Optimal. Leaf size=145 \[ \frac {\left (a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac {2 a b (2 A+3 C) \sin (c+d x)}{3 d}+\frac {a A b \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]
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Rubi [A] time = 0.38, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4095, 4074, 4047, 2637, 4045, 8} \[ \frac {\left (a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac {2 a b (2 A+3 C) \sin (c+d x)}{3 d}+\frac {a A b \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2637
Rule 4045
Rule 4047
Rule 4074
Rule 4095
Rubi steps
\begin {align*} \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (2 A b+a (3 A+4 C) \sec (c+d x)+b (A+4 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a A b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac {1}{12} \int \cos ^2(c+d x) \left (-3 \left (2 A b^2+a^2 (3 A+4 C)\right )-8 a b (2 A+3 C) \sec (c+d x)-3 b^2 (A+4 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a A b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac {1}{12} \int \cos ^2(c+d x) \left (-3 \left (2 A b^2+a^2 (3 A+4 C)\right )-3 b^2 (A+4 C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{3} (2 a b (2 A+3 C)) \int \cos (c+d x) \, dx\\ &=\frac {2 a b (2 A+3 C) \sin (c+d x)}{3 d}+\frac {\left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac {1}{8} \left (-4 b^2 (A+2 C)-a^2 (3 A+4 C)\right ) \int 1 \, dx\\ &=\frac {1}{8} \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac {2 a b (2 A+3 C) \sin (c+d x)}{3 d}+\frac {\left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.41, size = 104, normalized size = 0.72 \[ \frac {12 (c+d x) \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+24 \left (a^2 (A+C)+A b^2\right ) \sin (2 (c+d x))+3 a^2 A \sin (4 (c+d x))+48 a b (3 A+4 C) \sin (c+d x)+16 a A b \sin (3 (c+d x))}{96 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 104, normalized size = 0.72 \[ \frac {3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \, {\left (A + 2 \, C\right )} b^{2}\right )} d x + {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 16 \, A a b \cos \left (d x + c\right )^{2} + 16 \, {\left (2 \, A + 3 \, C\right )} a b + 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \, A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.23, size = 378, normalized size = 2.61 \[ \frac {3 \, {\left (3 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 8 \, C b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 144 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 144 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.17, size = 140, normalized size = 0.97 \[ \frac {a^{2} A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 A a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \sin \left (d x +c \right )+b^{2} C \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 130, normalized size = 0.90 \[ \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} + 96 \, {\left (d x + c\right )} C b^{2} + 192 \, C a b \sin \left (d x + c\right )}{96 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.54, size = 145, normalized size = 1.00 \[ \frac {3\,A\,a^2\,x}{8}+\frac {A\,b^2\,x}{2}+\frac {C\,a^2\,x}{2}+C\,b^2\,x+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {A\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,A\,a\,b\,\sin \left (c+d\,x\right )}{2\,d}+\frac {2\,C\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{6\,d} \]
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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