Integrand size = 41, antiderivative size = 405 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 b^2 \left (5 a^2 A b^2-4 A b^4-4 a^3 b B+3 a b^3 B+3 a^4 C-2 a^2 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {\left (8 A b^3-a^3 B-6 a b^2 B+2 a^2 b (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 a^5 d}-\frac {\left (12 A b^4+6 a^3 b B-9 a b^3 B-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac {\left (4 A b^3+a^3 B-3 a b^2 B-2 a^2 b (A-C)\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}-\frac {\left (4 A b^2-3 a b B-a^2 (A-3 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:
2*b^2*(5*A*a^2*b^2-4*A*b^4-4*B*a^3*b+3*B*a*b^3+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 -1/2*(8*A*b^3-B*a^3-6*B*a*b^2+2*a^2*b*(A+2*C))*arctanh(sin(d*x+c))/a^5/d-1 /3*(12*A*b^4+6*B*a^3*b-9*B*a*b^3-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+1/2*(4*A*b^3+B*a^3-3*B*a*b^2-2*a^2*b*(A-C))*sec(d*x+c)* tan(d*x+c)/a^3/(a^2-b^2)/d-1/3*(4*A*b^2-3*B*a*b-a^2*(A-3*C))*sec(d*x+c)^2* tan(d*x+c)/a^2/(a^2-b^2)/d+(A*b^2-a*(B*b-C*a))*sec(d*x+c)^2*tan(d*x+c)/a/( a^2-b^2)/d/(a+b*cos(d*x+c))
Time = 4.88 (sec) , antiderivative size = 519, normalized size of antiderivative = 1.28 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {-\frac {24 b^2 \left (4 A b^4+4 a^3 b B-3 a b^3 B-3 a^4 C+a^2 b^2 (-5 A+2 C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+6 \left (8 A b^3-a^3 B-6 a b^2 B+2 a^2 b (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \left (-8 A b^3+a^3 B+6 a b^2 B-2 a^2 b (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a \left (8 a^5 A+4 a^3 A b^2-12 a A b^4-9 a^4 b B+9 a^2 b^3 B+6 a^5 C-6 a^3 b^2 C+\left (-36 A b^5+6 a^5 B-24 a^3 b^2 B+27 a b^4 B+a^2 b^3 (29 A-18 C)+a^4 (-2 A b+9 b C)\right ) \cos (c+d x)+a \left (a^2-b^2\right ) \left (12 A b^2-9 a b B+a^2 (4 A+6 C)\right ) \cos (2 (c+d x))+2 a^4 A b \cos (3 (c+d x))+7 a^2 A b^3 \cos (3 (c+d x))-12 A b^5 \cos (3 (c+d x))-6 a^3 b^2 B \cos (3 (c+d x))+9 a b^4 B \cos (3 (c+d x))+3 a^4 b C \cos (3 (c+d x))-6 a^2 b^3 C \cos (3 (c+d x))\right ) \sec ^2(c+d x) \tan (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}}{12 a^5 d} \] Input:
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b* Cos[c + d*x])^2,x]
Output:
((-24*b^2*(4*A*b^4 + 4*a^3*b*B - 3*a*b^3*B - 3*a^4*C + a^2*b^2*(-5*A + 2*C ))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2 ) + 6*(8*A*b^3 - a^3*B - 6*a*b^2*B + 2*a^2*b*(A + 2*C))*Log[Cos[(c + d*x)/ 2] - Sin[(c + d*x)/2]] + 6*(-8*A*b^3 + a^3*B + 6*a*b^2*B - 2*a^2*b*(A + 2* C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a*(8*a^5*A + 4*a^3*A*b^2 - 12*a*A*b^4 - 9*a^4*b*B + 9*a^2*b^3*B + 6*a^5*C - 6*a^3*b^2*C + (-36*A*b^5 + 6*a^5*B - 24*a^3*b^2*B + 27*a*b^4*B + a^2*b^3*(29*A - 18*C) + a^4*(-2*A *b + 9*b*C))*Cos[c + d*x] + a*(a^2 - b^2)*(12*A*b^2 - 9*a*b*B + a^2*(4*A + 6*C))*Cos[2*(c + d*x)] + 2*a^4*A*b*Cos[3*(c + d*x)] + 7*a^2*A*b^3*Cos[3*( c + d*x)] - 12*A*b^5*Cos[3*(c + d*x)] - 6*a^3*b^2*B*Cos[3*(c + d*x)] + 9*a *b^4*B*Cos[3*(c + d*x)] + 3*a^4*b*C*Cos[3*(c + d*x)] - 6*a^2*b^3*C*Cos[3*( c + d*x)])*Sec[c + d*x]^2*Tan[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d*x])) )/(12*a^5*d)
Time = 2.79 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.439, Rules used = {3042, 3534, 25, 3042, 3534, 25, 3042, 3534, 25, 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.
\(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\int -\frac {\left (-\left ((A-3 C) a^2\right )-3 b B a+(A b+C b-a B) \cos (c+d x) a+4 A b^2-3 \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {\left (-\left ((A-3 C) a^2\right )-3 b B a+(A b+C b-a B) \cos (c+d x) a+4 A b^2-3 \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((A-3 C) a^2\right )-3 b B a+(A b+C b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a+4 A b^2-3 \left (A b^2-a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\int -\frac {\left (-2 b \left (-\left ((A-3 C) a^2\right )-3 b B a+4 A b^2\right ) \cos ^2(c+d x)+a \left ((2 A+3 C) a^2-3 b B a+A b^2\right ) \cos (c+d x)+3 \left (B a^3-2 b (A-C) a^2-3 b^2 B a+4 A b^3\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}+\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\int \frac {\left (-2 b \left (-\left ((A-3 C) a^2\right )-3 b B a+4 A b^2\right ) \cos ^2(c+d x)+a \left ((2 A+3 C) a^2-3 b B a+A b^2\right ) \cos (c+d x)+3 \left (B a^3-2 b (A-C) a^2-3 b^2 B a+4 A b^3\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\int \frac {-2 b \left (-\left ((A-3 C) a^2\right )-3 b B a+4 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left ((2 A+3 C) a^2-3 b B a+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (B a^3-2 b (A-C) a^2-3 b^2 B a+4 A b^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {\int -\frac {\left (-3 b \left (B a^3-2 b (A-C) a^2-3 b^2 B a+4 A b^3\right ) \cos ^2(c+d x)+a \left (-3 B a^3+2 b (A+3 C) a^2-3 b^2 B a+4 A b^3\right ) \cos (c+d x)+2 \left (-\left ((2 A+3 C) a^4\right )+6 b B a^3-b^2 (7 A-6 C) a^2-9 b^3 B a+12 A b^4\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\int \frac {\left (-3 b \left (B a^3-2 b (A-C) a^2-3 b^2 B a+4 A b^3\right ) \cos ^2(c+d x)+a \left (-3 B a^3+2 b (A+3 C) a^2-3 b^2 B a+4 A b^3\right ) \cos (c+d x)+2 \left (-\left ((2 A+3 C) a^4\right )+6 b B a^3-b^2 (7 A-6 C) a^2-9 b^3 B a+12 A b^4\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\int \frac {-3 b \left (B a^3-2 b (A-C) a^2-3 b^2 B a+4 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left (-3 B a^3+2 b (A+3 C) a^2-3 b^2 B a+4 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left (-\left ((2 A+3 C) a^4\right )+6 b B a^3-b^2 (7 A-6 C) a^2-9 b^3 B a+12 A b^4\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\frac {\int \frac {3 \left (\left (a^2-b^2\right ) \left (-B a^3+2 b (A+2 C) a^2-6 b^2 B a+8 A b^3\right )-a b \left (B a^3-2 b (A-C) a^2-3 b^2 B a+4 A b^3\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 \tan (c+d x) \left (-\left (a^4 (2 A+3 C)\right )+6 a^3 b B-a^2 b^2 (7 A-6 C)-9 a b^3 B+12 A b^4\right )}{a d}}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\frac {3 \int \frac {\left (\left (a^2-b^2\right ) \left (-B a^3+2 b (A+2 C) a^2-6 b^2 B a+8 A b^3\right )-a b \left (B a^3-2 b (A-C) a^2-3 b^2 B a+4 A b^3\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 \tan (c+d x) \left (-\left (a^4 (2 A+3 C)\right )+6 a^3 b B-a^2 b^2 (7 A-6 C)-9 a b^3 B+12 A b^4\right )}{a d}}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\frac {3 \int \frac {\left (a^2-b^2\right ) \left (-B a^3+2 b (A+2 C) a^2-6 b^2 B a+8 A b^3\right )-a b \left (B a^3-2 b (A-C) a^2-3 b^2 B a+4 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {2 \tan (c+d x) \left (-\left (a^4 (2 A+3 C)\right )+6 a^3 b B-a^2 b^2 (7 A-6 C)-9 a b^3 B+12 A b^4\right )}{a d}}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3480 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (a^3 (-B)+2 a^2 b (A+2 C)-6 a b^2 B+8 A b^3\right ) \int \sec (c+d x)dx}{a}+\frac {2 b^2 \left (-3 a^4 C+4 a^3 b B-a^2 b^2 (5 A-2 C)-3 a b^3 B+4 A b^4\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}+\frac {2 \tan (c+d x) \left (-\left (a^4 (2 A+3 C)\right )+6 a^3 b B-a^2 b^2 (7 A-6 C)-9 a b^3 B+12 A b^4\right )}{a d}}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (a^3 (-B)+2 a^2 b (A+2 C)-6 a b^2 B+8 A b^3\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {2 b^2 \left (-3 a^4 C+4 a^3 b B-a^2 b^2 (5 A-2 C)-3 a b^3 B+4 A b^4\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}+\frac {2 \tan (c+d x) \left (-\left (a^4 (2 A+3 C)\right )+6 a^3 b B-a^2 b^2 (7 A-6 C)-9 a b^3 B+12 A b^4\right )}{a d}}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (a^3 (-B)+2 a^2 b (A+2 C)-6 a b^2 B+8 A b^3\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {4 b^2 \left (-3 a^4 C+4 a^3 b B-a^2 b^2 (5 A-2 C)-3 a b^3 B+4 A b^4\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}+\frac {2 \tan (c+d x) \left (-\left (a^4 (2 A+3 C)\right )+6 a^3 b B-a^2 b^2 (7 A-6 C)-9 a b^3 B+12 A b^4\right )}{a d}}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (a^3 (-B)+2 a^2 b (A+2 C)-6 a b^2 B+8 A b^3\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}+\frac {4 b^2 \left (-3 a^4 C+4 a^3 b B-a^2 b^2 (5 A-2 C)-3 a b^3 B+4 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}+\frac {2 \tan (c+d x) \left (-\left (a^4 (2 A+3 C)\right )+6 a^3 b B-a^2 b^2 (7 A-6 C)-9 a b^3 B+12 A b^4\right )}{a d}}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\tan (c+d x) \sec ^2(c+d x) \left (-\left (a^2 (A-3 C)\right )-3 a b B+4 A b^2\right )}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^3 B-2 a^2 b (A-C)-3 a b^2 B+4 A b^3\right )}{2 a d}-\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (a^3 (-B)+2 a^2 b (A+2 C)-6 a b^2 B+8 A b^3\right ) \text {arctanh}(\sin (c+d x))}{a d}+\frac {4 b^2 \left (-3 a^4 C+4 a^3 b B-a^2 b^2 (5 A-2 C)-3 a b^3 B+4 A b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}+\frac {2 \tan (c+d x) \left (-\left (a^4 (2 A+3 C)\right )+6 a^3 b B-a^2 b^2 (7 A-6 C)-9 a b^3 B+12 A b^4\right )}{a d}}{2 a}}{3 a}}{a \left (a^2-b^2\right )}\) |
Input:
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + b*Cos[c + d*x])^2,x]
Output:
((A*b^2 - a*(b*B - a*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 - 3*a*b*B - a^2*(A - 3*C))*Sec[c + d*x]^2* Tan[c + d*x])/(3*a*d) - ((3*(4*A*b^3 + a^3*B - 3*a*b^2*B - 2*a^2*b*(A - C) )*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - ((3*((4*b^2*(4*A*b^4 + 4*a^3*b*B - 3*a*b^3*B - 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) + ((a^2 - b^2)*(8*A*b^3 - a^3*B - 6*a*b^2*B + 2*a^2*b*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(a*d)))/a + (2*(12*A*b^4 + 6*a^3*b*B - 9*a*b^3*B - a^2*b^2*(7*A - 6*C) - a^4*(2*A + 3*C))*Tan[c + d*x])/(a*d))/(2*a))/(3*a))/(a*(a^2 - b^2))
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[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.98 (sec) , antiderivative size = 488, normalized size of antiderivative = 1.20
method | result | size |
derivativedivides | \(\frac {\frac {2 b^{2} \left (-\frac {a \left (A \,b^{2}-B a b +a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (5 A \,a^{2} b^{2}-4 A \,b^{4}-4 B \,a^{3} b +3 B a \,b^{3}+3 a^{4} C -2 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5}}-\frac {A}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a A +2 A b -B a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (2 A \,a^{2} b +8 A \,b^{3}-B \,a^{3}-6 B a \,b^{2}+4 a^{2} b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{5}}-\frac {2 A \,a^{2}+2 a A b +6 A \,b^{2}-B \,a^{2}-4 B a b +2 a^{2} C}{2 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a A -2 A b +B a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (-2 A \,a^{2} b -8 A \,b^{3}+B \,a^{3}+6 B a \,b^{2}-4 a^{2} b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{5}}-\frac {2 A \,a^{2}+2 a A b +6 A \,b^{2}-B \,a^{2}-4 B a b +2 a^{2} C}{2 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(488\) |
default | \(\frac {\frac {2 b^{2} \left (-\frac {a \left (A \,b^{2}-B a b +a^{2} C \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (5 A \,a^{2} b^{2}-4 A \,b^{4}-4 B \,a^{3} b +3 B a \,b^{3}+3 a^{4} C -2 C \,a^{2} b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5}}-\frac {A}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a A +2 A b -B a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (2 A \,a^{2} b +8 A \,b^{3}-B \,a^{3}-6 B a \,b^{2}+4 a^{2} b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{5}}-\frac {2 A \,a^{2}+2 a A b +6 A \,b^{2}-B \,a^{2}-4 B a b +2 a^{2} C}{2 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{3 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a A -2 A b +B a}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (-2 A \,a^{2} b -8 A \,b^{3}+B \,a^{3}+6 B a \,b^{2}-4 a^{2} b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{5}}-\frac {2 A \,a^{2}+2 a A b +6 A \,b^{2}-B \,a^{2}-4 B a b +2 a^{2} C}{2 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(488\) |
risch | \(\text {Expression too large to display}\) | \(2208\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x,meth od=_RETURNVERBOSE)
Output:
1/d*(2*b^2/a^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)+(5*A*a^2*b^2-4*A*b^4-4*B*a ^3*b+3*B*a*b^3+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)))-1/3*A/a^2/(tan(1/2*d*x+1/2 *c)-1)^3-1/2*(A*a+2*A*b-B*a)/a^3/(tan(1/2*d*x+1/2*c)-1)^2+1/2*(2*A*a^2*b+8 *A*b^3-B*a^3-6*B*a*b^2+4*C*a^2*b)/a^5*ln(tan(1/2*d*x+1/2*c)-1)-1/2*(2*A*a^ 2+2*A*a*b+6*A*b^2-B*a^2-4*B*a*b+2*C*a^2)/a^4/(tan(1/2*d*x+1/2*c)-1)-1/3*A/ a^2/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(-A*a-2*A*b+B*a)/a^3/(tan(1/2*d*x+1/2*c)+ 1)^2+1/2/a^5*(-2*A*a^2*b-8*A*b^3+B*a^3+6*B*a*b^2-4*C*a^2*b)*ln(tan(1/2*d*x +1/2*c)+1)-1/2*(2*A*a^2+2*A*a*b+6*A*b^2-B*a^2-4*B*a*b+2*C*a^2)/a^4/(tan(1/ 2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 863 vs. \(2 (383) = 766\).
Time = 48.09 (sec) , antiderivative size = 1795, normalized size of antiderivative = 4.43 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2, x, algorithm="fricas")
Output:
[-1/12*(6*((3*C*a^4*b^3 - 4*B*a^3*b^4 + (5*A - 2*C)*a^2*b^5 + 3*B*a*b^6 - 4*A*b^7)*cos(d*x + c)^4 + (3*C*a^5*b^2 - 4*B*a^4*b^3 + (5*A - 2*C)*a^3*b^4 + 3*B*a^2*b^5 - 4*A*a*b^6)*cos(d*x + c)^3)*sqrt(-a^2 + b^2)*log((2*a*b*co s(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*((B*a^7*b - 2*(A + 2*C)*a^6*b^2 + 4*B*a^5*b^3 - 4*(A - 2* C)*a^4*b^4 - 11*B*a^3*b^5 + 2*(7*A - 2*C)*a^2*b^6 + 6*B*a*b^7 - 8*A*b^8)*c os(d*x + c)^4 + (B*a^8 - 2*(A + 2*C)*a^7*b + 4*B*a^6*b^2 - 4*(A - 2*C)*a^5 *b^3 - 11*B*a^4*b^4 + 2*(7*A - 2*C)*a^3*b^5 + 6*B*a^2*b^6 - 8*A*a*b^7)*cos (d*x + c)^3)*log(sin(d*x + c) + 1) + 3*((B*a^7*b - 2*(A + 2*C)*a^6*b^2 + 4 *B*a^5*b^3 - 4*(A - 2*C)*a^4*b^4 - 11*B*a^3*b^5 + 2*(7*A - 2*C)*a^2*b^6 + 6*B*a*b^7 - 8*A*b^8)*cos(d*x + c)^4 + (B*a^8 - 2*(A + 2*C)*a^7*b + 4*B*a^6 *b^2 - 4*(A - 2*C)*a^5*b^3 - 11*B*a^4*b^4 + 2*(7*A - 2*C)*a^3*b^5 + 6*B*a^ 2*b^6 - 8*A*a*b^7)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(2*A*a^8 - 4 *A*a^6*b^2 + 2*A*a^4*b^4 + 2*((2*A + 3*C)*a^7*b - 6*B*a^6*b^2 + (5*A - 9*C )*a^5*b^3 + 15*B*a^4*b^4 - (19*A - 6*C)*a^3*b^5 - 9*B*a^2*b^6 + 12*A*a*b^7 )*cos(d*x + c)^3 + (2*(2*A + 3*C)*a^8 - 9*B*a^7*b + 4*(A - 3*C)*a^6*b^2 + 18*B*a^5*b^3 - 2*(10*A - 3*C)*a^4*b^4 - 9*B*a^3*b^5 + 12*A*a^2*b^6)*cos(d* x + c)^2 + (3*B*a^8 - 4*A*a^7*b - 6*B*a^6*b^2 + 8*A*a^5*b^3 + 3*B*a^4*b^4 - 4*A*a^3*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9*b - 2*a^7*b^3 + a^5*b^...
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c))* *2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2, x, algorithm="maxima")
Output:
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 = 0.18 (sec) , antiderivative size = 619, normalized size of antiderivative = 1.53 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2, x, algorithm="giac")
Output:
-1/6*(12*(3*C*a^4*b^2 - 4*B*a^3*b^3 + 5*A*a^2*b^4 - 2*C*a^2*b^4 + 3*B*a*b^ 5 - 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)) + 12*(C*a^2*b^3*tan(1/2*d*x + 1/2*c) - B*a*b^4* 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*(B*a^3 - 2*A* a^2*b - 4*C*a^2*b + 6*B*a*b^2 - 8*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1) )/a^5 + 3*(B*a^3 - 2*A*a^2*b - 4*C*a^2*b + 6*B*a*b^2 - 8*A*b^3)*log(abs(ta n(1/2*d*x + 1/2*c) - 1))/a^5 + 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2 *tan(1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*a*b*tan(1/2 *d*x + 1/2*c)^5 - 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 18*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*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 - 36*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6 *A*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*tan(1/2*d*x + 1/2*c) + 6*C*a^2*tan(1 /2*d*x + 1/2*c) - 6*A*a*b*tan(1/2*d*x + 1/2*c) - 12*B*a*b*tan(1/2*d*x + 1/ 2*c) + 18*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 = 7.87 (sec) , antiderivative size = 11677, normalized size of antiderivative = 28.83 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + b*cos(c + d*x))^2),x)
Output:
((tan(c/2 + (d*x)/2)*(2*A*a^5 - 8*A*b^5 + B*a^5 + 2*C*a^5 + 6*A*a^2*b^3 + 2*A*a^3*b^2 + 3*B*a^2*b^3 - 5*B*a^3*b^2 - 4*C*a^2*b^3 - 2*C*a^3*b^2 - 4*A* a*b^4 + 6*B*a*b^4 - 3*B*a^4*b + 2*C*a^4*b))/(a^4*(a + b)*(a - b)) + (tan(c /2 + (d*x)/2)^3*(2*A*a^5 + 72*A*b^5 + 3*B*a^5 - 6*C*a^5 - 38*A*a^2*b^3 - 1 4*A*a^3*b^2 - 9*B*a^2*b^3 + 33*B*a^3*b^2 + 36*C*a^2*b^3 + 6*C*a^3*b^2 + 12 *A*a*b^4 - 16*A*a^4*b - 54*B*a*b^4 + 9*B*a^4*b - 18*C*a^4*b))/(3*a^4*(a + b)*(a - b)) + (tan(c/2 + (d*x)/2)^5*(2*A*a^5 - 72*A*b^5 - 3*B*a^5 - 6*C*a^ 5 + 38*A*a^2*b^3 - 14*A*a^3*b^2 - 9*B*a^2*b^3 - 33*B*a^3*b^2 - 36*C*a^2*b^ 3 + 6*C*a^3*b^2 + 12*A*a*b^4 + 16*A*a^4*b + 54*B*a*b^4 + 9*B*a^4*b + 18*C* a^4*b))/(3*a^4*(a + b)*(a - b)) + (tan(c/2 + (d*x)/2)^7*(2*A*a^5 + 8*A*b^5 - B*a^5 + 2*C*a^5 - 6*A*a^2*b^3 + 2*A*a^3*b^2 + 3*B*a^2*b^3 + 5*B*a^3*b^2 + 4*C*a^2*b^3 - 2*C*a^3*b^2 - 4*A*a*b^4 - 6*B*a*b^4 - 3*B*a^4*b - 2*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(((((((8*(2*B*a^18 + 16*A*a^10*b^8 - 8*A*a^11*b ^7 - 36*A*a^12*b^6 + 16*A*a^13*b^5 + 20*A*a^14*b^4 - 4*A*a^15*b^3 - 12*B*a ^11*b^7 + 6*B*a^12*b^6 + 28*B*a^13*b^5 - 14*B*a^14*b^4 - 16*B*a^15*b^3 + 6 *B*a^16*b^2 + 8*C*a^12*b^6 - 4*C*a^13*b^5 - 20*C*a^14*b^4 + 12*C*a^15*b^3 + 12*C*a^16*b^2 - 4*A*a^17*b - 8*C*a^17*b))/(a^14*b + a^15 - a^12*b^3 - a^ 13*b^2) - (8*tan(c/2 + (d*x)/2)*(4*A*b^3 - (B*a^3)/2 + a^2*(A*b + 2*C*b...
Time = 0.21 (sec) , antiderivative size = 3534, normalized size of antiderivative = 8.73 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^2,x)
Output:
(36*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**4*b**2*c + 12*sqrt(a**2 - b* *2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos( c + d*x)*sin(c + d*x)**2*a**3*b**4 - 24*sqrt(a**2 - b**2)*atan((tan((c + d *x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x )**2*a**2*b**4*c - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a*b**6 - 36*s qrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**4*b**2*c - 12*sqrt(a**2 - b**2)*atan((tan((c + d* x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**3*b**4 + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a **2 - b**2))*cos(c + d*x)*a**2*b**4*c + 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a*b**6 - 36*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( a**2 - b**2))*sin(c + d*x)**4*a**3*b**3*c - 12*sqrt(a**2 - b**2)*atan((tan ((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**4*a **2*b**5 + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2 )*b)/sqrt(a**2 - b**2))*sin(c + d*x)**4*a*b**5*c + 12*sqrt(a**2 - b**2)*at an((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d* x)**4*b**7 + 72*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d...