Optimal. Leaf size=235 \[ -\frac {3003 i a^8 \sec (c+d x)}{16 d}-\frac {3003 a^8 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}-\frac {1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac {13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac {429 i a^2 \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^3}{40 d}-\frac {143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d} \]
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Rubi [A] time = 0.20, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3496, 3498, 3486, 3770} \[ -\frac {3003 i a^8 \sec (c+d x)}{16 d}-\frac {3003 a^8 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac {429 i a^2 \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^3}{40 d}-\frac {143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac {1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac {1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d} \]
Antiderivative was successfully verified.
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Rule 3486
Rule 3496
Rule 3498
Rule 3770
Rubi steps
\begin {align*} \int \cos (c+d x) (a+i a \tan (c+d x))^8 \, dx &=-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\left (13 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^6 \, dx\\ &=-\frac {13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac {1}{6} \left (143 a^3\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx\\ &=-\frac {13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac {143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac {1}{10} \left (429 a^4\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^4 \, dx\\ &=-\frac {429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac {13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac {143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac {1}{40} \left (3003 a^5\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=-\frac {429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac {13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac {143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac {1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac {1}{8} \left (1001 a^6\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=-\frac {429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac {13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac {143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac {1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac {1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}-\frac {1}{16} \left (3003 a^7\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx\\ &=-\frac {3003 i a^8 \sec (c+d x)}{16 d}-\frac {429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac {13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac {143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac {1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac {1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}-\frac {1}{16} \left (3003 a^8\right ) \int \sec (c+d x) \, dx\\ &=-\frac {3003 a^8 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {3003 i a^8 \sec (c+d x)}{16 d}-\frac {429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac {13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac {143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac {1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac {1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}\\ \end {align*}
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Mathematica [A] time = 3.38, size = 205, normalized size = 0.87 \[ \frac {a^8 (\cos (8 c)-i \sin (8 c)) \cos ^2(c+d x) (\tan (c+d x)-i)^8 \left (-658944 i \cos (c+d x)+5 (12870 \sin (c+d x)+22165 \sin (3 (c+d x))+10959 \sin (5 (c+d x))+1536 \sin (7 (c+d x))-73216 i \cos (3 (c+d x))-19968 i \cos (5 (c+d x))-1536 i \cos (7 (c+d x)))+720720 \cos ^6(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3840 d (\cos (d x)+i \sin (d x))^8} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 378, normalized size = 1.61 \[ \frac {-30720 i \, a^{8} e^{\left (13 i \, d x + 13 i \, c\right )} - 309270 i \, a^{8} e^{\left (11 i \, d x + 11 i \, c\right )} - 953810 i \, a^{8} e^{\left (9 i \, d x + 9 i \, c\right )} - 1446588 i \, a^{8} e^{\left (7 i \, d x + 7 i \, c\right )} - 1189188 i \, a^{8} e^{\left (5 i \, d x + 5 i \, c\right )} - 510510 i \, a^{8} e^{\left (3 i \, d x + 3 i \, c\right )} - 90090 i \, a^{8} e^{\left (i \, d x + i \, c\right )} - 45045 \, {\left (a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{8}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) + 45045 \, {\left (a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{8}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{240 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 5.36, size = 924, normalized size = 3.93 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.54, size = 464, normalized size = 1.97 \[ -\frac {7 a^{8} \left (\sin ^{7}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{4}}+\frac {21 a^{8} \left (\sin ^{7}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{8} \left (\sin ^{9}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}-\frac {a^{8} \left (\sin ^{9}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {5 a^{8} \left (\sin ^{9}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}-\frac {8 i a^{8} \left (\sin ^{8}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}-\frac {328 i a^{8} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{5 d}-\frac {8 i a^{8} \left (\sin ^{8}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}-\frac {56 i a^{8} \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}-\frac {2152 i a^{8} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{15 d}+\frac {8 i a^{8} \left (\sin ^{8}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{3}}+\frac {56 i a^{8} \left (\sin ^{6}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}-\frac {56 i a^{8} \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}-\frac {8 i a^{8} \cos \left (d x +c \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{d}-\frac {4424 i a^{8} \cos \left (d x +c \right )}{15 d}+\frac {5 a^{8} \left (\sin ^{7}\left (d x +c \right )\right )}{16 d}+\frac {175 a^{8} \left (\sin ^{5}\left (d x +c \right )\right )}{16 d}+\frac {2555 a^{8} \left (\sin ^{3}\left (d x +c \right )\right )}{48 d}-\frac {3003 a^{8} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {35 a^{8} \left (\sin ^{5}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {3019 a^{8} \sin \left (d x +c \right )}{16 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.49, size = 396, normalized size = 1.69 \[ -\frac {5 \, a^{8} {\left (\frac {2 \, {\left (87 \, \sin \left (d x + c\right )^{5} - 136 \, \sin \left (d x + c\right )^{3} + 57 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + 105 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 96 \, \sin \left (d x + c\right )\right )} + 840 \, a^{8} {\left (\frac {2 \, {\left (9 \, \sin \left (d x + c\right )^{3} - 7 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, \sin \left (d x + c\right )\right )} + 8400 \, a^{8} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )} + 26880 i \, a^{8} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 8960 i \, a^{8} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} + 768 i \, a^{8} {\left (\frac {15 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 1}{\cos \left (d x + c\right )^{5}} + 5 \, \cos \left (d x + c\right )\right )} + 6720 \, a^{8} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 3840 i \, a^{8} \cos \left (d x + c\right ) - 480 \, a^{8} \sin \left (d x + c\right )}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.30, size = 399, normalized size = 1.70 \[ \frac {\frac {3019\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{8}+\frac {a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,2891{}\mathrm {i}}{8}-\frac {52795\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{24}-\frac {a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,45115{}\mathrm {i}}{24}+\frac {22415\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{4}+\frac {a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,43757{}\mathrm {i}}{12}-\frac {97811\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{12}-\frac {a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,12977{}\mathrm {i}}{4}+\frac {167237\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{24}+\frac {a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,160729{}\mathrm {i}}{120}-\frac {127113\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{40}-\frac {a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,25499{}\mathrm {i}}{120}+\frac {8848\,a^8}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,1{}\mathrm {i}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,6{}\mathrm {i}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,15{}\mathrm {i}-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,20{}\mathrm {i}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,15{}\mathrm {i}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,6{}\mathrm {i}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}-\frac {3003\,a^8\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.94, size = 320, normalized size = 1.36 \[ \frac {3003 a^{8} \left (\frac {\log {\left (e^{i d x} - i e^{- i c} \right )}}{16} - \frac {\log {\left (e^{i d x} + i e^{- i c} \right )}}{16}\right )}{d} + \frac {62475 i a^{8} e^{11 i c} e^{11 i d x} + 246505 i a^{8} e^{9 i c} e^{9 i d x} + 416094 i a^{8} e^{7 i c} e^{7 i d x} + 364194 i a^{8} e^{5 i c} e^{5 i d x} + 163095 i a^{8} e^{3 i c} e^{3 i d x} + 29685 i a^{8} e^{i c} e^{i d x}}{- 120 d e^{12 i c} e^{12 i d x} - 720 d e^{10 i c} e^{10 i d x} - 1800 d e^{8 i c} e^{8 i d x} - 2400 d e^{6 i c} e^{6 i d x} - 1800 d e^{4 i c} e^{4 i d x} - 720 d e^{2 i c} e^{2 i d x} - 120 d} + \begin {cases} - \frac {128 i a^{8} e^{i c} e^{i d x}}{d} & \text {for}\: d \neq 0 \\128 a^{8} x e^{i c} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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