Integrand size = 41, antiderivative size = 407 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) \text {arctanh}(\sin (c+d x))}{2 b^5 d}+\frac {2 a^2 \left (2 a^2 A b^2-3 A b^4-3 a^3 b B+4 a b^3 B+4 a^4 C-5 a^2 b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^5 (a+b)^{3/2} d}-\frac {\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \tan (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac {\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \sec ^3(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
1/2*(6*B*a^2*b+B*b^3-8*a^3*C-2*a*b^2*(2*A+C))*arctanh(sin(d*x+c))/b^5/d+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)*arctanh( (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))*tan(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)*sec(d*x+c) *tan(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)*sec(d*x+c) ^2*tan(d*x+c)/b^2/(a^2-b^2)/d-(A*b^2-a*(B*b-C*a))*sec(d*x+c)^3*tan(d*x+c)/ b/(a^2-b^2)/d/(a+b*sec(d*x+c))
Time = 8.45 (sec) , antiderivative size = 605, normalized size of antiderivative = 1.49 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {(b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {24 a^2 \left (-3 A b^4-3 a^3 b B+4 a b^3 B+a^2 b^2 (2 A-5 C)+4 a^4 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))}{\left (a^2-b^2\right )^{3/2}}+6 \left (-6 a^2 b B-b^3 B+8 a^3 C+2 a b^2 (2 A+C)\right ) (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b \left (-6 a^2 A b^3+6 A b^5+9 a^3 b^2 B-9 a b^4 B-12 a^4 b C+4 a^2 b^3 C+8 b^5 C+\left (27 a^4 b B-24 a^2 b^3 B+6 b^5 B+a b^4 (9 A-2 C)-36 a^5 C+a^3 b^2 (-18 A+29 C)\right ) \cos (c+d x)+b \left (-a^2+b^2\right ) \left (6 A b^2-9 a b B+12 a^2 C+4 b^2 C\right ) \cos (2 (c+d x))-6 a^3 A b^2 \cos (3 (c+d x))+3 a A b^4 \cos (3 (c+d x))+9 a^4 b B \cos (3 (c+d x))-6 a^2 b^3 B \cos (3 (c+d x))-12 a^5 C \cos (3 (c+d x))+7 a^3 b^2 C \cos (3 (c+d x))+2 a b^4 C \cos (3 (c+d x))\right ) \sec ^2(c+d x) \tan (c+d x)}{-a^2+b^2}\right )}{6 b^5 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^2} \]
((b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-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]]*(b + a*Cos[c + d*x]))/(a^2 - b ^2)^(3/2) + 6*(-6*a^2*b*B - b^3*B + 8*a^3*C + 2*a*b^2*(2*A + C))*(b + a*Co s[c + d*x])*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*(6*a^2*b*B + b^3* B - 8*a^3*C - 2*a*b^2*(2*A + C))*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b*(-6*a^2*A*b^3 + 6*A*b^5 + 9*a^3*b^2*B - 9*a*b^4* B - 12*a^4*b*C + 4*a^2*b^3*C + 8*b^5*C + (27*a^4*b*B - 24*a^2*b^3*B + 6*b^ 5*B + a*b^4*(9*A - 2*C) - 36*a^5*C + a^3*b^2*(-18*A + 29*C))*Cos[c + d*x] + b*(-a^2 + b^2)*(6*A*b^2 - 9*a*b*B + 12*a^2*C + 4*b^2*C)*Cos[2*(c + d*x)] - 6*a^3*A*b^2*Cos[3*(c + d*x)] + 3*a*A*b^4*Cos[3*(c + d*x)] + 9*a^4*b*B*C os[3*(c + d*x)] - 6*a^2*b^3*B*Cos[3*(c + d*x)] - 12*a^5*C*Cos[3*(c + d*x)] + 7*a^3*b^2*C*Cos[3*(c + d*x)] + 2*a*b^4*C*Cos[3*(c + d*x)])*Sec[c + d*x] ^2*Tan[c + d*x])/(-a^2 + b^2)))/(6*b^5*d*(A + 2*C + 2*B*Cos[c + d*x] + A*C os[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)
Time = 2.85 (sec) , antiderivative size = 414, 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, 4586, 3042, 4590, 25, 3042, 4580, 3042, 4570, 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.
\(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4586 |
\(\displaystyle -\frac {\int \frac {\sec ^3(c+d x) \left (-\left (\left (4 C a^2-3 b B a+3 A b^2-b^2 C\right ) \sec ^2(c+d x)\right )+b (b B-a (A+C)) \sec (c+d x)+3 \left (A b^2-a (b B-a C)\right )\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (\left (-4 C a^2+3 b B a-3 A b^2+b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (b B-a (A+C)) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left (A b^2-a (b B-a C)\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4590 |
\(\displaystyle -\frac {\frac {\int -\frac {\sec ^2(c+d x) \left (3 \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right ) \sec ^2(c+d x)-b \left (C a^2-3 b B a+3 A b^2+2 b^2 C\right ) \sec (c+d x)+2 a \left (4 C a^2-3 b B a+3 A b^2-b^2 C\right )\right )}{a+b \sec (c+d x)}dx}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {-\frac {\int \frac {\sec ^2(c+d x) \left (3 \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right ) \sec ^2(c+d x)-b \left (C a^2-3 b B a+3 A b^2+2 b^2 C\right ) \sec (c+d x)+2 a \left (4 C a^2-3 b B a+3 A b^2-b^2 C\right )\right )}{a+b \sec (c+d x)}dx}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (3 \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (C a^2-3 b B a+3 A b^2+2 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a \left (4 C a^2-3 b B a+3 A b^2-b^2 C\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4580 |
\(\displaystyle -\frac {-\frac {\frac {\int \frac {\sec (c+d x) \left (-2 \left (-12 C a^4+9 b B a^3-b^2 (6 A-7 C) a^2-6 b^3 B a+b^4 (3 A+2 C)\right ) \sec ^2(c+d x)-b \left (-4 C a^3+3 b B a^2-2 b^2 (3 A+C) a+3 b^3 B\right ) \sec (c+d x)+3 a \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )\right )}{a+b \sec (c+d x)}dx}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-2 \left (-12 C a^4+9 b B a^3-b^2 (6 A-7 C) a^2-6 b^3 B a+b^4 (3 A+2 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (-4 C a^3+3 b B a^2-2 b^2 (3 A+C) a+3 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 a \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4570 |
\(\displaystyle -\frac {-\frac {\frac {\frac {\int \frac {3 \sec (c+d x) \left (a b \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )+\left (a^2-b^2\right ) \left (-8 C a^3+6 b B a^2-2 b^2 (2 A+C) a+b^3 B\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {-\frac {\frac {\frac {3 \int \frac {\sec (c+d x) \left (a b \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )+\left (a^2-b^2\right ) \left (-8 C a^3+6 b B a^2-2 b^2 (2 A+C) a+b^3 B\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\frac {\frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b \left (-4 C a^3+3 b B a^2-2 b^2 (A-C) a-b^3 B\right )+\left (a^2-b^2\right ) \left (-8 C a^3+6 b B a^2-2 b^2 (2 A+C) a+b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4486 |
\(\displaystyle -\frac {-\frac {\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right ) \int \sec (c+d x)dx}{b}-\frac {2 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}\right )}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {2 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -\frac {-\frac {\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle -\frac {-\frac {\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}\right )}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}\right )}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle -\frac {-\frac {\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}\right )}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}+\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}}{3 b}-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {-\frac {\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (-4 a^3 C+3 a^2 b B-2 a b^2 (A-C)-b^3 B\right )}{2 b d}+\frac {\frac {3 \left (\frac {\left (a^2-b^2\right ) \left (-8 a^3 C+6 a^2 b B-2 a b^2 (2 A+C)+b^3 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^2 \left (-4 a^4 C+3 a^3 b B-a^2 b^2 (2 A-5 C)-4 a b^3 B+3 A b^4\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\right )}{b}-\frac {2 \tan (c+d x) \left (-12 a^4 C+9 a^3 b B-a^2 b^2 (6 A-7 C)-6 a b^3 B+b^4 (3 A+2 C)\right )}{b d}}{2 b}}{3 b}}{b \left (a^2-b^2\right )}\) |
-(((A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^3*Tan[c + d*x])/(b*(a^2 - b^2)*d*( a + b*Sec[c + d*x]))) - (-1/3*((3*A*b^2 - 3*a*b*B + 4*a^2*C - b^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(b*d) - ((3*(3*a^2*b*B - b^3*B - 2*a*b^2*(A - C) - 4*a^3*C)*Sec[c + d*x]*Tan[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))*ArcTanh[Sin[c + d*x]])/(b*d) - (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)*Arc Tanh[(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))*Tan[c + d*x])/(b*d))/(2*b))/(3*b))/(b*(a^2 - b^2))
3.10.8.3.1 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[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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*b*B + 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*(b*B - a*C)*(n - 1) + b*(a*A - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(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, B, C }, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1 )*((d*Csc[e + f*x])^(n - 1)/(b*f*(m + n + 1))), x] + Simp[d/(b*(m + n + 1)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + ( A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C*n)*Csc [e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 0]
Time = 1.02 (sec) , antiderivative size = 492, normalized size of antiderivative = 1.21
method | result | size |
derivativedivides | \(\frac {-\frac {C}{3 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {B b -2 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (-4 a A \,b^{2}+6 B \,a^{2} b +B \,b^{3}-8 a^{3} C -2 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{5}}-\frac {2 A \,b^{2}-4 B a b -B \,b^{2}+6 C \,a^{2}+2 C a b +2 C \,b^{2}}{2 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{2} \left (\frac {a b \left (A \,b^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 A \,a^{2} b^{2}-3 A \,b^{4}-3 B \,a^{3} b +4 B a \,b^{3}+4 a^{4} C -5 C \,a^{2} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{5}}-\frac {C}{3 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-B b +2 C a +C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (4 a A \,b^{2}-6 B \,a^{2} b -B \,b^{3}+8 a^{3} C +2 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{5}}-\frac {2 A \,b^{2}-4 B a b -B \,b^{2}+6 C \,a^{2}+2 C a b +2 C \,b^{2}}{2 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(492\) |
default | \(\frac {-\frac {C}{3 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {B b -2 C a -C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (-4 a A \,b^{2}+6 B \,a^{2} b +B \,b^{3}-8 a^{3} C -2 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{5}}-\frac {2 A \,b^{2}-4 B a b -B \,b^{2}+6 C \,a^{2}+2 C a b +2 C \,b^{2}}{2 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{2} \left (\frac {a b \left (A \,b^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 A \,a^{2} b^{2}-3 A \,b^{4}-3 B \,a^{3} b +4 B a \,b^{3}+4 a^{4} C -5 C \,a^{2} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{5}}-\frac {C}{3 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-B b +2 C a +C b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (4 a A \,b^{2}-6 B \,a^{2} b -B \,b^{3}+8 a^{3} C +2 C a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{5}}-\frac {2 A \,b^{2}-4 B a b -B \,b^{2}+6 C \,a^{2}+2 C a b +2 C \,b^{2}}{2 b^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(492\) |
risch | \(\text {Expression too large to display}\) | \(2208\) |
1/d*(-1/3*C/b^2/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(B*b-2*C*a-C*b)/b^3/(tan(1/2* d*x+1/2*c)+1)^2+1/2/b^5*(-4*A*a*b^2+6*B*a^2*b+B*b^3-8*C*a^3-2*C*a*b^2)*ln( tan(1/2*d*x+1/2*c)+1)-1/2*(2*A*b^2-4*B*a*b-B*b^2+6*C*a^2+2*C*a*b+2*C*b^2)/ b^4/(tan(1/2*d*x+1/2*c)+1)-2*a^2/b^5*(a*b*(A*b^2-B*a*b+C*a^2)/(a^2-b^2)*ta n(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)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))-1/3* C/b^2/(tan(1/2*d*x+1/2*c)-1)^3-1/2*(-B*b+2*C*a+C*b)/b^3/(tan(1/2*d*x+1/2*c )-1)^2+1/2*(4*A*a*b^2-6*B*a^2*b-B*b^3+8*C*a^3+2*C*a*b^2)/b^5*ln(tan(1/2*d* x+1/2*c)-1)-1/2*(2*A*b^2-4*B*a*b-B*b^2+6*C*a^2+2*C*a*b+2*C*b^2)/b^4/(tan(1 /2*d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 889 vs. \(2 (389) = 778\).
Time = 48.41 (sec) , antiderivative size = 1835, normalized size of antiderivative = 4.51 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2, x, algorithm="fricas")
[1/12*(6*((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)*cos(d*x + c)^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)^3)*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)) - 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)*cos( d*x + c)^4 + (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)*cos(d* x + c)^3)*log(sin(d*x + c) + 1) + 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)*cos(d*x + c)^4 + (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)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) + 2*(2*C*a^4*b^4 - 4*C*a^2*b^6 + 2*C*b^8 + 2*(12*C*a^7*b - 9*B*a^6*b^2 + (6*A - 19*C)*a^5*b^3 + 15*B*a^4*b^4 - (9*A - 5*C)*a^3*b^5 - 6*B*a^2*b^6 + (3*A + 2*C)*a*b^7)*c os(d*x + c)^3 + (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)^2 - (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))*sin(d*x + c))/((a^5*b^5 - 2*a^3*b^7 + a*b^9)*...
\[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4/(a + b*s ec(c + d*x))**2, x)
Exception generated. \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(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*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.36 (sec) , antiderivative size = 627, normalized size of antiderivative = 1.54 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (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 )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{5}} + \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 )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\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(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(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*ta n(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)*log(abs(tan(1/2*d*x + 1/2*c) + 1 ))/b^5 + 3*(8*C*a^3 - 6*B*a^2*b + 4*A*a*b^2 + 2*C*a*b^2 - B*b^3)*log(abs(t an(1/2*d*x + 1/2*c) - 1))/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*ta n(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 = 31.05 (sec) , antiderivative size = 11687, normalized size of antiderivative = 28.71 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
(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^11 + 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)/2 - 4*C*a^3 - b^2*(2*A*a + C*a) + 3*B*a^2*b)*(8*a*b^15 - 8*a^2*b^14 - 16*a^3*b^13 + 16*a^4*b^12 + 8*a^5*b^11 - 8*a^6*b^10))/(b^5* (a*b^10 + b^11 - a^2*b^9 - a^3*b^8)))*((B*b^3)/2 - 4*C*a^3 - b^2*(2*A*a + C*a) + 3*B*a^2*b))/b^5 - (8*tan(c/2 + (d*x)/2)*(B^2*b^12 + 128*C^2*a^12 - 2*B^2*a*b^11 - 128*C^2*a^11*b + 16*A^2*a^2*b^10 - 32*A^2*a^3*b^9 + 20*A^2* a^4*b^8 + 64*A^2*a^5*b^7 - 64*A^2*a^6*b^6 - 32*A^2*a^7*b^5 + 32*A^2*a^8*b^ 4 + 11*B^2*a^2*b^10 - 20*B^2*a^3*b^9 + 23*B^2*a^4*b^8 - 26*B^2*a^5*b^7 + 1 7*B^2*a^6*b^6 + 120*B^2*a^7*b^5 - 120*B^2*a^8*b^4 - 72*B^2*a^9*b^3 + 72*B^ 2*a^10*b^2 + 4*C^2*a^2*b^10 - 8*C^2*a^3*b^9 + 28*C^2*a^4*b^8 - 48*C^2*a^5* b^7 + 28*C^2*a^6*b^6 - 8*C^2*a^7*b^5 + 8*C^2*a^8*b^4 + 192*C^2*a^9*b^3 - 1 92*C^2*a^10*b^2 - 8*A*B*a*b^11 - 4*B*C*a*b^11 - 192*B*C*a^11*b + 16*A*B*a^ 2*b^10 - 40*A*B*a^3*b^9 + 64*A*B*a^4*b^8 - 40*A*B*a^5*b^7 - 176*A*B*a^6*b^ 6 + 176*A*B*a^7*b^5 + 96*A*B*a^8*b^4 - 96*A*B*a^9*b^3 + 16*A*C*a^2*b^10 - 32*A*C*a^3*b^9 + 48*A*C*a^4*b^8 - 64*A*C*a^5*b^7 + 40*A*C*a^6*b^6 + 224*A* C*a^7*b^5 - 224*A*C*a^8*b^4 - 128*A*C*a^9*b^3 + 128*A*C*a^10*b^2 + 8*B*...