Optimal. Leaf size=232 \[ \frac {\tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{15 d}+\frac {\left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{15 d}+\frac {\tan (c+d x) \sec (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )}{8 d}+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]
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Rubi [A] time = 0.51, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3047, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {\tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{15 d}+\frac {\left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{15 d}+\frac {\tan (c+d x) \sec (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )}{8 d}+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 3021
Rule 3031
Rule 3047
Rule 3767
Rule 3768
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^2 \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))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+b \cos (c+d x)) \left (2 A b+5 a B+(4 a A+5 b B+5 a C) \cos (c+d x)+b (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{20} \int \left (-4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right )-5 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x)-4 b^2 (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{60} \int \left (-15 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right )-4 \left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{4} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int \sec ^3(c+d x) \, dx-\frac {1}{15} \left (-20 a b B-5 b^2 (2 A+3 C)-2 a^2 (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{8} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int \sec (c+d x) \, dx-\frac {\left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end {align*}
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Mathematica [A] time = 2.51, size = 167, normalized size = 0.72 \[ \frac {15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 \tan ^2(c+d x) \left (a^2 (2 A+C)+2 a b B+A b^2\right )+15 \left (a^2 (A+C)+2 a b B+b^2 (A+C)\right )+3 a^2 A \tan ^4(c+d x)\right )+15 \sec (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )+30 a (a B+2 A b) \sec ^3(c+d x)\right )}{120 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 243, normalized size = 1.05 \[ \frac {15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{2} + 20 \, B a b + 5 \, {\left (2 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, A a^{2} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.50, size = 766, normalized size = 3.30 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.50, size = 404, normalized size = 1.74 \[ \frac {8 a^{2} A \tan \left (d x +c \right )}{15 d}+\frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d}+\frac {a^{2} B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 a^{2} C \tan \left (d x +c \right )}{3 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a A b \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {4 B a b \tan \left (d x +c \right )}{3 d}+\frac {2 B a b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {C a b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 A \,b^{2} \tan \left (d x +c \right )}{3 d}+\frac {A \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {b^{2} B \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {b^{2} C \tan \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 357, normalized size = 1.54 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 15 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, A a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, C a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, C b^{2} \tan \left (d x + c\right )}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.17, size = 455, normalized size = 1.96 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^2}{8}+\frac {B\,b^2}{2}+\frac {3\,A\,a\,b}{4}+C\,a\,b\right )}{\frac {3\,B\,a^2}{2}+2\,B\,b^2+3\,A\,a\,b+4\,C\,a\,b}\right )\,\left (\frac {3\,B\,a^2}{4}+B\,b^2+\frac {3\,A\,a\,b}{2}+2\,C\,a\,b\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-\frac {5\,B\,a^2}{4}-B\,b^2+2\,C\,a^2+2\,C\,b^2-\frac {5\,A\,a\,b}{2}+4\,B\,a\,b-2\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B\,a^2}{2}-\frac {16\,A\,b^2}{3}-\frac {8\,A\,a^2}{3}+2\,B\,b^2-\frac {16\,C\,a^2}{3}-8\,C\,b^2+A\,a\,b-\frac {32\,B\,a\,b}{3}+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^2}{15}+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+12\,C\,b^2+\frac {40\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^2}{3}-\frac {16\,A\,b^2}{3}-\frac {B\,a^2}{2}-2\,B\,b^2-\frac {16\,C\,a^2}{3}-8\,C\,b^2-A\,a\,b-\frac {32\,B\,a\,b}{3}-4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+\frac {5\,B\,a^2}{4}+B\,b^2+2\,C\,a^2+2\,C\,b^2+\frac {5\,A\,a\,b}{2}+4\,B\,a\,b+2\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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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