x) \left (6 b \left (a^2-b^2\right )^2 C \sec ^2(c+d x)-a \left (3 C a^4-b^2 (3 A+10 C) a^2+4 b^4 (2 A+3 C)\right ) \sec (c+d x)+2 b \left (-3 C a^4+b^2 (3 A+8 C) a^2+2 A b^4\right )\right )}{(a+b \sec (c+d x))^2}dx}{2 b^2 \left (a^2-b^2\right )}+\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\int \frac {\sec (c+d x) \left (6 b \left (a^2-b^2\right )^2 C \sec ^2(c+d x)-a \left (3 C a^4-b^2 (3 A+10 C) a^2+4 b^4 (2 A+3 C)\right ) \sec (c+d x)+2 b \left (-3 C a^4+b^2 (3 A+8 C) a^2+2 A b^4\right )\right )}{(a+b \sec (c+d x))^2}dx}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (6 b \left (a^2-b^2\right )^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2-a \left (3 C a^4-b^2 (3 A+10 C) a^2+4 b^4 (2 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 b \left (-3 C a^4+b^2 (3 A+8 C) a^2+2 A b^4\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 4568 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {-\frac {\int -\frac {3 \sec (c+d x) \left (2 b C \sec (c+d x) \left (a^2-b^2\right )^3+a b^2 \left (C a^4+b^2 (A-2 C) a^2+2 b^4 (2 A+3 C)\right )\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {3 \int \frac {\sec (c+d x) \left (2 b C \sec (c+d x) \left (a^2-b^2\right )^3+a b^2 \left (C a^4+b^2 (A-2 C) a^2+2 b^4 (2 A+3 C)\right )\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 b C \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-b^2\right )^3+a b^2 \left (C a^4+b^2 (A-2 C) a^2+2 b^4 (2 A+3 C)\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {3 \left (2 C \left (a^2-b^2\right )^3 \int \sec (c+d x)dx+a \left (-2 a^6 C+7 a^4 b^2 C+a^2 b^4 (A-8 C)+4 b^6 (A+2 C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx\right )}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {3 \left (2 C \left (a^2-b^2\right )^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a \left (-2 a^6 C+7 a^4 b^2 C+a^2 b^4 (A-8 C)+4 b^6 (A+2 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {3 \left (a \left (-2 a^6 C+7 a^4 b^2 C+a^2 b^4 (A-8 C)+4 b^6 (A+2 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 C \left (a^2-b^2\right )^3 \text {arctanh}(\sin (c+d x))}{d}\right )}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {3 \left (\frac {a \left (-2 a^6 C+7 a^4 b^2 C+a^2 b^4 (A-8 C)+4 b^6 (A+2 C)\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b}+\frac {2 C \left (a^2-b^2\right )^3 \text {arctanh}(\sin (c+d x))}{d}\right )}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {3 \left (\frac {a \left (-2 a^6 C+7 a^4 b^2 C+a^2 b^4 (A-8 C)+4 b^6 (A+2 C)\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b}+\frac {2 C \left (a^2-b^2\right )^3 \text {arctanh}(\sin (c+d x))}{d}\right )}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle -\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {3 \left (\frac {2 a \left (-2 a^6 C+7 a^4 b^2 C+a^2 b^4 (A-8 C)+4 b^6 (A+2 C)\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}+\frac {2 C \left (a^2-b^2\right )^3 \text {arctanh}(\sin (c+d x))}{d}\right )}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}-\frac {\frac {a \left (-3 a^4 C+a^2 b^2 (3 A+8 C)+2 A b^4\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {3 \left (\frac {2 C \left (a^2-b^2\right )^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a \left (-2 a^6 C+7 a^4 b^2 C+a^2 b^4 (A-8 C)+4 b^6 (A+2 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}\right )}{b \left (a^2-b^2\right )}-\frac {\left (9 a^6 C-a^4 b^2 (3 A+28 C)+2 a^2 b^4 (7 A+17 C)+4 A b^6\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}\) |
-1/3*((A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b *Sec[c + d*x])^3) - ((a*(2*A*b^4 - 3*a^4*C + a^2*b^2*(3*A + 8*C))*Tan[c + d*x])/(2*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - ((3*((2*(a^2 - b^2)^3 *C*ArcTanh[Sin[c + d*x]])/d + (2*a*(a^2*b^4*(A - 8*C) - 2*a^6*C + 7*a^4*b^ 2*C + 4*b^6*(A + 2*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]] )/(Sqrt[a - b]*Sqrt[a + b]*d)))/(b*(a^2 - b^2)) - ((4*A*b^6 + 9*a^6*C + 2* a^2*b^4*(7*A + 17*C) - a^4*b^2*(3*A + 28*C))*Tan[c + d*x])/((a^2 - b^2)*d* (a + b*Sec[c + d*x])))/(2*b^2*(a^2 - b^2)))/(3*b*(a^2 - b^2))
3.7.99.3.1 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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e + f*x] )^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[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))*Csc[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[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x ])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Simp[1/(b^2*(m + 1)*(a^2 - b^ 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b *B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1 ]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-d) *(A*b^2 + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x] )^(n - 1)/(b*f*(a^2 - b^2)*(m + 1))), x] + Simp[d/(b*(a^2 - b^2)*(m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) + a^2*C*(n - 1) + a*b*(A + C)*(m + 1)*Csc[e + f*x] - (A*b^2*(m + n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]
Time = 1.12 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.60
| method | result | size |
| derivativedivides | \(\frac {\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}-\frac {2 \left (\frac {-\frac {\left (A \,a^{3} b^{3}+6 a^{2} A \,b^{4}+2 A a \,b^{5}+2 A \,b^{6}+2 a^{6} C -a^{5} C b -6 a^{4} b^{2} C +4 C \,a^{3} b^{3}+12 C \,a^{2} b^{4}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (7 a^{2} A \,b^{4}+3 A \,b^{6}+3 a^{6} C -11 a^{4} b^{2} C +18 C \,a^{2} b^{4}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (A \,a^{3} b^{3}-6 a^{2} A \,b^{4}+2 A a \,b^{5}-2 A \,b^{6}-2 a^{6} C -a^{5} C b +6 a^{4} b^{2} C +4 C \,a^{3} b^{3}-12 C \,a^{2} b^{4}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}-\frac {a \left (a^{2} A \,b^{4}+4 A \,b^{6}-2 a^{6} C +7 a^{4} b^{2} C -8 C \,a^{2} b^{4}+8 C \,b^{6}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}}{d}\) | \(502\) |
| default | \(\frac {\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}-\frac {2 \left (\frac {-\frac {\left (A \,a^{3} b^{3}+6 a^{2} A \,b^{4}+2 A a \,b^{5}+2 A \,b^{6}+2 a^{6} C -a^{5} C b -6 a^{4} b^{2} C +4 C \,a^{3} b^{3}+12 C \,a^{2} b^{4}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (7 a^{2} A \,b^{4}+3 A \,b^{6}+3 a^{6} C -11 a^{4} b^{2} C +18 C \,a^{2} b^{4}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (A \,a^{3} b^{3}-6 a^{2} A \,b^{4}+2 A a \,b^{5}-2 A \,b^{6}-2 a^{6} C -a^{5} C b +6 a^{4} b^{2} C +4 C \,a^{3} b^{3}-12 C \,a^{2} b^{4}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}-\frac {a \left (a^{2} A \,b^{4}+4 A \,b^{6}-2 a^{6} C +7 a^{4} b^{2} C -8 C \,a^{2} b^{4}+8 C \,b^{6}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}}{d}\) | \(502\) |
| risch | \(\text {Expression too large to display}\) | \(1714\) |
1/d*(C/b^4*ln(tan(1/2*d*x+1/2*c)+1)-2/b^4*((-1/2*(A*a^3*b^3+6*A*a^2*b^4+2* A*a*b^5+2*A*b^6+2*C*a^6-C*a^5*b-6*C*a^4*b^2+4*C*a^3*b^3+12*C*a^2*b^4)*b/(a -b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5+2/3*(7*A*a^2*b^4+3*A*b^ 6+3*C*a^6-11*C*a^4*b^2+18*C*a^2*b^4)*b/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan (1/2*d*x+1/2*c)^3+1/2*(A*a^3*b^3-6*A*a^2*b^4+2*A*a*b^5-2*A*b^6-2*C*a^6-C*a ^5*b+6*C*a^4*b^2+4*C*a^3*b^3-12*C*a^2*b^4)*b/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^ 3)*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) ^3-1/2*a*(A*a^2*b^4+4*A*b^6-2*C*a^6+7*C*a^4*b^2-8*C*a^2*b^4+8*C*b^6)/(a^6- 3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2 *c)/((a+b)*(a-b))^(1/2)))-C/b^4*ln(tan(1/2*d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 1049 vs. \(2 (301) = 602\).
Time = 12.96 (sec) , antiderivative size = 2156, normalized size of antiderivative = 6.89 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]
[1/12*(3*(2*C*a^7*b^3 - 7*C*a^5*b^5 - (A - 8*C)*a^3*b^7 - 4*(A + 2*C)*a*b^ 9 + (2*C*a^10 - 7*C*a^8*b^2 - (A - 8*C)*a^6*b^4 - 4*(A + 2*C)*a^4*b^6)*cos (d*x + c)^3 + 3*(2*C*a^9*b - 7*C*a^7*b^3 - (A - 8*C)*a^5*b^5 - 4*(A + 2*C) *a^3*b^7)*cos(d*x + c)^2 + 3*(2*C*a^8*b^2 - 7*C*a^6*b^4 - (A - 8*C)*a^4*b^ 6 - 4*(A + 2*C)*a^2*b^8)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + 6*(C*a^8*b^3 - 4*C*a^6*b^5 + 6*C*a^4*b^7 - 4*C*a^2*b^9 + C*b^11 + (C*a^11 - 4*C*a^9*b^2 + 6*C*a^7*b^4 - 4*C*a^5*b^6 + C*a^3*b^8)*cos(d*x + c )^3 + 3*(C*a^10*b - 4*C*a^8*b^3 + 6*C*a^6*b^5 - 4*C*a^4*b^7 + C*a^2*b^9)*c os(d*x + c)^2 + 3*(C*a^9*b^2 - 4*C*a^7*b^4 + 6*C*a^5*b^6 - 4*C*a^3*b^8 + C *a*b^10)*cos(d*x + c))*log(sin(d*x + c) + 1) - 6*(C*a^8*b^3 - 4*C*a^6*b^5 + 6*C*a^4*b^7 - 4*C*a^2*b^9 + C*b^11 + (C*a^11 - 4*C*a^9*b^2 + 6*C*a^7*b^4 - 4*C*a^5*b^6 + C*a^3*b^8)*cos(d*x + c)^3 + 3*(C*a^10*b - 4*C*a^8*b^3 + 6 *C*a^6*b^5 - 4*C*a^4*b^7 + C*a^2*b^9)*cos(d*x + c)^2 + 3*(C*a^9*b^2 - 4*C* a^7*b^4 + 6*C*a^5*b^6 - 4*C*a^3*b^8 + C*a*b^10)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(11*C*a^8*b^3 - (A + 43*C)*a^6*b^5 + (11*A + 68*C)*a^4*b^7 - 4*(A + 9*C)*a^2*b^9 - 6*A*b^11 + (6*C*a^10*b - 23*C*a^8*b^3 + (13*A + 43 *C)*a^6*b^5 - (11*A + 26*C)*a^4*b^7 - 2*A*a^2*b^9)*cos(d*x + c)^2 + 3*(5*C *a^9*b^2 - (A + 20*C)*a^7*b^4 + 5*(2*A + 7*C)*a^5*b^6 - (7*A + 20*C)*a^...
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]
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*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 876 vs. \(2 (301) = 602\).
Time = 0.42 (sec) , antiderivative size = 876, normalized size of antiderivative = 2.80 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]
-1/3*(3*(2*C*a^7 - 7*C*a^5*b^2 - A*a^3*b^4 + 8*C*a^3*b^4 - 4*A*a*b^6 - 8*C *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^6*b^4 - 3*a^4*b^6 + 3*a^2*b^8 - b^10)*sqrt(-a^2 + b^2)) - 3*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + 3*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - (6*C*a^ 8*tan(1/2*d*x + 1/2*c)^5 - 15*C*a^7*b*tan(1/2*d*x + 1/2*c)^5 - 6*C*a^6*b^2 *tan(1/2*d*x + 1/2*c)^5 + 3*A*a^5*b^3*tan(1/2*d*x + 1/2*c)^5 + 45*C*a^5*b^ 3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^4*b^4*tan(1/2*d*x + 1/2*c)^5 - 6*C*a^4*b ^4*tan(1/2*d*x + 1/2*c)^5 - 27*A*a^3*b^5*tan(1/2*d*x + 1/2*c)^5 - 60*C*a^3 *b^5*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 + 36*C*a ^2*b^6*tan(1/2*d*x + 1/2*c)^5 - 6*A*a*b^7*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^8 *tan(1/2*d*x + 1/2*c)^5 - 12*C*a^8*tan(1/2*d*x + 1/2*c)^3 + 56*C*a^6*b^2*t an(1/2*d*x + 1/2*c)^3 - 28*A*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 - 116*C*a^4*b^ 4*tan(1/2*d*x + 1/2*c)^3 + 16*A*a^2*b^6*tan(1/2*d*x + 1/2*c)^3 + 72*C*a^2* b^6*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^8*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^8*tan (1/2*d*x + 1/2*c) + 15*C*a^7*b*tan(1/2*d*x + 1/2*c) - 6*C*a^6*b^2*tan(1/2* d*x + 1/2*c) - 3*A*a^5*b^3*tan(1/2*d*x + 1/2*c) - 45*C*a^5*b^3*tan(1/2*d*x + 1/2*c) + 12*A*a^4*b^4*tan(1/2*d*x + 1/2*c) - 6*C*a^4*b^4*tan(1/2*d*x + 1/2*c) + 27*A*a^3*b^5*tan(1/2*d*x + 1/2*c) + 60*C*a^3*b^5*tan(1/2*d*x + 1/ 2*c) + 12*A*a^2*b^6*tan(1/2*d*x + 1/2*c) + 36*C*a^2*b^6*tan(1/2*d*x + 1...
Time = 34.14 (sec) , antiderivative size = 9753, normalized size of antiderivative = 31.16 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]
- ((tan(c/2 + (d*x)/2)*(2*A*b^6 + 2*C*a^6 + 6*A*a^2*b^4 - A*a^3*b^3 + 12*C *a^2*b^4 - 4*C*a^3*b^3 - 6*C*a^4*b^2 - 2*A*a*b^5 + C*a^5*b))/((a + b)*(3*a *b^5 - b^6 - 3*a^2*b^4 + a^3*b^3)) - (4*tan(c/2 + (d*x)/2)^3*(3*A*b^6 + 3* C*a^6 + 7*A*a^2*b^4 + 18*C*a^2*b^4 - 11*C*a^4*b^2))/(3*(a + b)^2*(b^5 - 2* a*b^4 + a^2*b^3)) + (tan(c/2 + (d*x)/2)^5*(2*A*b^6 + 2*C*a^6 + 6*A*a^2*b^4 + A*a^3*b^3 + 12*C*a^2*b^4 + 4*C*a^3*b^3 - 6*C*a^4*b^2 + 2*A*a*b^5 - C*a^ 5*b))/((a*b^3 - b^4)*(a + b)^3))/(d*(tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2 *b - 3*a^3 + 3*b^3) - tan(c/2 + (d*x)/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 - 3* b^3) + 3*a*b^2 + 3*a^2*b + a^3 + b^3 - tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a ^2*b + a^3 - b^3))) - (C*atan(((C*((8*tan(c/2 + (d*x)/2)*(8*C^2*a^14 + 4*C ^2*b^14 - 8*C^2*a*b^13 - 8*C^2*a^13*b + 16*A^2*a^2*b^12 + 8*A^2*a^4*b^10 + A^2*a^6*b^8 + 44*C^2*a^2*b^12 + 48*C^2*a^3*b^11 - 92*C^2*a^4*b^10 - 120*C ^2*a^5*b^9 + 156*C^2*a^6*b^8 + 160*C^2*a^7*b^7 - 164*C^2*a^8*b^6 - 120*C^2 *a^9*b^5 + 117*C^2*a^10*b^4 + 48*C^2*a^11*b^3 - 48*C^2*a^12*b^2 + 64*A*C*a ^2*b^12 - 48*A*C*a^4*b^10 + 40*A*C*a^6*b^8 - 2*A*C*a^8*b^6 - 4*A*C*a^10*b^ 4))/(a*b^16 + b^17 - 5*a^2*b^15 - 5*a^3*b^14 + 10*a^4*b^13 + 10*a^5*b^12 - 10*a^6*b^11 - 10*a^7*b^10 + 5*a^8*b^9 + 5*a^9*b^8 - a^10*b^7 - a^11*b^6) + (C*((8*(4*C*b^21 + 8*A*a^2*b^19 + 22*A*a^3*b^18 - 22*A*a^4*b^17 - 18*A*a ^5*b^16 + 18*A*a^6*b^15 + 2*A*a^7*b^14 - 2*A*a^8*b^13 + 2*A*a^9*b^12 - 2*A *a^10*b^11 - 12*C*a^2*b^19 + 64*C*a^3*b^18 + 20*C*a^4*b^17 - 110*C*a^5*...