Optimal. Leaf size=86 \[ \frac {a (2 A+3 C) \tan (c+d x)}{3 d}+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A b \tan (c+d x) \sec (c+d x)}{2 d} \]
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Rubi [A] time = 0.17, antiderivative size = 86, 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 = {3032, 3021, 2748, 3767, 8, 3770} \[ \frac {a (2 A+3 C) \tan (c+d x)}{3 d}+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A b \tan (c+d x) \sec (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 3021
Rule 3032
Rule 3767
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int \left (3 A b+a (2 A+3 C) \cos (c+d x)+3 b C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {A b \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int (2 a (2 A+3 C)+3 b (A+2 C) \cos (c+d x)) \sec ^2(c+d x) \, dx\\ &=\frac {A b \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} (b (A+2 C)) \int \sec (c+d x) \, dx+\frac {1}{3} (a (2 A+3 C)) \int \sec ^2(c+d x) \, dx\\ &=\frac {b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A b \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {(a (2 A+3 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (2 A+3 C) \tan (c+d x)}{3 d}+\frac {A b \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 0.34, size = 59, normalized size = 0.69 \[ \frac {\tan (c+d x) \left (2 a A \tan ^2(c+d x)+6 a (A+C)+3 A b \sec (c+d x)\right )+3 b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{6 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.86, size = 107, normalized size = 1.24 \[ \frac {3 \, {\left (A + 2 \, C\right )} b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + 2 \, C\right )} b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, A b \cos \left (d x + c\right ) + 2 \, A a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.42, size = 184, normalized size = 2.14 \[ \frac {3 \, {\left (A b + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A b + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.34, size = 108, normalized size = 1.26 \[ \frac {2 a A \tan \left (d x +c \right )}{3 d}+\frac {a A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a C \tan \left (d x +c \right )}{d}+\frac {A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.67, size = 107, normalized size = 1.24 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a - 3 \, A b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a \tan \left (d x + c\right )}{12 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.29, size = 137, normalized size = 1.59 \[ \frac {b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+2\,C\right )}{d}-\frac {\left (2\,A\,a-A\,b+2\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A\,a}{3}-4\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+A\,b+2\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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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