Optimal. Leaf size=233 \[ \frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (20 a b B+5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {\left (5 A b^2+10 a b B+2 a^2 C+4 b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {b (5 b B+2 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {C \sec ^2(c+d x) (a+b \sec (c+d x))^2 \tan (c+d x)}{5 d} \]
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Rubi [A]
time = 0.42, antiderivative size = 281, normalized size of antiderivative = 1.21, 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 = {4177, 4167,
4087, 4082, 3872, 3855, 3852, 8} \begin {gather*} \frac {\left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \left (2 a^2 C-5 a b B+20 A b^2+16 b^2 C\right ) (a+b \sec (c+d x))^2}{60 b^2 d}-\frac {\tan (c+d x) \sec (c+d x) \left (-4 a^3 C+10 a^2 b B-2 a b^2 (20 A+13 C)-45 b^3 B\right )}{120 b d}-\frac {\tan (c+d x) \left (-2 a^4 C+5 a^3 b B-4 a^2 b^2 (5 A+3 C)-40 a b^3 B-4 b^4 (5 A+4 C)\right )}{30 b^2 d}+\frac {(5 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b^2 d}+\frac {C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 3852
Rule 3855
Rule 3872
Rule 4082
Rule 4087
Rule 4167
Rule 4177
Rubi steps
\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \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))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (a C+b (5 A+4 C) \sec (c+d x)+(5 b B-2 a C) \sec ^2(c+d x)\right ) \, dx}{5 b}\\ &=\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (b (15 b B-2 a C)+\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) \sec (c+d x)\right ) \, dx}{20 b^2}\\ &=\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (40 A b^2+35 a b B-2 a^2 C+32 b^2 C\right )-\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x)\right ) \, dx}{60 b^2}\\ &=-\frac {\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) \left (15 b^2 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right )-4 \left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{120 b^2}\\ &=-\frac {\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {1}{8} \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \int \sec (c+d x) \, dx-\frac {\left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \int \sec ^2(c+d x) \, dx}{30 b^2}\\ &=\frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 b^2 d}\\ &=\frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {\left (5 a^3 b B-40 a b^3 B-2 a^4 C-4 a^2 b^2 (5 A+3 C)-4 b^4 (5 A+4 C)\right ) \tan (c+d x)}{30 b^2 d}-\frac {\left (10 a^2 b B-45 b^3 B-4 a^3 C-2 a b^2 (20 A+13 C)\right ) \sec (c+d x) \tan (c+d x)}{120 b d}+\frac {\left (20 A b^2-5 a b B+2 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b^2 d}+\frac {(5 b B-2 a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}\\ \end {align*}
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Mathematica [A]
time = 2.26, size = 371, normalized size = 1.59 \begin {gather*} \frac {\left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \sec ^5(c+d x) \left (-120 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+2 \left (180 a^2 A+200 A b^2+400 a b B+200 a^2 C+256 b^2 C+15 \left (24 a A b+12 a^2 B+17 b^2 B+34 a b C\right ) \cos (c+d x)+48 \left (10 a b B+5 a^2 (A+C)+b^2 (5 A+4 C)\right ) \cos (2 (c+d x))+120 a A b \cos (3 (c+d x))+60 a^2 B \cos (3 (c+d x))+45 b^2 B \cos (3 (c+d x))+90 a b C \cos (3 (c+d x))+60 a^2 A \cos (4 (c+d x))+40 A b^2 \cos (4 (c+d x))+80 a b B \cos (4 (c+d x))+40 a^2 C \cos (4 (c+d x))+32 b^2 C \cos (4 (c+d x))\right ) \sin (c+d x)\right )}{480 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.12, size = 292, normalized size = 1.25 Too large to display
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 357, normalized size = 1.53 \begin {gather*} \frac {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} + 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 )} C b^{2} - 30 \, C 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 )} - 15 \, B 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 )} - 60 \, B a^{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 )} - 120 \, A 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 )} + 240 \, A a^{2} \tan \left (d x + c\right )}{240 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 4.58, size = 243, normalized size = 1.04 \begin {gather*} \frac {15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, 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 (5 \, {\left (3 \, A + 2 \, C\right )} a^{2} + 20 \, B a b + 2 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, C b^{2} + 8 \, {\left (5 \, C a^{2} + 10 \, B a b + {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end {gather*}
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 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 766 vs.
\(2 (221) = 442\).
time = 0.53, size = 766, normalized size = 3.29 \begin {gather*} \frac {15 \, {\left (4 \, B a^{2} + 8 \, A a b + 6 \, C a b + 3 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, B a^{2} + 8 \, A a b + 6 \, C a b + 3 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 240 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 150 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 480 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 320 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 240 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 640 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 320 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 720 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 480 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 640 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 240 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 150 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 7.48, size = 455, normalized size = 1.95 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^2}{2}+\frac {3\,B\,b^2}{8}+A\,a\,b+\frac {3\,C\,a\,b}{4}\right )}{2\,B\,a^2+\frac {3\,B\,b^2}{2}+4\,A\,a\,b+3\,C\,a\,b}\right )\,\left (B\,a^2+\frac {3\,B\,b^2}{4}+2\,A\,a\,b+\frac {3\,C\,a\,b}{2}\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-B\,a^2-\frac {5\,B\,b^2}{4}+2\,C\,a^2+2\,C\,b^2-2\,A\,a\,b+4\,B\,a\,b-\frac {5\,C\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,B\,a^2-\frac {16\,A\,b^2}{3}-8\,A\,a^2+\frac {B\,b^2}{2}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}+4\,A\,a\,b-\frac {32\,B\,a\,b}{3}+C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,A\,a^2+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+\frac {116\,C\,b^2}{15}+\frac {40\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,A\,a^2-\frac {16\,A\,b^2}{3}-2\,B\,a^2-\frac {B\,b^2}{2}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-4\,A\,a\,b-\frac {32\,B\,a\,b}{3}-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+B\,a^2+\frac {5\,B\,b^2}{4}+2\,C\,a^2+2\,C\,b^2+2\,A\,a\,b+4\,B\,a\,b+\frac {5\,C\,a\,b}{2}\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 )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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