Optimal. Leaf size=116 \[ \frac {\left (2 a A b+a^2 B+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\left (2 a^2 A+3 A b^2+6 a b B\right ) \tan (c+d x)}{3 d}+\frac {a (2 A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
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Rubi [A]
time = 0.18, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3067, 3100,
2827, 3852, 8, 3855} \begin {gather*} \frac {\left (2 a^2 A+6 a b B+3 A b^2\right ) \tan (c+d x)}{3 d}+\frac {\left (a^2 B+2 a A b+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {a (a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 2827
Rule 3067
Rule 3100
Rule 3852
Rule 3855
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx &=\frac {a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{3} \int \left (-3 a (2 A b+a B)-\left (2 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)-3 b^2 B \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a (2 A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-2 \left (2 a^2 A+3 A b^2+6 a b B\right )-3 \left (2 a A b+a^2 B+2 b^2 B\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a (2 A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{3} \left (-2 a^2 A-3 A b^2-6 a b B\right ) \int \sec ^2(c+d x) \, dx-\frac {1}{2} \left (-2 a A b-a^2 B-2 b^2 B\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (2 a A b+a^2 B+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (2 A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {\left (2 a^2 A+3 A b^2+6 a b B\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {\left (2 a A b+a^2 B+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\left (2 a^2 A+3 A b^2+6 a b B\right ) \tan (c+d x)}{3 d}+\frac {a (2 A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A]
time = 0.52, size = 92, normalized size = 0.79 \begin {gather*} \frac {3 \left (2 a A b+a^2 B+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 a (2 A b+a B) \sec (c+d x)+2 \left (3 a^2 A+3 A b^2+6 a b B+a^2 A \tan ^2(c+d x)\right )\right )}{6 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.24, size = 143, normalized size = 1.23
method | result | size |
derivativedivides | \(\frac {-a^{2} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 A a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 B a b \tan \left (d x +c \right )+A \,b^{2} \tan \left (d x +c \right )+B \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(143\) |
default | \(\frac {-a^{2} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 A a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 B a b \tan \left (d x +c \right )+A \,b^{2} \tan \left (d x +c \right )+B \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(143\) |
risch | \(-\frac {i \left (6 A a b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-6 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-12 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-6 A a b \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-4 a^{2} A -6 A \,b^{2}-12 B a b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}\) | \(298\) |
norman | \(\frac {-\frac {2 \left (2 a^{2} A -2 A a b -2 A \,b^{2}-B \,a^{2}-4 B a b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 a^{2} A -2 A a b +2 A \,b^{2}-B \,a^{2}+4 B a b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (2 a^{2} A +2 A a b -2 A \,b^{2}+B \,a^{2}-4 B a b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 a^{2} A +2 A a b +2 A \,b^{2}+B \,a^{2}+4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (14 a^{2} A -18 A a b +6 A \,b^{2}-9 B \,a^{2}+12 B a b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (14 a^{2} A +18 A a b +6 A \,b^{2}+9 B \,a^{2}+12 B a b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {\left (2 A a b +B \,a^{2}+2 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 A a b +B \,a^{2}+2 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(368\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.27, size = 172, normalized size = 1.48 \begin {gather*} \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{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 )} + 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 )} + 24 \, B a b \tan \left (d x + c\right ) + 12 \, A b^{2} \tan \left (d x + c\right )}{12 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 150, normalized size = 1.29 \begin {gather*} \frac {3 \, {\left (B a^{2} + 2 \, A 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 \, A 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 (2 \, A 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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 294 vs.
\(2 (108) = 216\).
time = 0.45, size = 294, normalized size = 2.53 \begin {gather*} \frac {3 \, {\left (B a^{2} + 2 \, A 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 + 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 \, 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} - 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 \, 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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.66, size = 227, normalized size = 1.96 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^2}{2}+A\,a\,b+B\,b^2\right )}{2\,B\,a^2+4\,A\,a\,b+4\,B\,b^2}\right )\,\left (B\,a^2+2\,A\,a\,b+2\,B\,b^2\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-B\,a^2-2\,A\,a\,b+4\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A\,a^2}{3}-8\,B\,a\,b-4\,A\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+B\,a^2+2\,A\,a\,b+4\,B\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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