Optimal. Leaf size=222 \[ \frac {a^2 (10 A+9 B+8 C) \tan ^3(c+d x)}{15 d}+\frac {a^2 (10 A+9 B+8 C) \tan (c+d x)}{5 d}+\frac {a^2 (14 A+12 B+11 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^2 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac {a^2 (14 A+12 B+11 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {(3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{15 d}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]
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Rubi [A] time = 0.43, antiderivative size = 222, 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 = {4088, 4018, 3997, 3787, 3768, 3770, 3767} \[ \frac {a^2 (10 A+9 B+8 C) \tan ^3(c+d x)}{15 d}+\frac {a^2 (10 A+9 B+8 C) \tan (c+d x)}{5 d}+\frac {a^2 (14 A+12 B+11 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^2 (10 A+12 B+9 C) \tan (c+d x) \sec ^3(c+d x)}{40 d}+\frac {a^2 (14 A+12 B+11 C) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {(3 B+C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{15 d}+\frac {C \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]
Antiderivative was successfully verified.
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Rule 3767
Rule 3768
Rule 3770
Rule 3787
Rule 3997
Rule 4018
Rule 4088
Rubi steps
\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {\int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (3 a (2 A+C)+2 a (3 B+C) \sec (c+d x)) \, dx}{6 a}\\ &=\frac {C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {\int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (3 a^2 (10 A+6 B+7 C)+3 a^2 (10 A+12 B+9 C) \sec (c+d x)\right ) \, dx}{30 a}\\ &=\frac {a^2 (10 A+12 B+9 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {\int \sec ^3(c+d x) \left (15 a^3 (14 A+12 B+11 C)+24 a^3 (10 A+9 B+8 C) \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac {a^2 (10 A+12 B+9 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {1}{5} \left (a^2 (10 A+9 B+8 C)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{8} \left (a^2 (14 A+12 B+11 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {a^2 (14 A+12 B+11 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^2 (10 A+12 B+9 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {1}{16} \left (a^2 (14 A+12 B+11 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (10 A+9 B+8 C)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {a^2 (14 A+12 B+11 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^2 (10 A+9 B+8 C) \tan (c+d x)}{5 d}+\frac {a^2 (14 A+12 B+11 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^2 (10 A+12 B+9 C) \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {C \sec ^3(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {(3 B+C) \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{15 d}+\frac {a^2 (10 A+9 B+8 C) \tan ^3(c+d x)}{15 d}\\ \end {align*}
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Mathematica [A] time = 3.60, size = 359, normalized size = 1.62 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (15 (14 A+12 B+11 C) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) \cos ^5(c+d x) (15 \sin (c) (14 A+12 B+11 C)+32 (10 A+9 B+8 C) \sin (d x))-\sec (c) \cos ^4(c+d x) (16 \sin (c) (10 A+9 B+8 C)+15 (14 A+12 B+11 C) \sin (d x))-2 \sec (c) \cos ^3(c+d x) (5 \sin (c) (6 A+12 B+11 C)+8 (10 A+9 B+8 C) \sin (d x))-2 \sec (c) \cos ^2(c+d x) (5 (6 A+12 B+11 C) \sin (d x)+24 (B+2 C) \sin (c))-8 \sec (c) \cos (c+d x) (6 (B+2 C) \sin (d x)+5 C \sin (c))-40 C \sec (c) \sin (d x)\right )}{480 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 203, normalized size = 0.91 \[ \frac {15 \, {\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, {\left (10 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 15 \, {\left (14 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 16 \, {\left (10 \, A + 9 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (6 \, A + 12 \, B + 11 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 40 \, C a^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.34, size = 392, normalized size = 1.77 \[ \frac {15 \, {\left (14 \, A a^{2} + 12 \, B a^{2} + 11 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (14 \, A a^{2} + 12 \, B a^{2} + 11 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (210 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 180 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 165 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1190 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1020 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 935 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2580 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2568 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1986 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3180 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2808 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3006 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2330 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1860 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1305 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 750 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 780 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 795 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.67, size = 386, normalized size = 1.74 \[ \frac {7 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {7 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {6 a^{2} B \tan \left (d x +c \right )}{5 d}+\frac {3 a^{2} B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {11 a^{2} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {11 a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {11 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {4 a^{2} A \tan \left (d x +c \right )}{3 d}+\frac {2 a^{2} A \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {16 a^{2} C \tan \left (d x +c \right )}{15 d}+\frac {2 a^{2} C \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {8 a^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{2} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 477, normalized size = 2.15 \[ \frac {320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{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 a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 64 \, {\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^{2} - 5 \, C a^{2} {\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 )} - 30 \, A 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 )} - 60 \, 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 \, C 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 )} - 120 \, A 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 )}}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.59, size = 337, normalized size = 1.52 \[ \frac {a^2\,\mathrm {atanh}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (14\,A+12\,B+11\,C\right )}{4\,\left (\frac {7\,A\,a^2}{2}+3\,B\,a^2+\frac {11\,C\,a^2}{4}\right )}\right )\,\left (14\,A+12\,B+11\,C\right )}{8\,d}-\frac {\left (\frac {7\,A\,a^2}{4}+\frac {3\,B\,a^2}{2}+\frac {11\,C\,a^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (-\frac {119\,A\,a^2}{12}-\frac {17\,B\,a^2}{2}-\frac {187\,C\,a^2}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {43\,A\,a^2}{2}+\frac {107\,B\,a^2}{5}+\frac {331\,C\,a^2}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-\frac {53\,A\,a^2}{2}-\frac {117\,B\,a^2}{5}-\frac {501\,C\,a^2}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {233\,A\,a^2}{12}+\frac {31\,B\,a^2}{2}+\frac {87\,C\,a^2}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-\frac {25\,A\,a^2}{4}-\frac {13\,B\,a^2}{2}-\frac {53\,C\,a^2}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
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 \[ a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{7}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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