Integrand size = 33, antiderivative size = 226 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a b (4 A+3 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d} \]
[Out]
Time = 0.58 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4178, 4167, 4087, 4082, 3872, 3855, 3852, 8} \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{30 b^2 d}+\frac {a \left (2 a^2 C+20 A b^2+13 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{60 b d}+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac {a b (4 A+3 C) \text {arctanh}(\sin (c+d x))}{4 d}-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{10 b^2 d}+\frac {C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \]
[In]
[Out]
Rule 8
Rule 3852
Rule 3855
Rule 3872
Rule 4082
Rule 4087
Rule 4167
Rule 4178
Rubi steps \begin{align*} \text {integral}& = \frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (a C+b (5 A+4 C) \sec (c+d x)-2 a C \sec ^2(c+d x)\right ) \, dx}{5 b} \\ & = -\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (-2 a b C+2 \left (a^2 C+2 b^2 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{20 b^2} \\ & = \frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (2 b \left (20 A b^2-a^2 C+16 b^2 C\right )+2 a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x)\right ) \, dx}{60 b^2} \\ & = \frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) \left (30 a b^3 (4 A+3 C)+8 \left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{120 b^2} \\ & = \frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {1}{4} (a b (4 A+3 C)) \int \sec (c+d x) \, dx+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \int \sec ^2(c+d x) \, dx}{15 b^2} \\ & = \frac {a b (4 A+3 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}-\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 b^2 d} \\ & = \frac {a b (4 A+3 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d} \\ \end{align*}
Time = 2.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.52 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 a b (4 A+3 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (60 \left (a^2+b^2\right ) (A+C)+15 a b (4 A+3 C) \sec (c+d x)+30 a b C \sec ^3(c+d x)+20 \left (A b^2+\left (a^2+2 b^2\right ) C\right ) \tan ^2(c+d x)+12 b^2 C \tan ^4(c+d x)\right )}{60 d} \]
[In]
[Out]
Time = 1.16 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.79
method | result | size |
parts | \(-\frac {\left (A \,b^{2}+C \,a^{2}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {C \,b^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {2 C a b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {2 a A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(178\) |
derivativedivides | \(\frac {a^{2} A \tan \left (d x +c \right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 C a b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \,b^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(184\) |
default | \(\frac {a^{2} A \tan \left (d x +c \right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 C a b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \,b^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(184\) |
parallelrisch | \(\frac {-3 a b \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 a b \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (9 A +10 C \right ) a^{2}+10 b^{2} \left (A +\frac {4 C}{5}\right )\right ) \sin \left (3 d x +3 c \right )+\left (a^{2} \left (3 A +2 C \right )+2 b^{2} \left (A +\frac {4 C}{5}\right )\right ) \sin \left (5 d x +5 c \right )+12 a b \left (A +\frac {7 C}{4}\right ) \sin \left (2 d x +2 c \right )+6 a b \left (A +\frac {3 C}{4}\right ) \sin \left (4 d x +4 c \right )+6 \sin \left (d x +c \right ) \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 b^{2} \left (A +2 C \right )}{3}\right )}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(266\) |
norman | \(\frac {-\frac {4 \left (45 a^{2} A +25 A \,b^{2}+25 C \,a^{2}+29 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {\left (4 a^{2} A -4 a A b +4 A \,b^{2}+4 C \,a^{2}-5 C a b +4 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 d}-\frac {\left (4 a^{2} A +4 a A b +4 A \,b^{2}+4 C \,a^{2}+5 C a b +4 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {\left (24 a^{2} A -12 a A b +16 A \,b^{2}+16 C \,a^{2}-3 C a b +8 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {\left (24 a^{2} A +12 a A b +16 A \,b^{2}+16 C \,a^{2}+3 C a b +8 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a b \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {a b \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(314\) |
risch | \(-\frac {i \left (60 a A b \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C a b \,{\mathrm e}^{9 i \left (d x +c \right )}-60 A \,a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+120 A a b \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C a b \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-120 A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-120 C \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-360 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-280 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-280 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-320 C \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-120 a A b \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C a b \,{\mathrm e}^{3 i \left (d x +c \right )}-240 a^{2} A \,{\mathrm e}^{2 i \left (d x +c \right )}-200 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-200 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-160 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-60 a A b \,{\mathrm e}^{i \left (d x +c \right )}-45 C b a \,{\mathrm e}^{i \left (d x +c \right )}-60 a^{2} A -40 A \,b^{2}-40 C \,a^{2}-32 C \,b^{2}\right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{4 d}-\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{4 d}\) | \(421\) |
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.80 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, {\left (3 \, A + 2 \, C\right )} a^{2} + 2 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, C a b \cos \left (d x + c\right ) + 12 \, C b^{2} + 4 \, {\left (5 \, C a^{2} + {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]
[In]
[Out]
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \]
[In]
[Out]
none
Time = 0.22 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.96 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} + 8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{2} - 15 \, C a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, A a^{2} \tan \left (d x + c\right )}{120 \, d} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (214) = 428\).
Time = 0.36 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.35 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A a b + 3 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a b + 3 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 240 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 232 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 240 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, 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 )}^{5}}}{60 \, d} \]
[In]
[Out]
Time = 19.63 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.42 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{2\,d}-\frac {\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2-2\,A\,a\,b-\frac {5\,C\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (4\,A\,a\,b-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-8\,A\,a^2+C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,A\,a^2+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+\frac {116\,C\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,A\,a^2-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-4\,A\,a\,b-C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2+2\,A\,a\,b+\frac {5\,C\,a\,b}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
[In]
[Out]