Integrand size = 39, antiderivative size = 121 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^2 (2 A+B) x+\frac {a^2 (2 A+4 B+3 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {a^2 (2 A-2 B-3 C) \tan (c+d x)}{2 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d} \]
[Out]
Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4171, 4002, 3999, 3852, 8, 3855} \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (2 A+4 B+3 C) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {a^2 (2 A-2 B-3 C) \tan (c+d x)}{2 d}+a^2 x (2 A+B)-\frac {(2 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^2}{d} \]
[In]
[Out]
Rule 8
Rule 3852
Rule 3855
Rule 3999
Rule 4002
Rule 4171
Rubi steps \begin{align*} \text {integral}& = \frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}+\frac {\int (a+a \sec (c+d x))^2 (a (2 A+B)-a (2 A-C) \sec (c+d x)) \, dx}{a} \\ & = \frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}+\frac {\int (a+a \sec (c+d x)) \left (2 a^2 (2 A+B)-a^2 (2 A-2 B-3 C) \sec (c+d x)\right ) \, dx}{2 a} \\ & = a^2 (2 A+B) x+\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}-\frac {1}{2} \left (a^2 (2 A-2 B-3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (a^2 (2 A+4 B+3 C)\right ) \int \sec (c+d x) \, dx \\ & = a^2 (2 A+B) x+\frac {a^2 (2 A+4 B+3 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}+\frac {\left (a^2 (2 A-2 B-3 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d} \\ & = a^2 (2 A+B) x+\frac {a^2 (2 A+4 B+3 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {a^2 (2 A-2 B-3 C) \tan (c+d x)}{2 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d} \\ \end{align*}
Time = 2.17 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.07 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=2 a^2 A x+a^2 B x+\frac {a^2 A \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a^2 B \text {arctanh}(\sin (c+d x))}{d}+\frac {3 a^2 C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 A \sin (c+d x)}{d}+\frac {a^2 B \tan (c+d x)}{d}+\frac {2 a^2 C \tan (c+d x)}{d}+\frac {a^2 C \sec (c+d x) \tan (c+d x)}{2 d} \]
[In]
[Out]
Time = 0.46 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.28
method | result | size |
derivativedivides | \(\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \tan \left (d x +c \right )+C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} A \left (d x +c \right )+2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 C \,a^{2} \tan \left (d x +c \right )+a^{2} A \sin \left (d x +c \right )+B \,a^{2} \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(155\) |
default | \(\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \tan \left (d x +c \right )+C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} A \left (d x +c \right )+2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 C \,a^{2} \tan \left (d x +c \right )+a^{2} A \sin \left (d x +c \right )+B \,a^{2} \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(155\) |
parallelrisch | \(\frac {a^{2} \left (-2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +2 B +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +2 B +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 x d \left (A +\frac {B}{2}\right ) \cos \left (2 d x +2 c \right )+\left (2 B +4 C \right ) \sin \left (2 d x +2 c \right )+A \sin \left (3 d x +3 c \right )+\sin \left (d x +c \right ) \left (A +2 C \right )+4 x d \left (A +\frac {B}{2}\right )\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(156\) |
risch | \(2 a^{2} A x +a^{2} B x -\frac {i a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{2} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{2} \left (C \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-4 C \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{i \left (d x +c \right )}-2 B -4 C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) | \(260\) |
norman | \(\frac {\left (-2 a^{2} A -B \,a^{2}\right ) x +\left (-4 a^{2} A -2 B \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (2 a^{2} A +B \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (4 a^{2} A +2 B \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {a^{2} \left (2 A -2 B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {a^{2} \left (2 B +3 C +6 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {a^{2} \left (2 A +2 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a^{2} \left (6 A -2 B -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a^{2} \left (2 A +4 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (2 A +4 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(298\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.18 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (2 \, A + B\right )} a^{2} d x \cos \left (d x + c\right )^{2} + {\left (2 \, A + 4 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + 4 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
[In]
[Out]
\[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{2} \left (\int A \cos {\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.59 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (d x + c\right )} A a^{2} + 4 \, {\left (d x + c\right )} B a^{2} - C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{2} \sin \left (d x + c\right ) + 4 \, B a^{2} \tan \left (d x + c\right ) + 8 \, C a^{2} \tan \left (d x + c\right )}{4 \, d} \]
[In]
[Out]
none
Time = 0.35 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.69 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (2 \, A a^{2} + B a^{2}\right )} {\left (d x + c\right )} + {\left (2 \, A a^{2} + 4 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a^{2} + 4 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
[In]
[Out]
Time = 16.98 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.02 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+C\,a^2\,\sin \left (2\,c+2\,d\,x\right )+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{4}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {2\,\left (-2\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}-B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+\frac {C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{2}\right )}{d} \]
[In]
[Out]