Optimal. Leaf size=278 \[ \frac {a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{30 d}+\frac {\tan (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{15 d}+\frac {\left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{40 d}+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d} \]
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Rubi [A] time = 0.92, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac {\tan (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{15 d}+\frac {\left (3 a^2 b (3 A+4 C)+3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{30 d}+\frac {\tan (c+d x) \sec (c+d x) \left (15 a^2 b (3 A+4 C)+15 a^3 B+50 a b^2 B+6 A b^3\right )}{40 d}+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 3021
Rule 3031
Rule 3047
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac {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 (3 A b+5 a B+(4 a A+5 b B+5 a C) \cos (c+d x)+b (A+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {(3 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+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 \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right )+\left (15 a^2 B+20 b^2 B+a b (29 A+40 C)\right ) \cos (c+d x)+b (7 A b+5 a B+20 b C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 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+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{60} \int \left (-3 \left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right )-4 \left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \cos (c+d x)-3 b^2 (7 A b+5 a B+20 b C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 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+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{120} \int \left (-8 \left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right )-15 \left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 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+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{8} \left (-3 a^3 B-12 a b^2 B-4 b^3 (A+2 C)-3 a^2 b (3 A+4 C)\right ) \int \sec (c+d x) \, dx-\frac {1}{15} \left (-30 a^2 b B-15 b^3 B-15 a b^2 (2 A+3 C)-2 a^3 (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {\left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 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+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {\left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {\left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 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+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 = 4.86, size = 204, normalized size = 0.73 \[ \frac {15 \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (30 a^2 (a B+3 A b) \sec ^3(c+d x)+8 \left (3 a^3 A \tan ^4(c+d x)+5 a \tan ^2(c+d x) \left (a^2 (2 A+C)+3 a b B+3 A b^2\right )+15 \left (a^3 (A+C)+3 a^2 b B+3 a b^2 (A+C)+b^3 B\right )\right )+15 \sec (c+d x) \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 A b^3\right )\right )}{120 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 292, normalized size = 1.05 \[ \frac {15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{3} + 30 \, B a^{2} b + 15 \, {\left (2 \, A + 3 \, C\right )} a b^{2} + 15 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, A a^{3} + 15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.70, size = 989, normalized size = 3.56 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.50, size = 504, normalized size = 1.81 \[ \frac {8 A \,a^{3} \tan \left (d x +c \right )}{15 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 C \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {3 A \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 A \,a^{2} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {9 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 a^{2} b B \tan \left (d x +c \right )}{d}+\frac {a^{2} b B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 C \,a^{2} b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {3 C \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 A a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {A a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 B a \,b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 C a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {A \,b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {b^{3} B \tan \left (d x +c \right )}{d}+\frac {b^{3} C \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 = 0.36, size = 452, normalized size = 1.63 \[ \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} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} - 15 \, B 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 )} - 45 \, A a^{2} 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 \, C a^{2} b {\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 \, B a 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 )} - 60 \, 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 \, C b^{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 b^{2} \tan \left (d x + c\right ) + 240 \, B b^{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 = 3.88, size = 601, normalized size = 2.16 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A\,b^3}{2}+\frac {3\,B\,a^3}{8}+C\,b^3+\frac {9\,A\,a^2\,b}{8}+\frac {3\,B\,a\,b^2}{2}+\frac {3\,C\,a^2\,b}{2}\right )}{2\,A\,b^3+\frac {3\,B\,a^3}{2}+4\,C\,b^3+\frac {9\,A\,a^2\,b}{2}+6\,B\,a\,b^2+6\,C\,a^2\,b}\right )\,\left (A\,b^3+\frac {3\,B\,a^3}{4}+2\,C\,b^3+\frac {9\,A\,a^2\,b}{4}+3\,B\,a\,b^2+3\,C\,a^2\,b\right )}{d}-\frac {\left (2\,A\,a^3-A\,b^3-\frac {5\,B\,a^3}{4}+2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2-\frac {15\,A\,a^2\,b}{4}-3\,B\,a\,b^2+6\,B\,a^2\,b+6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,b^3-\frac {8\,A\,a^3}{3}+\frac {B\,a^3}{2}-8\,B\,b^3-\frac {16\,C\,a^3}{3}-16\,A\,a\,b^2+\frac {3\,A\,a^2\,b}{2}+6\,B\,a\,b^2-16\,B\,a^2\,b-24\,C\,a\,b^2+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^3}{15}+12\,B\,b^3+\frac {20\,C\,a^3}{3}+20\,A\,a\,b^2+20\,B\,a^2\,b+36\,C\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^3}{3}-2\,A\,b^3-\frac {B\,a^3}{2}-8\,B\,b^3-\frac {16\,C\,a^3}{3}-16\,A\,a\,b^2-\frac {3\,A\,a^2\,b}{2}-6\,B\,a\,b^2-16\,B\,a^2\,b-24\,C\,a\,b^2-6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+A\,b^3+\frac {5\,B\,a^3}{4}+2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2+\frac {15\,A\,a^2\,b}{4}+3\,B\,a\,b^2+6\,B\,a^2\,b+6\,C\,a\,b^2+3\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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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