Optimal. Leaf size=267 \[ \frac {a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac {a \left (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {\left (8 a^4 A+40 a^3 b B+60 a^2 A b^2+60 a b^3 B+15 A b^4\right ) \tan (c+d x)}{15 d}+\frac {\left (3 a^4 B+12 a^3 A b+24 a^2 b^2 B+16 a A b^3+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d} \]
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Rubi [A] time = 0.72, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2989, 3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac {\left (60 a^2 A b^2+8 a^4 A+40 a^3 b B+60 a b^3 B+15 A b^4\right ) \tan (c+d x)}{15 d}+\frac {\left (12 a^3 A b+24 a^2 b^2 B+3 a^4 B+16 a A b^3+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac {a \left (60 a^2 A b+15 a^3 B+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2989
Rule 3021
Rule 3031
Rule 3047
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx &=\frac {a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+b \cos (c+d x))^2 \left (a (8 A b+5 a B)+\left (4 a^2 A+5 A b^2+10 a b B\right ) \cos (c+d x)+b (a A+5 b B) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int (a+b \cos (c+d x)) \left (2 a \left (8 a^2 A+18 A b^2+25 a b B\right )+\left (44 a^2 A b+20 A b^3+15 a^3 B+60 a b^2 B\right ) \cos (c+d x)+b \left (12 a A b+5 a^2 B+20 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{60} \int \left (-3 a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right )-4 \left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \cos (c+d x)-3 b^2 \left (12 a A b+5 a^2 B+20 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{120} \int \left (-8 \left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right )-15 \left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{15} \left (-8 a^4 A-60 a^2 A b^2-15 A b^4-40 a^3 b B-60 a b^3 B\right ) \int \sec ^2(c+d x) \, dx-\frac {1}{8} \left (-12 a^3 A b-16 a A b^3-3 a^4 B-24 a^2 b^2 B-8 b^4 B\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {\left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {\left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \tan (c+d x)}{15 d}+\frac {a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \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 = 4.25, size = 198, normalized size = 0.74 \[ \frac {15 \left (3 a^4 B+12 a^3 A b+24 a^2 b^2 B+16 a A b^3+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (24 a^4 A \tan ^4(c+d x)+30 a^3 (a B+4 A b) \sec ^3(c+d x)+80 a^2 \left (a^2 A+2 a b B+3 A b^2\right ) \tan ^2(c+d x)+15 a \left (3 a^3 B+12 a^2 A b+24 a b^2 B+16 A b^3\right ) \sec (c+d x)+120 \left (a^4 A+4 a^3 b B+6 a^2 A b^2+4 a b^3 B+A b^4\right )\right )}{120 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 281, normalized size = 1.05 \[ \frac {15 \, {\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 8 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 8 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, A a^{4} + 8 \, {\left (8 \, A a^{4} + 40 \, B a^{3} b + 60 \, A a^{2} b^{2} + 60 \, B a b^{3} + 15 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 16 \, {\left (2 \, A a^{4} + 10 \, B a^{3} b + 15 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.54, size = 850, normalized size = 3.18 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 431, normalized size = 1.61 \[ \frac {8 A \,a^{4} \tan \left (d x +c \right )}{15 d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {a^{4} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{4} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {A \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {3 A \,a^{3} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {8 B \,a^{3} b \tan \left (d x +c \right )}{3 d}+\frac {4 B \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {4 A \,a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {2 A \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 B \,a^{2} b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {3 B \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 A a \,b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {2 A a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 B a \,b^{3} \tan \left (d x +c \right )}{d}+\frac {A \,b^{4} \tan \left (d x +c \right )}{d}+\frac {B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.59, size = 386, normalized size = 1.45 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{4} + 320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} b + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} b^{2} - 15 \, B a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, B a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, A a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, B b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 960 \, B a b^{3} \tan \left (d x + c\right ) + 240 \, A b^{4} \tan \left (d x + c\right )}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.88, size = 555, normalized size = 2.08 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^4}{8}+\frac {3\,A\,a^3\,b}{2}+3\,B\,a^2\,b^2+2\,A\,a\,b^3+B\,b^4\right )}{\frac {3\,B\,a^4}{2}+6\,A\,a^3\,b+12\,B\,a^2\,b^2+8\,A\,a\,b^3+4\,B\,b^4}\right )\,\left (\frac {3\,B\,a^4}{4}+3\,A\,a^3\,b+6\,B\,a^2\,b^2+4\,A\,a\,b^3+2\,B\,b^4\right )}{d}-\frac {\left (2\,A\,a^4+2\,A\,b^4-\frac {5\,B\,a^4}{4}+12\,A\,a^2\,b^2-6\,B\,a^2\,b^2-4\,A\,a\,b^3-5\,A\,a^3\,b+8\,B\,a\,b^3+8\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B\,a^4}{2}-8\,A\,b^4-\frac {8\,A\,a^4}{3}-32\,A\,a^2\,b^2+12\,B\,a^2\,b^2+8\,A\,a\,b^3+2\,A\,a^3\,b-32\,B\,a\,b^3-\frac {64\,B\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^4}{15}+\frac {80\,B\,a^3\,b}{3}+40\,A\,a^2\,b^2+48\,B\,a\,b^3+12\,A\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^4}{3}-8\,A\,b^4-\frac {B\,a^4}{2}-32\,A\,a^2\,b^2-12\,B\,a^2\,b^2-8\,A\,a\,b^3-2\,A\,a^3\,b-32\,B\,a\,b^3-\frac {64\,B\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^4+2\,A\,b^4+\frac {5\,B\,a^4}{4}+12\,A\,a^2\,b^2+6\,B\,a^2\,b^2+4\,A\,a\,b^3+5\,A\,a^3\,b+8\,B\,a\,b^3+8\,B\,a^3\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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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