Optimal. Leaf size=80 \[ \frac {1}{2} x \left (a^2 A+4 a b B+2 A b^2\right )+\frac {a^2 A \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a (a B+2 A b) \sin (c+d x)}{d}+\frac {b^2 B \tanh ^{-1}(\sin (c+d x))}{d} \]
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Rubi [A] time = 0.17, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4024, 4047, 8, 4045, 3770} \[ \frac {1}{2} x \left (a^2 A+4 a b B+2 A b^2\right )+\frac {a^2 A \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a (a B+2 A b) \sin (c+d x)}{d}+\frac {b^2 B \tanh ^{-1}(\sin (c+d x))}{d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3770
Rule 4024
Rule 4045
Rule 4047
Rubi steps
\begin {align*} \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac {a^2 A \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} \int \cos (c+d x) \left (-2 a (2 A b+a B)+\left (A \left (-a^2-2 b^2\right )-4 a b B\right ) \sec (c+d x)-2 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 A \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} \int \cos (c+d x) \left (-2 a (2 A b+a B)-2 b^2 B \sec ^2(c+d x)\right ) \, dx-\frac {1}{2} \left (-a^2 A-2 A b^2-4 a b B\right ) \int 1 \, dx\\ &=\frac {1}{2} \left (a^2 A+2 A b^2+4 a b B\right ) x+\frac {a (2 A b+a B) \sin (c+d x)}{d}+\frac {a^2 A \cos (c+d x) \sin (c+d x)}{2 d}+\left (b^2 B\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} \left (a^2 A+2 A b^2+4 a b B\right ) x+\frac {b^2 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a (2 A b+a B) \sin (c+d x)}{d}+\frac {a^2 A \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 = 1.50 \[ \frac {2 (c+d x) \left (a^2 A+4 a b B+2 A b^2\right )+a^2 A \sin (2 (c+d x))+4 a (a B+2 A b) \sin (c+d x)-4 b^2 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 87, normalized size = 1.09 \[ \frac {B b^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B b^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} d x + {\left (A a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2} + 4 \, A a b\right )} \sin \left (d x + c\right )}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.45, size = 178, normalized size = 2.22 \[ \frac {2 \, B b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, B b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.71, size = 120, normalized size = 1.50 \[ \frac {a^{2} A \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a^{2} A x}{2}+\frac {A \,a^{2} c}{2 d}+\frac {B \,a^{2} \sin \left (d x +c \right )}{d}+\frac {2 A a b \sin \left (d x +c \right )}{d}+2 B x a b +\frac {2 B a b c}{d}+A x \,b^{2}+\frac {A \,b^{2} c}{d}+\frac {b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.88, size = 99, normalized size = 1.24 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 8 \, {\left (d x + c\right )} B a b + 4 \, {\left (d x + c\right )} A b^{2} + 2 \, B b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} \sin \left (d x + c\right ) + 8 \, A a b \sin \left (d x + c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.38, size = 169, normalized size = 2.11 \[ \frac {B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,A\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,B\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]
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 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \cos ^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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