Optimal. Leaf size=381 \[ \frac {\tan (c+d x) \left (2 a^2 C-6 a b B+30 A b^2+25 b^2 C\right ) (a+b \sec (c+d x))^3}{120 b^2 d}+\frac {\left (8 a^3 B+6 a^2 b (4 A+3 C)+18 a b^2 B+b^3 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {\tan (c+d x) \left (-2 a^3 C+6 a^2 b B-3 a b^2 (10 A+7 C)-32 b^3 B\right ) (a+b \sec (c+d x))^2}{120 b^2 d}-\frac {\tan (c+d x) \sec (c+d x) \left (-4 a^4 C+12 a^3 b B-12 a^2 b^2 (5 A+3 C)-142 a b^3 B-15 b^4 (6 A+5 C)\right )}{240 b d}-\frac {\tan (c+d x) \left (-2 a^5 C+6 a^4 b B-a^3 b^2 (30 A+17 C)-104 a^2 b^3 B-24 a b^4 (5 A+4 C)-32 b^5 B\right )}{60 b^2 d}+\frac {(3 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^4}{15 b^2 d}+\frac {C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^4}{6 b d} \]
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Rubi [A] time = 0.87, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {4092, 4082, 4002, 3997, 3787, 3770, 3767, 8} \[ -\frac {\tan (c+d x) \left (-a^3 b^2 (30 A+17 C)-104 a^2 b^3 B+6 a^4 b B-2 a^5 C-24 a b^4 (5 A+4 C)-32 b^5 B\right )}{60 b^2 d}+\frac {\left (6 a^2 b (4 A+3 C)+8 a^3 B+18 a b^2 B+b^3 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\tan (c+d x) \left (2 a^2 C-6 a b B+30 A b^2+25 b^2 C\right ) (a+b \sec (c+d x))^3}{120 b^2 d}-\frac {\tan (c+d x) \left (6 a^2 b B-2 a^3 C-3 a b^2 (10 A+7 C)-32 b^3 B\right ) (a+b \sec (c+d x))^2}{120 b^2 d}-\frac {\tan (c+d x) \sec (c+d x) \left (-12 a^2 b^2 (5 A+3 C)+12 a^3 b B-4 a^4 C-142 a b^3 B-15 b^4 (6 A+5 C)\right )}{240 b d}+\frac {(3 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^4}{15 b^2 d}+\frac {C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^4}{6 b d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3787
Rule 3997
Rule 4002
Rule 4082
Rule 4092
Rubi steps
\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (a C+b (6 A+5 C) \sec (c+d x)+2 (3 b B-a C) \sec ^2(c+d x)\right ) \, dx}{6 b}\\ &=\frac {(3 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (3 b (8 b B-a C)+\left (30 A b^2-6 a b B+2 a^2 C+25 b^2 C\right ) \sec (c+d x)\right ) \, dx}{30 b^2}\\ &=\frac {\left (30 A b^2-6 a b B+2 a^2 C+25 b^2 C\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}+\frac {(3 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (3 b \left (30 A b^2+26 a b B-2 a^2 C+25 b^2 C\right )-3 \left (6 a^2 b B-32 b^3 B-2 a^3 C-3 a b^2 (10 A+7 C)\right ) \sec (c+d x)\right ) \, dx}{120 b^2}\\ &=-\frac {\left (6 a^2 b B-32 b^3 B-2 a^3 C-3 a b^2 (10 A+7 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac {\left (30 A b^2-6 a b B+2 a^2 C+25 b^2 C\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}+\frac {(3 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (3 b \left (66 a^2 b B+64 b^3 B-2 a^3 C+3 a b^2 (50 A+39 C)\right )-3 \left (12 a^3 b B-142 a b^3 B-4 a^4 C-12 a^2 b^2 (5 A+3 C)-15 b^4 (6 A+5 C)\right ) \sec (c+d x)\right ) \, dx}{360 b^2}\\ &=-\frac {\left (12 a^3 b B-142 a b^3 B-4 a^4 C-12 a^2 b^2 (5 A+3 C)-15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}-\frac {\left (6 a^2 b B-32 b^3 B-2 a^3 C-3 a b^2 (10 A+7 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac {\left (30 A b^2-6 a b B+2 a^2 C+25 b^2 C\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}+\frac {(3 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) \left (45 b^2 \left (8 a^3 B+18 a b^2 B+6 a^2 b (4 A+3 C)+b^3 (6 A+5 C)\right )-12 \left (6 a^4 b B-104 a^2 b^3 B-32 b^5 B-2 a^5 C-24 a b^4 (5 A+4 C)-a^3 b^2 (30 A+17 C)\right ) \sec (c+d x)\right ) \, dx}{720 b^2}\\ &=-\frac {\left (12 a^3 b B-142 a b^3 B-4 a^4 C-12 a^2 b^2 (5 A+3 C)-15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}-\frac {\left (6 a^2 b B-32 b^3 B-2 a^3 C-3 a b^2 (10 A+7 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac {\left (30 A b^2-6 a b B+2 a^2 C+25 b^2 C\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}+\frac {(3 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac {1}{16} \left (8 a^3 B+18 a b^2 B+6 a^2 b (4 A+3 C)+b^3 (6 A+5 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (6 a^4 b B-104 a^2 b^3 B-32 b^5 B-2 a^5 C-24 a b^4 (5 A+4 C)-a^3 b^2 (30 A+17 C)\right ) \int \sec ^2(c+d x) \, dx}{60 b^2}\\ &=\frac {\left (8 a^3 B+18 a b^2 B+6 a^2 b (4 A+3 C)+b^3 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {\left (12 a^3 b B-142 a b^3 B-4 a^4 C-12 a^2 b^2 (5 A+3 C)-15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}-\frac {\left (6 a^2 b B-32 b^3 B-2 a^3 C-3 a b^2 (10 A+7 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac {\left (30 A b^2-6 a b B+2 a^2 C+25 b^2 C\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}+\frac {(3 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}+\frac {\left (6 a^4 b B-104 a^2 b^3 B-32 b^5 B-2 a^5 C-24 a b^4 (5 A+4 C)-a^3 b^2 (30 A+17 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{60 b^2 d}\\ &=\frac {\left (8 a^3 B+18 a b^2 B+6 a^2 b (4 A+3 C)+b^3 (6 A+5 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac {\left (6 a^4 b B-104 a^2 b^3 B-32 b^5 B-2 a^5 C-24 a b^4 (5 A+4 C)-a^3 b^2 (30 A+17 C)\right ) \tan (c+d x)}{60 b^2 d}-\frac {\left (12 a^3 b B-142 a b^3 B-4 a^4 C-12 a^2 b^2 (5 A+3 C)-15 b^4 (6 A+5 C)\right ) \sec (c+d x) \tan (c+d x)}{240 b d}-\frac {\left (6 a^2 b B-32 b^3 B-2 a^3 C-3 a b^2 (10 A+7 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b^2 d}+\frac {\left (30 A b^2-6 a b B+2 a^2 C+25 b^2 C\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b^2 d}+\frac {(3 b B-a C) (a+b \sec (c+d x))^4 \tan (c+d x)}{15 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^4 \tan (c+d x)}{6 b d}\\ \end {align*}
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Mathematica [A] time = 3.03, size = 384, normalized size = 1.01 \[ -\frac {\sec ^5(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (-b \left (5 \sin (2 (c+d x)) \left (18 a^2 C+18 a b B+6 A b^2+5 b^2 C\right )+48 b (3 a C+b B) \sin (c+d x)+40 b^2 C \tan (c+d x)\right )-16 \sin (c+d x) \cos ^4(c+d x) \left (5 a^3 (3 A+2 C)+30 a^2 b B+6 a b^2 (5 A+4 C)+8 b^3 B\right )-15 \sin (c+d x) \cos ^3(c+d x) \left (8 a^3 B+6 a^2 b (4 A+3 C)+18 a b^2 B+b^3 (6 A+5 C)\right )-16 \sin (c+d x) \cos ^2(c+d x) \left (5 a^3 C+15 a^2 b B+3 a b^2 (5 A+4 C)+4 b^3 B\right )+15 \cos ^5(c+d x) \left (8 a^3 B+6 a^2 b (4 A+3 C)+18 a b^2 B+b^3 (6 A+5 C)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{120 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 342, normalized size = 0.90 \[ \frac {15 \, {\left (8 \, B a^{3} + 6 \, {\left (4 \, A + 3 \, C\right )} a^{2} b + 18 \, B a b^{2} + {\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, B a^{3} + 6 \, {\left (4 \, A + 3 \, C\right )} a^{2} b + 18 \, B a b^{2} + {\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, {\left (3 \, A + 2 \, C\right )} a^{3} + 30 \, B a^{2} b + 6 \, {\left (5 \, A + 4 \, C\right )} a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{5} + 15 \, {\left (8 \, B a^{3} + 6 \, {\left (4 \, A + 3 \, C\right )} a^{2} b + 18 \, B a b^{2} + {\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} + 40 \, C b^{3} + 16 \, {\left (5 \, C a^{3} + 15 \, B a^{2} b + 3 \, {\left (5 \, A + 4 \, C\right )} a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (18 \, C a^{2} b + 18 \, B a b^{2} + {\left (6 \, A + 5 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 48 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.43, size = 1370, normalized size = 3.60 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.74, size = 644, normalized size = 1.69 \[ \frac {3 B a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {2 A a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {3 A \,b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {5 b^{3} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {9 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {9 B a \,b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {4 b^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {5 b^{3} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {A \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {b^{3} C \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 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 \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 C a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{5 d}+\frac {3 C \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {A a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {5 b^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {8 b^{3} B \tan \left (d x +c \right )}{15 d}+\frac {9 C \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {8 C a \,b^{2} \tan \left (d x +c \right )}{5 d}+\frac {3 A \,a^{2} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {9 C \,a^{2} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {2 a^{2} b B \tan \left (d x +c \right )}{d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {b^{3} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {2 a^{3} C \tan \left (d x +c \right )}{3 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.37, size = 565, normalized size = 1.48 \[ \frac {160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} + 96 \, {\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 )} C a b^{2} + 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 b^{3} - 5 \, C b^{3} {\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 )} - 90 \, C 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 )} - 90 \, B a 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 )} - 30 \, A b^{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 )} - 120 \, B a^{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 )} - 360 \, A 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 )} + 480 \, A a^{3} \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 = 6.25, size = 769, normalized size = 2.02 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,b^3}{8}+\frac {B\,a^3}{2}+\frac {5\,C\,b^3}{16}+\frac {3\,A\,a^2\,b}{2}+\frac {9\,B\,a\,b^2}{8}+\frac {9\,C\,a^2\,b}{8}\right )}{\frac {3\,A\,b^3}{2}+2\,B\,a^3+\frac {5\,C\,b^3}{4}+6\,A\,a^2\,b+\frac {9\,B\,a\,b^2}{2}+\frac {9\,C\,a^2\,b}{2}}\right )\,\left (\frac {3\,A\,b^3}{4}+B\,a^3+\frac {5\,C\,b^3}{8}+3\,A\,a^2\,b+\frac {9\,B\,a\,b^2}{4}+\frac {9\,C\,a^2\,b}{4}\right )}{d}+\frac {\left (\frac {5\,A\,b^3}{4}-2\,A\,a^3+B\,a^3-2\,B\,b^3-2\,C\,a^3+\frac {11\,C\,b^3}{8}-6\,A\,a\,b^2+3\,A\,a^2\,b+\frac {15\,B\,a\,b^2}{4}-6\,B\,a^2\,b-6\,C\,a\,b^2+\frac {15\,C\,a^2\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (10\,A\,a^3-\frac {7\,A\,b^3}{4}-3\,B\,a^3+\frac {14\,B\,b^3}{3}+\frac {22\,C\,a^3}{3}+\frac {5\,C\,b^3}{24}+22\,A\,a\,b^2-9\,A\,a^2\,b-\frac {21\,B\,a\,b^2}{4}+22\,B\,a^2\,b+14\,C\,a\,b^2-\frac {21\,C\,a^2\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {A\,b^3}{2}-20\,A\,a^3+2\,B\,a^3-\frac {52\,B\,b^3}{5}-12\,C\,a^3+\frac {15\,C\,b^3}{4}-36\,A\,a\,b^2+6\,A\,a^2\,b+\frac {3\,B\,a\,b^2}{2}-36\,B\,a^2\,b-\frac {156\,C\,a\,b^2}{5}+\frac {3\,C\,a^2\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (20\,A\,a^3+\frac {A\,b^3}{2}+2\,B\,a^3+\frac {52\,B\,b^3}{5}+12\,C\,a^3+\frac {15\,C\,b^3}{4}+36\,A\,a\,b^2+6\,A\,a^2\,b+\frac {3\,B\,a\,b^2}{2}+36\,B\,a^2\,b+\frac {156\,C\,a\,b^2}{5}+\frac {3\,C\,a^2\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,C\,b^3}{24}-\frac {7\,A\,b^3}{4}-3\,B\,a^3-\frac {14\,B\,b^3}{3}-\frac {22\,C\,a^3}{3}-10\,A\,a^3-22\,A\,a\,b^2-9\,A\,a^2\,b-\frac {21\,B\,a\,b^2}{4}-22\,B\,a^2\,b-14\,C\,a\,b^2-\frac {21\,C\,a^2\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+\frac {5\,A\,b^3}{4}+B\,a^3+2\,B\,b^3+2\,C\,a^3+\frac {11\,C\,b^3}{8}+6\,A\,a\,b^2+3\,A\,a^2\,b+\frac {15\,B\,a\,b^2}{4}+6\,B\,a^2\,b+6\,C\,a\,b^2+\frac {15\,C\,a^2\,b}{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 )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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