Optimal. Leaf size=126 \[ \frac {1}{2} x \left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right )-\frac {b \sin (c+d x) (2 a A-2 a C-b B)}{d}+\frac {a (a B+2 A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d} \]
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Rubi [A] time = 0.32, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3047, 3033, 3023, 2735, 3770} \[ \frac {1}{2} x \left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right )-\frac {b \sin (c+d x) (2 a A-2 a C-b B)}{d}+\frac {a (a B+2 A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 2735
Rule 3023
Rule 3033
Rule 3047
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+\int (a+b \cos (c+d x)) \left (2 A b+a B+(b B+a C) \cos (c+d x)-b (2 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+\frac {1}{2} \int \left (2 a (2 A b+a B)+\left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \cos (c+d x)-2 b (2 a A-b B-2 a C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b (2 a A-b B-2 a C) \sin (c+d x)}{d}-\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+\frac {1}{2} \int \left (2 a (2 A b+a B)+\left (2 A b^2+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 (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) x-\frac {b (2 a A-b B-2 a C) \sin (c+d x)}{d}-\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+(a (2 A b+a B)) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} \left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) x+\frac {a (2 A b+a B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b (2 a A-b B-2 a C) \sin (c+d x)}{d}-\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}\\ \end {align*}
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Mathematica [A] time = 1.07, size = 155, normalized size = 1.23 \[ \frac {2 (c+d x) \left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right )+\tan (c+d x) \left (4 a^2 A+4 b (2 a C+b B) \cos (c+d x)+b^2 C \cos (2 (c+d x))+b^2 C\right )-4 a (a B+2 A b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a (a B+2 A b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 147, normalized size = 1.17 \[ \frac {{\left (2 \, C a^{2} + 4 \, B a b + {\left (2 \, A + C\right )} b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C b^{2} \cos \left (d x + c\right )^{2} + 2 \, A a^{2} + 2 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 229, normalized size = 1.82 \[ -\frac {\frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (2 \, C a^{2} + 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 171, normalized size = 1.36 \[ \frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+a^{2} C x +\frac {C \,a^{2} c}{d}+\frac {2 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+2 B x a b +\frac {2 B a b c}{d}+\frac {2 C a b \sin \left (d x +c \right )}{d}+A x \,b^{2}+\frac {A \,b^{2} c}{d}+\frac {b^{2} B \sin \left (d x +c \right )}{d}+\frac {b^{2} C \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {b^{2} C x}{2}+\frac {b^{2} C c}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 148, normalized size = 1.17 \[ \frac {4 \, {\left (d x + c\right )} C a^{2} + 8 \, {\left (d x + c\right )} B a b + 4 \, {\left (d x + c\right )} A b^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a b \sin \left (d x + c\right ) + 4 \, B b^{2} \sin \left (d x + c\right ) + 4 \, A a^{2} \tan \left (d x + c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.38, size = 274, normalized size = 2.17 \[ \frac {B\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {2\,C\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {C\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}-\frac {B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{d}-\frac {A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d}-\frac {B\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d} \]
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 \[ \int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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