Integrand size = 24, antiderivative size = 82 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {4 i (a+i a \tan (c+d x))^5}{5 a^3 d}+\frac {2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac {i (a+i a \tan (c+d x))^7}{7 a^5 d} \]
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Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {i (a+i a \tan (c+d x))^7}{7 a^5 d}+\frac {2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac {4 i (a+i a \tan (c+d x))^5}{5 a^3 d} \]
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Rule 45
Rule 3568
Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int (a-x)^2 (a+x)^4 \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i \text {Subst}\left (\int \left (4 a^2 (a+x)^4-4 a (a+x)^5+(a+x)^6\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {4 i (a+i a \tan (c+d x))^5}{5 a^3 d}+\frac {2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac {i (a+i a \tan (c+d x))^7}{7 a^5 d} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 \sec ^7(c+d x) (7+22 \cos (2 (c+d x))-20 i \sin (2 (c+d x))) (-i \cos (5 (c+d x))+\sin (5 (c+d x)))}{105 d} \]
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Time = 34.89 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {128 i a^{2} \left (35 \,{\mathrm e}^{8 i \left (d x +c \right )}+35 \,{\mathrm e}^{6 i \left (d x +c \right )}+21 \,{\mathrm e}^{4 i \left (d x +c \right )}+7 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\) | \(69\) |
derivativedivides | \(\frac {-a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{3 \cos \left (d x +c \right )^{6}}-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(113\) |
default | \(\frac {-a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{3 \cos \left (d x +c \right )^{6}}-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(113\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (64) = 128\).
Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {128 \, {\left (-35 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 35 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 21 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 7 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2}\right )}}{105 \, {\left (d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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\[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=- a^{2} \left (\int \tan ^{2}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \left (- 2 i \tan {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx + \int \left (- \sec ^{6}{\left (c + d x \right )}\right )\, dx\right ) \]
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Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {15 \, a^{2} \tan \left (d x + c\right )^{7} - 35 i \, a^{2} \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} - 105 i \, a^{2} \tan \left (d x + c\right )^{4} - 35 \, a^{2} \tan \left (d x + c\right )^{3} - 105 i \, a^{2} \tan \left (d x + c\right )^{2} - 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \]
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Time = 0.54 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {15 \, a^{2} \tan \left (d x + c\right )^{7} - 35 i \, a^{2} \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} - 105 i \, a^{2} \tan \left (d x + c\right )^{4} - 35 \, a^{2} \tan \left (d x + c\right )^{3} - 105 i \, a^{2} \tan \left (d x + c\right )^{2} - 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \]
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Time = 3.70 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.61 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2\,\sin \left (c+d\,x\right )\,\left (105\,{\cos \left (c+d\,x\right )}^6+{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )\,105{}\mathrm {i}+35\,{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2+{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^3\,105{}\mathrm {i}-21\,{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^4+\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^5\,35{}\mathrm {i}-15\,{\sin \left (c+d\,x\right )}^6\right )}{105\,d\,{\cos \left (c+d\,x\right )}^7} \]
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