Integrand size = 21, antiderivative size = 65 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A+4 C) \tan ^3(c+d x)}{15 d} \]
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Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4131, 3852} \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]
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Rule 3852
Rule 4131
Rubi steps \begin{align*} \text {integral}& = \frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} (5 A+4 C) \int \sec ^4(c+d x) \, dx \\ & = \frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {(5 A+4 C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {(5 A+4 C) \tan (c+d x)}{5 d}+\frac {C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 A+4 C) \tan ^3(c+d x)}{15 d} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d}+\frac {C \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d} \]
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Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(58\) |
default | \(\frac {-A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(58\) |
parts | \(-\frac {A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(60\) |
parallelrisch | \(\frac {\left (50 A +40 C \right ) \sin \left (3 d x +3 c \right )+\left (10 A +8 C \right ) \sin \left (5 d x +5 c \right )+40 \sin \left (d x +c \right ) \left (A +2 C \right )}{15 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(85\) |
risch | \(\frac {4 i \left (15 A \,{\mathrm e}^{6 i \left (d x +c \right )}+35 A \,{\mathrm e}^{4 i \left (d x +c \right )}+40 C \,{\mathrm e}^{4 i \left (d x +c \right )}+25 A \,{\mathrm e}^{2 i \left (d x +c \right )}+20 C \,{\mathrm e}^{2 i \left (d x +c \right )}+5 A +4 C \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\) | \(87\) |
norman | \(\frac {-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {8 \left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {8 \left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {4 \left (25 A +29 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}\) | \(119\) |
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Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \]
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\[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]
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Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, C \tan \left (d x + c\right )^{5} + 5 \, {\left (A + 2 \, C\right )} \tan \left (d x + c\right )^{3} + 15 \, {\left (A + C\right )} \tan \left (d x + c\right )}{15 \, d} \]
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Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, C \tan \left (d x + c\right )^{5} + 5 \, A \tan \left (d x + c\right )^{3} + 10 \, C \tan \left (d x + c\right )^{3} + 15 \, A \tan \left (d x + c\right ) + 15 \, C \tan \left (d x + c\right )}{15 \, d} \]
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Time = 14.97 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {C\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+\left (\frac {A}{3}+\frac {2\,C}{3}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (A+C\right )\,\mathrm {tan}\left (c+d\,x\right )}{d} \]
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