Optimal. Leaf size=169 \[ \frac {a^2 (10 A+9 B) \tan ^3(c+d x)}{15 d}+\frac {a^2 (10 A+9 B) \tan (c+d x)}{5 d}+\frac {a^2 (7 A+6 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 (7 A+6 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {B \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]
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Rubi [A] time = 0.24, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4018, 3997, 3787, 3768, 3770, 3767} \[ \frac {a^2 (10 A+9 B) \tan ^3(c+d x)}{15 d}+\frac {a^2 (10 A+9 B) \tan (c+d x)}{5 d}+\frac {a^2 (7 A+6 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 (7 A+6 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {B \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]
Antiderivative was successfully verified.
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Rule 3767
Rule 3768
Rule 3770
Rule 3787
Rule 3997
Rule 4018
Rubi steps
\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) (a+a \sec (c+d x)) (a (5 A+3 B)+a (5 A+6 B) \sec (c+d x)) \, dx\\ &=\frac {a^2 (5 A+6 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{20} \int \sec ^3(c+d x) \left (5 a^2 (7 A+6 B)+4 a^2 (10 A+9 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^2 (5 A+6 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{4} \left (a^2 (7 A+6 B)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (a^2 (10 A+9 B)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {a^2 (7 A+6 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 A+6 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{8} \left (a^2 (7 A+6 B)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (10 A+9 B)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {a^2 (7 A+6 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (10 A+9 B) \tan (c+d x)}{5 d}+\frac {a^2 (7 A+6 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 A+6 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {a^2 (10 A+9 B) \tan ^3(c+d x)}{15 d}\\ \end {align*}
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Mathematica [A] time = 1.40, size = 280, normalized size = 1.66 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (7 A+6 B) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (2 A+B) \sin (2 c+d x)+80 (14 A+15 B) \sin (d x)+330 A \sin (c+2 d x)+330 A \sin (3 c+2 d x)+800 A \sin (2 c+3 d x)+105 A \sin (3 c+4 d x)+105 A \sin (5 c+4 d x)+160 A \sin (4 c+5 d x)+420 B \sin (c+2 d x)+420 B \sin (3 c+2 d x)+720 B \sin (2 c+3 d x)+90 B \sin (3 c+4 d x)+90 B \sin (5 c+4 d x)+144 B \sin (4 c+5 d x))\right )}{7680 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 165, normalized size = 0.98 \[ \frac {15 \, {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (10 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (10 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 24 \, B a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.73, size = 246, normalized size = 1.46 \[ \frac {15 \, {\left (7 \, A a^{2} + 6 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (7 \, A a^{2} + 6 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 90 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 490 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 420 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 800 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 864 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 790 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 540 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 375 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 390 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.40, size = 235, normalized size = 1.39 \[ \frac {7 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {7 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {6 a^{2} B \tan \left (d x +c \right )}{5 d}+\frac {3 a^{2} B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {4 a^{2} A \tan \left (d x +c \right )}{3 d}+\frac {2 a^{2} A \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{2} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 278, normalized size = 1.64 \[ \frac {160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 15 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.61, size = 224, normalized size = 1.33 \[ \frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (7\,A+6\,B\right )}{4\,d}-\frac {\left (\frac {7\,A\,a^2}{4}+\frac {3\,B\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {49\,A\,a^2}{6}-7\,B\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {40\,A\,a^2}{3}+\frac {72\,B\,a^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {79\,A\,a^2}{6}-9\,B\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {25\,A\,a^2}{4}+\frac {13\,B\,a^2}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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 \[ a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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