Integrand size = 21, antiderivative size = 70 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
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Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4131, 3853, 3855} \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]
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Rule 3853
Rule 3855
Rule 4131
Rubi steps \begin{align*} \text {integral}& = \frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} (4 A+3 C) \int \sec ^3(c+d x) \, dx \\ & = \frac {(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} (4 A+3 C) \int \sec (c+d x) \, dx \\ & = \frac {(4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 C \text {arctanh}(\sin (c+d x))}{8 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d}+\frac {3 C \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
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Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.21
method | result | size |
derivativedivides | \(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(85\) |
default | \(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(85\) |
parts | \(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(87\) |
parallelrisch | \(\frac {-8 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (4 A +3 C \right ) \sin \left (3 d x +3 c \right )+4 \left (A +\frac {11 C}{4}\right ) \sin \left (d x +c \right )}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(143\) |
norman | \(\frac {-\frac {\left (4 A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}-\frac {\left (4 A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {\left (4 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (4 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(157\) |
risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (4 A \,{\mathrm e}^{6 i \left (d x +c \right )}+3 C \,{\mathrm e}^{6 i \left (d x +c \right )}+4 A \,{\mathrm e}^{4 i \left (d x +c \right )}+11 C \,{\mathrm e}^{4 i \left (d x +c \right )}-4 A \,{\mathrm e}^{2 i \left (d x +c \right )}-11 C \,{\mathrm e}^{2 i \left (d x +c \right )}-4 A -3 C \right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}\) | \(199\) |
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Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, C\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
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\[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]
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Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.39 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (4 \, A + 3 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, A + 3 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (4 \, A + 3 \, C\right )} \sin \left (d x + c\right )^{3} - {\left (4 \, A + 5 \, C\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
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Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.40 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (4 \, A + 3 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (4 \, A + 3 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, A \sin \left (d x + c\right )^{3} + 3 \, C \sin \left (d x + c\right )^{3} - 4 \, A \sin \left (d x + c\right ) - 5 \, C \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
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Time = 15.55 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10 \[ \int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\sin \left (c+d\,x\right )\,\left (\frac {A}{2}+\frac {5\,C}{8}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {A}{2}+\frac {3\,C}{8}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {A}{2}+\frac {3\,C}{8}\right )}{d} \]
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