Optimal. Leaf size=232 \[ \frac {4 a^4 (18 A+25 C) \tan (c+d x)}{15 d}+\frac {7 a^4 (7 A+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 (417 A+550 C) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {(43 A+50 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{60 d}+\frac {(37 A+30 C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{120 d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^4}{6 d}+\frac {2 a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{15 d} \]
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Rubi [A] time = 0.76, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \frac {4 a^4 (18 A+25 C) \tan (c+d x)}{15 d}+\frac {7 a^4 (7 A+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 (417 A+550 C) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {(37 A+30 C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{120 d}+\frac {(43 A+50 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{60 d}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^4}{6 d}+\frac {2 a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{15 d} \]
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))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x))^4 (4 a A+a (A+6 C) \cos (c+d x)) \sec ^6(c+d x) \, dx}{6 a}\\ &=\frac {2 a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x))^3 \left (a^2 (37 A+30 C)+3 a^2 (3 A+10 C) \cos (c+d x)\right ) \sec ^5(c+d x) \, dx}{30 a}\\ &=\frac {(37 A+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x))^2 \left (6 a^3 (43 A+50 C)+a^3 (73 A+150 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{120 a}\\ &=\frac {(43 A+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x)) \left (3 a^4 (417 A+550 C)+3 a^4 (159 A+250 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{360 a}\\ &=\frac {(43 A+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int \left (3 a^5 (417 A+550 C)+\left (3 a^5 (159 A+250 C)+3 a^5 (417 A+550 C)\right ) \cos (c+d x)+3 a^5 (159 A+250 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{360 a}\\ &=\frac {a^4 (417 A+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {(43 A+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {\int \left (192 a^5 (18 A+25 C)+315 a^5 (7 A+10 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{720 a}\\ &=\frac {a^4 (417 A+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {(43 A+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{16} \left (7 a^4 (7 A+10 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{15} \left (4 a^4 (18 A+25 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {7 a^4 (7 A+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 (417 A+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {(43 A+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {\left (4 a^4 (18 A+25 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {7 a^4 (7 A+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {4 a^4 (18 A+25 C) \tan (c+d x)}{15 d}+\frac {a^4 (417 A+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {(43 A+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(37 A+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end {align*}
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Mathematica [A] time = 2.16, size = 358, normalized size = 1.54 \[ -\frac {a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (3360 (7 A+10 C) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-640 (18 A+25 C) \sin (c)+3750 A \sin (2 c+d x)+15360 A \sin (c+2 d x)-1920 A \sin (3 c+2 d x)+3845 A \sin (2 c+3 d x)+3845 A \sin (4 c+3 d x)+6912 A \sin (3 c+4 d x)+735 A \sin (4 c+5 d x)+735 A \sin (6 c+5 d x)+1152 A \sin (5 c+6 d x)+30 (125 A+62 C) \sin (d x)+1860 C \sin (2 c+d x)+17280 C \sin (c+2 d x)-6720 C \sin (3 c+2 d x)+2670 C \sin (2 c+3 d x)+2670 C \sin (4 c+3 d x)+8640 C \sin (3 c+4 d x)-960 C \sin (5 c+4 d x)+810 C \sin (4 c+5 d x)+810 C \sin (6 c+5 d x)+1600 C \sin (5 c+6 d x))\right )}{122880 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.84, size = 181, normalized size = 0.78 \[ \frac {105 \, {\left (7 \, A + 10 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (7 \, A + 10 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (64 \, {\left (18 \, A + 25 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} + 15 \, {\left (49 \, A + 54 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 64 \, {\left (9 \, A + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 10 \, {\left (41 \, A + 6 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 192 \, A a^{4} \cos \left (d x + c\right ) + 40 \, A a^{4}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.69, size = 280, normalized size = 1.21 \[ \frac {105 \, {\left (7 \, A a^{4} + 10 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, {\left (7 \, A a^{4} + 10 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (735 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1050 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 4165 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 5950 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 9702 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 13860 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 11802 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16860 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 7355 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10690 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2790 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.58, size = 258, normalized size = 1.11 \[ \frac {49 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {49 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {35 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {24 A \,a^{4} \tan \left (d x +c \right )}{5 d}+\frac {12 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{5 d}+\frac {20 a^{4} C \tan \left (d x +c \right )}{3 d}+\frac {41 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {27 a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {4 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 a^{4} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.51, size = 456, normalized size = 1.97 \[ \frac {128 \, {\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^{4} + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 5 \, A a^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \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} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, A a^{4} {\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 )} - 30 \, C a^{4} {\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 )} - 120 \, A a^{4} {\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 )} - 720 \, C a^{4} {\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 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 1920 \, C a^{4} \tan \left (d x + c\right )}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.60, size = 262, normalized size = 1.13 \[ \frac {\left (-\frac {49\,A\,a^4}{8}-\frac {35\,C\,a^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {833\,A\,a^4}{24}+\frac {595\,C\,a^4}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {1617\,A\,a^4}{20}-\frac {231\,C\,a^4}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {1967\,A\,a^4}{20}+\frac {281\,C\,a^4}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {1471\,A\,a^4}{24}-\frac {1069\,C\,a^4}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {207\,A\,a^4}{8}+\frac {93\,C\,a^4}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {7\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (7\,A+10\,C\right )}{8\,d} \]
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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