Optimal. Leaf size=86 \[ \frac {1}{2} x \left (2 a^2 C+4 a b B+b^2 C\right )+\frac {a^2 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b (3 a C+2 b B) \sin (c+d x)}{2 d}+\frac {b C \sin (c+d x) (a+b \cos (c+d x))}{2 d} \]
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Rubi [A] time = 0.24, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3029, 2990, 3023, 2735, 3770} \[ \frac {1}{2} x \left (2 a^2 C+4 a b B+b^2 C\right )+\frac {a^2 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b (3 a C+2 b B) \sin (c+d x)}{2 d}+\frac {b C \sin (c+d x) (a+b \cos (c+d x))}{2 d} \]
Antiderivative was successfully verified.
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Rule 2735
Rule 2990
Rule 3023
Rule 3029
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 ^2(c+d x) \, dx &=\int (a+b \cos (c+d x))^2 (B+C \cos (c+d x)) \sec (c+d x) \, dx\\ &=\frac {b C (a+b \cos (c+d x)) \sin (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a^2 B+\left (4 a b B+2 a^2 C+b^2 C\right ) \cos (c+d x)+b (2 b B+3 a C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {b (2 b B+3 a C) \sin (c+d x)}{2 d}+\frac {b C (a+b \cos (c+d x)) \sin (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a^2 B+\left (4 a b B+2 a^2 C+b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} \left (4 a b B+2 a^2 C+b^2 C\right ) x+\frac {b (2 b B+3 a C) \sin (c+d x)}{2 d}+\frac {b C (a+b \cos (c+d x)) \sin (c+d x)}{2 d}+\left (a^2 B\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} \left (4 a b B+2 a^2 C+b^2 C\right ) x+\frac {a^2 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b (2 b B+3 a C) \sin (c+d x)}{2 d}+\frac {b C (a+b \cos (c+d x)) \sin (c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.23, size = 120, normalized size = 1.40 \[ \frac {2 (c+d x) \left (2 a^2 C+4 a b B+b^2 C\right )-4 a^2 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 b (2 a C+b B) \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.73, size = 87, normalized size = 1.01 \[ \frac {B a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} d x + {\left (C b^{2} \cos \left (d x + c\right ) + 4 \, C a b + 2 \, B b^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.28, size = 178, normalized size = 2.07 \[ \frac {2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.21, size = 120, normalized size = 1.40 \[ a^{2} C x +\frac {C \,a^{2} c}{d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 C a b \sin \left (d x +c \right )}{d}+2 B x a b +\frac {2 B a b c}{d}+\frac {b^{2} C \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {b^{2} C x}{2}+\frac {b^{2} C c}{2 d}+\frac {b^{2} B \sin \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 99, normalized size = 1.15 \[ \frac {4 \, {\left (d x + c\right )} C a^{2} + 8 \, {\left (d x + c\right )} B a b + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a b \sin \left (d x + c\right ) + 4 \, B b^{2} \sin \left (d x + c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.96, size = 169, normalized size = 1.97 \[ \frac {B\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,C\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,B\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]
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 \[ \int \left (B + C \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{2} \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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