Optimal. Leaf size=226 \[ \frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{30 b^2 d}+\frac {a \left (2 a^2 C+20 A b^2+13 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{60 b d}+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac {a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{10 b^2 d}+\frac {C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \]
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Rubi [A] time = 0.50, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4093, 4082, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac {\left (2 a^2 b^2 (5 A+3 C)+a^4 C+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{30 b^2 d}+\frac {a \left (2 a^2 C+20 A b^2+13 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{60 b d}+\frac {a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{10 b^2 d}+\frac {C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3787
Rule 3997
Rule 4002
Rule 4082
Rule 4093
Rubi steps
\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+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)-2 a C \sec ^2(c+d x)\right ) \, dx}{5 b}\\ &=-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 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 (-2 a b C+2 \left (a^2 C+2 b^2 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{20 b^2}\\ &=\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 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 (2 b \left (20 A b^2-a^2 C+16 b^2 C\right )+2 a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x)\right ) \, dx}{60 b^2}\\ &=\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 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 (30 a b^3 (4 A+3 C)+8 \left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{120 b^2}\\ &=\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 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}{4} (a b (4 A+3 C)) \int \sec (c+d x) \, dx+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \int \sec ^2(c+d x) \, dx}{15 b^2}\\ &=\frac {a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 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 (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 b^2 d}\\ &=\frac {a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 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 = 2.52, size = 275, normalized size = 1.22 \[ -\frac {\sec ^5(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (60 a b (4 A+3 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sin (c+d x) \left (24 \left (5 a^2 (A+C)+b^2 (5 A+4 C)\right ) \cos (2 (c+d x))+30 a^2 A \cos (4 (c+d x))+90 a^2 A+20 a^2 C \cos (4 (c+d x))+100 a^2 C+15 a b (12 A+17 C) \cos (c+d x)+60 a A b \cos (3 (c+d x))+45 a b C \cos (3 (c+d x))+20 A b^2 \cos (4 (c+d x))+100 A b^2+16 b^2 C \cos (4 (c+d x))+128 b^2 C\right )\right )}{120 d (A \cos (2 (c+d x))+A+2 C)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 180, normalized size = 0.80 \[ \frac {15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, {\left (3 \, A + 2 \, C\right )} a^{2} + 2 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, C a b \cos \left (d x + c\right ) + 12 \, C b^{2} + 4 \, {\left (5 \, C a^{2} + {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.29, size = 532, normalized size = 2.35 \[ \frac {15 \, {\left (4 \, A a b + 3 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a b + 3 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 240 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 232 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 240 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, 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}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.85, size = 257, normalized size = 1.14 \[ \frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {2 a^{2} C \tan \left (d x +c \right )}{3 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C a b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 a b C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {2 A \,b^{2} \tan \left (d x +c \right )}{3 d}+\frac {A \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {8 b^{2} C \tan \left (d x +c \right )}{15 d}+\frac {b^{2} C \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 b^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 216, normalized size = 0.96 \[ \frac {40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} + 8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{2} - 15 \, C a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, A a^{2} \tan \left (d x + c\right )}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.55, size = 322, normalized size = 1.42 \[ \frac {a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{2\,d}-\frac {\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2-2\,A\,a\,b-\frac {5\,C\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (4\,A\,a\,b-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-8\,A\,a^2+C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,A\,a^2+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+\frac {116\,C\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,A\,a^2-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-4\,A\,a\,b-C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2+2\,A\,a\,b+\frac {5\,C\,a\,b}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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 \[ \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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