Optimal. Leaf size=116 \[ \frac {\left (2 a^2 B+6 a b C+3 b^2 B\right ) \tan (c+d x)}{3 d}+\frac {\left (a^2 C+2 a b B+2 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {a (a C+2 b B) \tan (c+d x) \sec (c+d x)}{2 d} \]
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Rubi [A] time = 0.36, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {3029, 2988, 3021, 2748, 3767, 8, 3770} \[ \frac {\left (2 a^2 B+6 a b C+3 b^2 B\right ) \tan (c+d x)}{3 d}+\frac {\left (a^2 C+2 a b B+2 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 B \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {a (a C+2 b B) \tan (c+d x) \sec (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2988
Rule 3021
Rule 3029
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\int (a+b \cos (c+d x))^2 (B+C \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac {a^2 B \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{3} \int \left (-3 a (2 b B+a C)-\left (2 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)-3 b^2 C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a (2 b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 B \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-2 \left (2 a^2 B+3 b^2 B+6 a b C\right )-3 \left (2 a b B+a^2 C+2 b^2 C\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a (2 b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 B \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{3} \left (-2 a^2 B-3 b^2 B-6 a b C\right ) \int \sec ^2(c+d x) \, dx-\frac {1}{2} \left (-2 a b B-a^2 C-2 b^2 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (2 a b B+a^2 C+2 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (2 b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 B \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {\left (2 a^2 B+3 b^2 B+6 a b C\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {\left (2 a b B+a^2 C+2 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\left (2 a^2 B+3 b^2 B+6 a b C\right ) \tan (c+d x)}{3 d}+\frac {a (2 b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 B \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 0.46, size = 92, normalized size = 0.79 \[ \frac {3 \left (a^2 C+2 a b B+2 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (2 \left (a^2 B \tan ^2(c+d x)+3 a^2 B+6 a b C+3 b^2 B\right )+3 a (a C+2 b B) \sec (c+d x)\right )}{6 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 150, normalized size = 1.29 \[ \frac {3 \, {\left (C a^{2} + 2 \, B a b + 2 \, C b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (C a^{2} + 2 \, B a b + 2 \, C b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B a^{2} + 2 \, {\left (2 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 294, normalized size = 2.53 \[ \frac {3 \, {\left (C a^{2} + 2 \, B a b + 2 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (C a^{2} + 2 \, B a b + 2 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.34, size = 174, normalized size = 1.50 \[ \frac {a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} B \tan \left (d x +c \right )}{3 d}+\frac {a^{2} B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {2 C a b \tan \left (d x +c \right )}{d}+\frac {B a b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} B \tan \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 172, normalized size = 1.48 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 3 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, B a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a b \tan \left (d x + c\right ) + 12 \, B b^{2} \tan \left (d x + c\right )}{12 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.09, size = 227, normalized size = 1.96 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {C\,a^2}{2}+B\,a\,b+C\,b^2\right )}{2\,C\,a^2+4\,B\,a\,b+4\,C\,b^2}\right )\,\left (C\,a^2+2\,B\,a\,b+2\,C\,b^2\right )}{d}-\frac {\left (2\,B\,a^2+2\,B\,b^2-C\,a^2-2\,B\,a\,b+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,B\,a^2}{3}-8\,C\,a\,b-4\,B\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a^2+2\,B\,b^2+C\,a^2+2\,B\,a\,b+4\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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