Integrand size = 31, antiderivative size = 140 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d} \]
[Out]
Time = 0.21 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4162, 4132, 3852, 4131, 3853, 3855} \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]
[In]
[Out]
Rule 3852
Rule 3853
Rule 3855
Rule 4131
Rule 4132
Rule 4162
Rubi steps \begin{align*} \text {integral}& = \frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (5 a A+a (5 A+4 C) \sec (c+d x)+5 a C \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (5 a A+5 a C \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} (a (5 A+4 C)) \int \sec ^4(c+d x) \, dx \\ & = \frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (a (4 A+3 C)) \int \sec ^3(c+d x) \, dx-\frac {(a (5 A+4 C)) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {1}{8} (a (4 A+3 C)) \int \sec (c+d x) \, dx \\ & = \frac {a (4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d} \\ \end{align*}
Time = 0.55 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (15 (4 A+3 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 (4 A+3 C) \sec (c+d x)+30 C \sec ^3(c+d x)+8 \left (15 (A+C)+5 (A+2 C) \tan ^2(c+d x)+3 C \tan ^4(c+d x)\right )\right )\right )}{120 d} \]
[In]
[Out]
Time = 0.58 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.01
method | result | size |
derivativedivides | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(141\) |
default | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(141\) |
parts | \(\frac {C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {C a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(149\) |
norman | \(\frac {-\frac {a \left (4 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {a \left (12 A +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (20 A +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {4 a \left (25 A +29 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {a \left (44 A +19 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(187\) |
parallelrisch | \(\frac {8 a \left (-\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}+\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}+\frac {3 \left (A +\frac {7 C}{4}\right ) \sin \left (2 d x +2 c \right )}{4}+\left (\frac {5 A}{4}+C \right ) \sin \left (3 d x +3 c \right )+\frac {3 \left (A +\frac {3 C}{4}\right ) \sin \left (4 d x +4 c \right )}{8}+\left (\frac {A}{4}+\frac {C}{5}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +2 C \right )\right )}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(211\) |
risch | \(-\frac {i a \left (60 A \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C \,{\mathrm e}^{9 i \left (d x +c \right )}+120 A \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A \,{\mathrm e}^{6 i \left (d x +c \right )}-560 A \,{\mathrm e}^{4 i \left (d x +c \right )}-640 C \,{\mathrm e}^{4 i \left (d x +c \right )}-120 A \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C \,{\mathrm e}^{3 i \left (d x +c \right )}-400 A \,{\mathrm e}^{2 i \left (d x +c \right )}-320 C \,{\mathrm e}^{2 i \left (d x +c \right )}-60 A \,{\mathrm e}^{i \left (d x +c \right )}-45 C \,{\mathrm e}^{i \left (d x +c \right )}-80 A -64 C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\) | \(265\) |
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{2} + 30 \, C a \cos \left (d x + c\right ) + 24 \, C a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
[In]
[Out]
\[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]
[In]
[Out]
none
Time = 0.21 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.25 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a - 15 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.56 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 200 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 130 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 440 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 190 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 180 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 195 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
[In]
[Out]
Time = 18.82 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.44 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{4\,d}-\frac {\left (A\,a+\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {10\,A\,a}{3}-\frac {13\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,a}{3}+\frac {116\,C\,a}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {22\,A\,a}{3}-\frac {19\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
[In]
[Out]