Optimal. Leaf size=168 \[ \frac {a \left (a^2 (A+2 C)+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {1}{2} b x \left (C \left (6 a^2+b^2\right )+2 A b^2\right )-\frac {3 a b^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {3 A b \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}-\frac {b^3 (4 A-C) \sin (c+d x) \cos (c+d x)}{2 d} \]
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Rubi [A] time = 0.58, antiderivative size = 168, 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 = {3048, 3047, 3033, 3023, 2735, 3770} \[ \frac {a \left (a^2 (A+2 C)+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {1}{2} b x \left (C \left (6 a^2+b^2\right )+2 A b^2\right )-\frac {3 a b^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {3 A b \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d}-\frac {b^3 (4 A-C) \sin (c+d x) \cos (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 2735
Rule 3023
Rule 3033
Rule 3047
Rule 3048
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x))^2 \left (3 A b+a (A+2 C) \cos (c+d x)-2 b (A-C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {3 A b (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x)) \left (6 A b^2+a^2 (A+2 C)-a b (A-4 C) \cos (c+d x)-2 b^2 (4 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b^3 (4 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 A b (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int \left (2 a \left (6 A b^2+a^2 (A+2 C)\right )+2 b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) \cos (c+d x)-6 a b^2 (3 A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {3 a b^2 (3 A-2 C) \sin (c+d x)}{2 d}-\frac {b^3 (4 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 A b (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int \left (2 a \left (6 A b^2+a^2 (A+2 C)\right )+2 b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) x-\frac {3 a b^2 (3 A-2 C) \sin (c+d x)}{2 d}-\frac {b^3 (4 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 A b (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a \left (6 A b^2+a^2 (A+2 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} b \left (2 A b^2+\left (6 a^2+b^2\right ) C\right ) x+\frac {a \left (6 A b^2+a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {3 a b^2 (3 A-2 C) \sin (c+d x)}{2 d}-\frac {b^3 (4 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 A b (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 1.51, size = 285, normalized size = 1.70 \[ \frac {\frac {a^3 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^3 A}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+2 b (c+d x) \left (C \left (6 a^2+b^2\right )+2 A b^2\right )-2 a \left (a^2 (A+2 C)+6 A b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a \left (a^2 (A+2 C)+6 A b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {12 a^2 A b \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {12 a^2 A b \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+12 a b^2 C \sin (c+d x)+b^3 C \sin (2 (c+d x))}{4 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.75, size = 171, normalized size = 1.02 \[ \frac {2 \, {\left (6 \, C a^{2} b + {\left (2 \, A + C\right )} b^{3}\right )} d x \cos \left (d x + c\right )^{2} + {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b^{3} \cos \left (d x + c\right )^{3} + 6 \, C a b^{2} \cos \left (d x + c\right )^{2} + 6 \, A a^{2} b \cos \left (d x + c\right ) + A a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.42, size = 385, normalized size = 2.29 \[ \frac {{\left (6 \, C a^{2} b + 2 \, A b^{3} + C b^{3}\right )} {\left (d x + c\right )} + {\left (A a^{3} + 2 \, C a^{3} + 6 \, A a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a^{3} + 2 \, C a^{3} + 6 \, A a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 196, normalized size = 1.17 \[ \frac {A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 A \,a^{2} b \tan \left (d x +c \right )}{d}+3 C \,a^{2} b x +\frac {3 C \,a^{2} b c}{d}+\frac {3 A a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 C a \,b^{2} \sin \left (d x +c \right )}{d}+A x \,b^{3}+\frac {A \,b^{3} c}{d}+\frac {b^{3} C \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {b^{3} C x}{2}+\frac {b^{3} C c}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.38, size = 179, normalized size = 1.07 \[ \frac {12 \, {\left (d x + c\right )} C a^{2} b + 4 \, {\left (d x + c\right )} A b^{3} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} - A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a b^{2} \sin \left (d x + c\right ) + 12 \, A a^{2} b \tan \left (d x + c\right )}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.99, size = 282, normalized size = 1.68 \[ \frac {2\,\left (\frac {A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+A\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+3\,A\,a\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+3\,C\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {C\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{8}+\frac {C\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{16}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{2}+\frac {3\,C\,a\,b^2\,\sin \left (c+d\,x\right )}{4}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,C\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]
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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