Optimal. Leaf size=194 \[ \frac {a^3 (38 A+55 C) \tan (c+d x)}{15 d}+\frac {a^3 (13 A+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{120 d}+\frac {(11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{30 d}+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{20 a d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]
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Rubi [A] time = 0.57, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3044, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac {a^3 (38 A+55 C) \tan (c+d x)}{15 d}+\frac {a^3 (13 A+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{120 d}+\frac {(11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{30 d}+\frac {3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{20 a d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2968
Rule 2975
Rule 3021
Rule 3044
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int (a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^3 (3 a A+a (A+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^2 \left (2 a^2 (11 A+10 C)+a^2 (7 A+20 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac {(11 A+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x)) \left (a^3 (109 A+140 C)+a^3 (43 A+80 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac {(11 A+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (a^4 (109 A+140 C)+\left (a^4 (43 A+80 C)+a^4 (109 A+140 C)\right ) \cos (c+d x)+a^4 (43 A+80 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac {a^3 (109 A+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(11 A+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (8 a^4 (38 A+55 C)+15 a^4 (13 A+20 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a}\\ &=\frac {a^3 (109 A+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(11 A+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (a^3 (13 A+20 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{15} \left (a^3 (38 A+55 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a^3 (13 A+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (109 A+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(11 A+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {\left (a^3 (38 A+55 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {a^3 (13 A+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (38 A+55 C) \tan (c+d x)}{15 d}+\frac {a^3 (109 A+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {(11 A+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end {align*}
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Mathematica [A] time = 1.49, size = 294, normalized size = 1.52 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (13 A+20 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (3 A+7 C) \sin (2 c+d x)+750 A \sin (c+2 d x)+750 A \sin (3 c+2 d x)+1520 A \sin (2 c+3 d x)+195 A \sin (3 c+4 d x)+195 A \sin (5 c+4 d x)+304 A \sin (4 c+5 d x)+80 (29 A+34 C) \sin (d x)+360 C \sin (c+2 d x)+360 C \sin (3 c+2 d x)+1840 C \sin (2 c+3 d x)-360 C \sin (4 c+3 d x)+180 C \sin (3 c+4 d x)+180 C \sin (5 c+4 d x)+440 C \sin (4 c+5 d x))\right )}{15360 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 161, normalized size = 0.83 \[ \frac {15 \, {\left (13 \, A + 20 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (13 \, A + 20 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (38 \, A + 55 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (13 \, A + 12 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (19 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 90 \, A a^{3} \cos \left (d x + c\right ) + 24 \, A a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.24, size = 246, normalized size = 1.27 \[ \frac {15 \, {\left (13 \, A a^{3} + 20 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (13 \, A a^{3} + 20 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (195 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 300 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 910 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1400 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1664 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2560 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1330 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 765 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 660 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.47, size = 212, normalized size = 1.09 \[ \frac {13 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {13 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {5 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {38 A \,a^{3} \tan \left (d x +c \right )}{15 d}+\frac {19 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {11 C \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {3 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 292, normalized size = 1.51 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 45 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 720 \, C a^{3} \tan \left (d x + c\right )}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.49, size = 224, normalized size = 1.15 \[ \frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (13\,A+20\,C\right )}{4\,d}-\frac {\left (\frac {13\,A\,a^3}{4}+5\,C\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {91\,A\,a^3}{6}-\frac {70\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {416\,A\,a^3}{15}+\frac {128\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {133\,A\,a^3}{6}-\frac {106\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,A\,a^3}{4}+11\,C\,a^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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