Optimal. Leaf size=324 \[ \frac {a^2 \left (25 a^2 A+72 a b B+48 A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{120 d}+\frac {a \left (4 a^3 B+16 a^2 A b+27 a b^2 B+13 A b^3\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac {\left (8 a^4 B+32 a^3 A b+60 a^2 b^2 B+40 a A b^3+15 b^4 B\right ) \tan (c+d x)}{15 d}+\frac {\left (5 a^4 A+24 a^3 b B+36 a^2 A b^2+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (5 a^4 A+24 a^3 b B+36 a^2 A b^2+32 a b^3 B+8 A b^4\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {a (2 a B+3 A b) \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{10 d}+\frac {a A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{6 d} \]
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Rubi [A] time = 0.80, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {2989, 3047, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {\left (32 a^3 A b+60 a^2 b^2 B+8 a^4 B+40 a A b^3+15 b^4 B\right ) \tan (c+d x)}{15 d}+\frac {\left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^2 \left (25 a^2 A+72 a b B+48 A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{120 d}+\frac {a \left (16 a^2 A b+4 a^3 B+27 a b^2 B+13 A b^3\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac {\left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {a (2 a B+3 A b) \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{10 d}+\frac {a A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{6 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2989
Rule 3021
Rule 3031
Rule 3047
Rule 3767
Rule 3768
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^7(c+d x) \, dx &=\frac {a A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{6} \int (a+b \cos (c+d x))^2 \left (3 a (3 A b+2 a B)+\left (5 a^2 A+6 A b^2+12 a b B\right ) \cos (c+d x)+2 b (a A+3 b B) \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac {a (3 A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{30} \int (a+b \cos (c+d x)) \left (a \left (25 a^2 A+48 A b^2+72 a b B\right )+\left (71 a^2 A b+30 A b^3+24 a^3 B+90 a b^2 B\right ) \cos (c+d x)+2 b \left (14 a A b+6 a^2 B+15 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (3 A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{120} \int \left (-24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right )-15 \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \cos (c+d x)-8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (3 A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{360} \int \left (-45 \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right )-24 \left (32 a^3 A b+40 a A b^3+8 a^4 B+60 a^2 b^2 B+15 b^4 B\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (3 A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{8} \left (-5 a^4 A-36 a^2 A b^2-8 A b^4-24 a^3 b B-32 a b^3 B\right ) \int \sec ^3(c+d x) \, dx-\frac {1}{15} \left (-32 a^3 A b-40 a A b^3-8 a^4 B-60 a^2 b^2 B-15 b^4 B\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (3 A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{16} \left (-5 a^4 A-36 a^2 A b^2-8 A b^4-24 a^3 b B-32 a b^3 B\right ) \int \sec (c+d x) \, dx-\frac {\left (32 a^3 A b+40 a A b^3+8 a^4 B+60 a^2 b^2 B+15 b^4 B\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (32 a^3 A b+40 a A b^3+8 a^4 B+60 a^2 b^2 B+15 b^4 B\right ) \tan (c+d x)}{15 d}+\frac {\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac {a (3 A b+2 a B) (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end {align*}
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Mathematica [A] time = 2.77, size = 244, normalized size = 0.75 \[ \frac {15 \left (5 a^4 A+24 a^3 b B+36 a^2 A b^2+32 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (40 a^4 A \sec ^5(c+d x)+48 a^3 (a B+4 A b) \tan ^4(c+d x)+10 a^2 \left (5 a^2 A+24 a b B+36 A b^2\right ) \sec ^3(c+d x)+160 a \left (a^3 B+4 a^2 A b+3 a b^2 B+2 A b^3\right ) \tan ^2(c+d x)+15 \left (5 a^4 A+24 a^3 b B+36 a^2 A b^2+32 a b^3 B+8 A b^4\right ) \sec (c+d x)+240 \left (a^4 B+4 a^3 A b+6 a^2 b^2 B+4 a A b^3+b^4 B\right )\right )}{240 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.10, size = 327, normalized size = 1.01 \[ \frac {15 \, {\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 60 \, B a^{2} b^{2} + 40 \, A a b^{3} + 15 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} + 40 \, A a^{4} + 15 \, {\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 32 \, {\left (2 \, B a^{4} + 8 \, A a^{3} b + 15 \, B a^{2} b^{2} + 10 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.93, size = 1186, normalized size = 3.66 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 550, normalized size = 1.70 \[ \frac {A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {4 A \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {2 B a \,b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {2 B \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {4 A a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {B \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {B \,b^{4} \tan \left (d x +c \right )}{d}+\frac {3 A \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {8 a^{4} B \tan \left (d x +c \right )}{15 d}+\frac {A \,b^{4} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {5 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 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} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {5 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {4 a^{4} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {32 A \,a^{3} b \tan \left (d x +c \right )}{15 d}+\frac {3 B \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {9 A \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {3 B \,a^{3} b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {9 A \,a^{2} b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{4 d}+\frac {16 A \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {5 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {4 B \,a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {8 A a \,b^{3} \tan \left (d x +c \right )}{3 d}+\frac {2 B a \,b^{3} \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 = 0.44, size = 474, normalized size = 1.46 \[ \frac {32 \, {\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 )} B a^{4} + 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^{3} b + 960 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b^{2} + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{3} - 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 )} - 120 \, B 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 )} - 180 \, A a^{2} b^{2} {\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 )} - 480 \, B 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 \, A b^{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 )} + 480 \, B b^{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 = 3.75, size = 706, normalized size = 2.18 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,A\,a^4}{16}+\frac {3\,B\,a^3\,b}{2}+\frac {9\,A\,a^2\,b^2}{4}+2\,B\,a\,b^3+\frac {A\,b^4}{2}\right )}{\frac {5\,A\,a^4}{4}+6\,B\,a^3\,b+9\,A\,a^2\,b^2+8\,B\,a\,b^3+2\,A\,b^4}\right )\,\left (\frac {5\,A\,a^4}{8}+3\,B\,a^3\,b+\frac {9\,A\,a^2\,b^2}{2}+4\,B\,a\,b^3+A\,b^4\right )}{d}+\frac {\left (\frac {11\,A\,a^4}{8}+A\,b^4-2\,B\,a^4-2\,B\,b^4+\frac {15\,A\,a^2\,b^2}{2}-12\,B\,a^2\,b^2-8\,A\,a\,b^3-8\,A\,a^3\,b+4\,B\,a\,b^3+5\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {5\,A\,a^4}{24}-3\,A\,b^4+\frac {14\,B\,a^4}{3}+10\,B\,b^4-\frac {21\,A\,a^2\,b^2}{2}+44\,B\,a^2\,b^2+\frac {88\,A\,a\,b^3}{3}+\frac {56\,A\,a^3\,b}{3}-12\,B\,a\,b^3-7\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {15\,A\,a^4}{4}+2\,A\,b^4-\frac {52\,B\,a^4}{5}-20\,B\,b^4+3\,A\,a^2\,b^2-72\,B\,a^2\,b^2-48\,A\,a\,b^3-\frac {208\,A\,a^3\,b}{5}+8\,B\,a\,b^3+2\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {15\,A\,a^4}{4}+2\,A\,b^4+\frac {52\,B\,a^4}{5}+20\,B\,b^4+3\,A\,a^2\,b^2+72\,B\,a^2\,b^2+48\,A\,a\,b^3+\frac {208\,A\,a^3\,b}{5}+8\,B\,a\,b^3+2\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,A\,a^4}{24}-3\,A\,b^4-\frac {14\,B\,a^4}{3}-10\,B\,b^4-\frac {21\,A\,a^2\,b^2}{2}-44\,B\,a^2\,b^2-\frac {88\,A\,a\,b^3}{3}-\frac {56\,A\,a^3\,b}{3}-12\,B\,a\,b^3-7\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {11\,A\,a^4}{8}+A\,b^4+2\,B\,a^4+2\,B\,b^4+\frac {15\,A\,a^2\,b^2}{2}+12\,B\,a^2\,b^2+8\,A\,a\,b^3+8\,A\,a^3\,b+4\,B\,a\,b^3+5\,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 )}^{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 )} \]
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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