Optimal. Leaf size=200 \[ \frac {\left (8 a b B+4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 b B+4 b^3 B-a^3 C+4 a b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 b d}+\frac {\left (12 A b^2+8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d} \]
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Rubi [A]
time = 0.25, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {4167, 4087,
4082, 3872, 3855, 3852, 8} \begin {gather*} \frac {\left (4 a^2 (2 A+C)+8 a b B+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec (c+d x) \left (-2 a^2 C+8 a b B+12 A b^2+9 b^2 C\right )}{24 d}+\frac {\tan (c+d x) \left (a^3 (-C)+4 a^2 b B+4 a b^2 (3 A+2 C)+4 b^3 B\right )}{6 b d}+\frac {(4 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 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
Rubi steps
\begin {align*} \int \sec (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 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 (b (4 A+3 C)+(4 b B-a C) \sec (c+d x)) \, dx}{4 b}\\ &=\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (b (12 a A+8 b B+7 a C)+\left (12 A b^2+8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x)\right ) \, dx}{12 b}\\ &=\frac {\left (12 A b^2+8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) \left (3 b \left (8 a b B+4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+4 \left (4 a^2 b B+4 b^3 B-a^3 C+4 a b^2 (3 A+2 C)\right ) \sec (c+d x)\right ) \, dx}{24 b}\\ &=\frac {\left (12 A b^2+8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\left (4 a^2 b B+4 b^3 B-a^3 C+4 a b^2 (3 A+2 C)\right ) \int \sec ^2(c+d x) \, dx}{6 b}+\frac {1}{8} \left (8 a b B+4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (8 a b B+4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (12 A b^2+8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}-\frac {\left (4 a^2 b B+4 b^3 B-a^3 C+4 a b^2 (3 A+2 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 b d}\\ &=\frac {\left (8 a b B+4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 b B+4 b^3 B-a^3 C+4 a b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 b d}+\frac {\left (12 A b^2+8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}\\ \end {align*}
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Mathematica [A]
time = 1.67, size = 300, normalized size = 1.50 \begin {gather*} -\frac {\left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \sec ^4(c+d x) \left (12 \left (8 a b B+4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \cos ^4(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 (12 A b^2+24 a b B+12 a^2 C+21 b^2 C+4 \left (9 a^2 B+10 b^2 B+2 a b (9 A+10 C)\right ) \cos (c+d x)+3 \left (4 A b^2+8 a b B+4 a^2 C+3 b^2 C\right ) \cos (2 (c+d x))+24 a A b \cos (3 (c+d x))+12 a^2 B \cos (3 (c+d x))+8 b^2 B \cos (3 (c+d x))+16 a b C \cos (3 (c+d x))\right ) \sin (c+d x)\right )}{48 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 = 0.10, size = 251, normalized size = 1.26 Too large to display
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 306, normalized size = 1.53 \begin {gather*} \frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{2} - 3 \, C 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 )} - 12 \, C 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 )} - 24 \, B 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 )} - 12 \, 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 )} + 48 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 48 \, B a^{2} \tan \left (d x + c\right ) + 96 \, A a b \tan \left (d x + c\right )}{48 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 3.21, size = 209, normalized size = 1.04 \begin {gather*} \frac {3 \, {\left (4 \, {\left (2 \, A + C\right )} a^{2} + 8 \, B a b + {\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, {\left (2 \, A + C\right )} a^{2} + 8 \, B a b + {\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 2 \, C\right )} a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, C b^{2} + 3 \, {\left (4 \, C a^{2} + 8 \, B a b + {\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \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 {\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. 630 vs.
\(2 (188) = 376\).
time = 0.51, size = 630, normalized size = 3.15 \begin {gather*} \frac {3 \, {\left (8 \, A a^{2} + 4 \, C a^{2} + 8 \, B a b + 4 \, A b^{2} + 3 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, A a^{2} + 4 \, C a^{2} + 8 \, B a b + 4 \, A b^{2} + 3 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 144 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 144 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, 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 )}^{4}}}{24 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 8.19, size = 389, normalized size = 1.94 \begin {gather*} \frac {\left (A\,b^2-2\,B\,a^2-2\,B\,b^2+C\,a^2+\frac {5\,C\,b^2}{4}-4\,A\,a\,b+2\,B\,a\,b-4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (6\,B\,a^2-A\,b^2+\frac {10\,B\,b^2}{3}-C\,a^2+\frac {3\,C\,b^2}{4}+12\,A\,a\,b-2\,B\,a\,b+\frac {20\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,C\,b^2}{4}-6\,B\,a^2-\frac {10\,B\,b^2}{3}-C\,a^2-A\,b^2-12\,A\,a\,b-2\,B\,a\,b-\frac {20\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,b^2+2\,B\,a^2+2\,B\,b^2+C\,a^2+\frac {5\,C\,b^2}{4}+4\,A\,a\,b+2\,B\,a\,b+4\,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 )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a^2+\frac {A\,b^2}{2}+\frac {C\,a^2}{2}+\frac {3\,C\,b^2}{8}+B\,a\,b\right )}{4\,A\,a^2+2\,A\,b^2+2\,C\,a^2+\frac {3\,C\,b^2}{2}+4\,B\,a\,b}\right )\,\left (2\,A\,a^2+A\,b^2+C\,a^2+\frac {3\,C\,b^2}{4}+2\,B\,a\,b\right )}{d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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