Optimal. Leaf size=167 \[ \frac {63 i a^5 \sec (c+d x)}{8 d}+\frac {63 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {21 i a \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d} \]
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Rubi [A] time = 0.13, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3498, 3486, 3770} \[ \frac {63 i a^5 \sec (c+d x)}{8 d}+\frac {63 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {21 i a \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d} \]
Antiderivative was successfully verified.
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Rule 3486
Rule 3498
Rule 3770
Rubi steps
\begin {align*} \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx &=\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {1}{5} (9 a) \int \sec (c+d x) (a+i a \tan (c+d x))^4 \, dx\\ &=\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {1}{20} \left (63 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=\frac {21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {1}{4} \left (21 a^3\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=\frac {21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac {1}{8} \left (63 a^4\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx\\ &=\frac {63 i a^5 \sec (c+d x)}{8 d}+\frac {21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac {1}{8} \left (63 a^5\right ) \int \sec (c+d x) \, dx\\ &=\frac {63 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {63 i a^5 \sec (c+d x)}{8 d}+\frac {21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}\\ \end {align*}
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Mathematica [A] time = 1.28, size = 115, normalized size = 0.69 \[ \frac {a^5 (\cos (5 d x)+i \sin (5 d x)) \left (5040 \tanh ^{-1}\left (\cos (c) \tan \left (\frac {d x}{2}\right )+\sin (c)\right )+i \sec ^5(c+d x) (450 i \sin (2 (c+d x))+325 i \sin (4 (c+d x))+1920 \cos (2 (c+d x))+640 \cos (4 (c+d x))+1344)\right )}{320 d (\cos (d x)+i \sin (d x))^5} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.52, size = 310, normalized size = 1.86 \[ \frac {1930 i \, a^{5} e^{\left (9 i \, d x + 9 i \, c\right )} + 4740 i \, a^{5} e^{\left (7 i \, d x + 7 i \, c\right )} + 5376 i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )} + 2940 i \, a^{5} e^{\left (3 i \, d x + 3 i \, c\right )} + 630 i \, a^{5} e^{\left (i \, d x + i \, c\right )} + 315 \, {\left (a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 315 \, {\left (a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{40 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.09, size = 189, normalized size = 1.13 \[ \frac {315 \, a^{5} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 315 \, a^{5} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {2 \, {\left (275 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 200 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 750 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1600 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3280 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 750 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2240 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 275 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 488 i \, a^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{40 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.34, size = 329, normalized size = 1.97 \[ -\frac {i a^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}-\frac {10 i a^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}+\frac {18 i a^{5} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5 d}+\frac {i a^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}+\frac {5 i a^{5}}{d \cos \left (d x +c \right )}+\frac {i a^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )}+\frac {5 a^{5} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {5 a^{5} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {5 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {55 a^{5} \sin \left (d x +c \right )}{8 d}+\frac {63 a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {36 i a^{5} \cos \left (d x +c \right )}{5 d}+\frac {10 i a^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )}+\frac {i a^{5} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{5 d}-\frac {5 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.76, size = 215, normalized size = 1.29 \[ \frac {75 \, a^{5} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 600 \, a^{5} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, a^{5} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac {1200 i \, a^{5}}{\cos \left (d x + c\right )} + \frac {800 i \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{5}}{\cos \left (d x + c\right )^{3}} + \frac {16 i \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 3\right )} a^{5}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.04, size = 228, normalized size = 1.37 \[ \frac {63\,a^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {55\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,10{}\mathrm {i}-\frac {75\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,80{}\mathrm {i}+a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,164{}\mathrm {i}+\frac {75\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,112{}\mathrm {i}-\frac {55\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {a^5\,122{}\mathrm {i}}{5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ i a^{5} \left (\int \left (- i \sec {\left (c + d x \right )}\right )\, dx + \int 5 \tan {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \left (- 10 \tan ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\right )\, dx + \int \tan ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 10 i \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \left (- 5 i \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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