Optimal. Leaf size=141 \[ \frac {\tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{3 d}+\frac {\left (a^2 B+2 a b (A+2 C)+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{6 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^2 C x \]
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Rubi [A] time = 0.37, antiderivative size = 141, 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, 3031, 3021, 2735, 3770} \[ \frac {\tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{3 d}+\frac {\left (a^2 B+2 a b (A+2 C)+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{6 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^2 C x \]
Antiderivative was successfully verified.
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Rule 2735
Rule 3021
Rule 3031
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 ^4(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+b \cos (c+d x)) \left (2 A b+3 a B+(2 a A+3 b B+3 a C) \cos (c+d x)+3 b C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-2 \left (2 A b^2+6 a b B+a^2 (2 A+3 C)\right )-3 \left (a^2 B+2 b^2 B+2 a b (A+2 C)\right ) \cos (c+d x)-6 b^2 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {\left (2 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac {a (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-3 \left (a^2 B+2 b^2 B+2 a b (A+2 C)\right )-6 b^2 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^2 C x+\frac {\left (2 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac {a (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{2} \left (-a^2 B-2 b^2 B-2 a b (A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=b^2 C x+\frac {\left (a^2 B+2 b^2 B+2 a b (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\left (2 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac {a (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 0.65, size = 104, normalized size = 0.74 \[ \frac {3 \left (a^2 B+2 a b (A+2 C)+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+3 \tan (c+d x) \left (2 a^2 (A+C)+a (a B+2 A b) \sec (c+d x)+4 a b B+2 A b^2\right )+2 a^2 A \tan ^3(c+d x)+6 b^2 C d x}{6 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 179, normalized size = 1.27 \[ \frac {12 \, C b^{2} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (B a^{2} + 2 \, {\left (A + 2 \, C\right )} a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a^{2} + 2 \, {\left (A + 2 \, C\right )} a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{2} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.52, size = 364, normalized size = 2.58 \[ \frac {6 \, {\left (d x + c\right )} C b^{2} + 3 \, {\left (B a^{2} + 2 \, A a b + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (B a^{2} + 2 \, A a b + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.37, size = 225, normalized size = 1.60 \[ \frac {2 a^{2} A \tan \left (d x +c \right )}{3 d}+\frac {a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a^{2} C \tan \left (d x +c \right )}{d}+\frac {a A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 B a b \tan \left (d x +c \right )}{d}+\frac {2 C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{2} \tan \left (d x +c \right )}{d}+\frac {b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+b^{2} C x +\frac {b^{2} C c}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 221, normalized size = 1.57 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 12 \, {\left (d x + c\right )} C b^{2} - 3 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a^{2} \tan \left (d x + c\right ) + 24 \, B a b \tan \left (d x + c\right ) + 12 \, A b^{2} \tan \left (d x + c\right )}{12 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.23, size = 512, normalized size = 3.63 \[ \frac {\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{6}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {A\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2}+\frac {A\,b^2\,\sin \left (c+d\,x\right )}{4}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{4}+\frac {A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,B\,a^2\,\cos \left (c+d\,x\right )\,\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 )}{4}+\frac {3\,B\,b^2\,\cos \left (c+d\,x\right )\,\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 )}{2}+\frac {3\,C\,b^2\,\cos \left (c+d\,x\right )\,\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 )}{2}+\frac {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 )\,\cos \left (3\,c+3\,d\,x\right )}{4}+\frac {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 )\,\cos \left (3\,c+3\,d\,x\right )}{2}+\frac {B\,a\,b\,\sin \left (c+d\,x\right )}{2}+\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 )\,\cos \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,A\,a\,b\,\cos \left (c+d\,x\right )\,\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 )}{2}+3\,C\,a\,b\,\cos \left (c+d\,x\right )\,\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 )+\frac {A\,a\,b\,\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 )\,\cos \left (3\,c+3\,d\,x\right )}{2}+C\,a\,b\,\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 )\,\cos \left (3\,c+3\,d\,x\right )}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\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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