Optimal. Leaf size=117 \[ \frac {a (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (3 A+4 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {b (2 A+3 C) \tan (c+d x)}{3 d}+\frac {A b \tan (c+d x) \sec ^2(c+d x)}{3 d} \]
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Rubi [A] time = 0.19, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3032, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {a (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (3 A+4 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {b (2 A+3 C) \tan (c+d x)}{3 d}+\frac {A b \tan (c+d x) \sec ^2(c+d x)}{3 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 3021
Rule 3032
Rule 3767
Rule 3768
Rule 3770
Rubi steps
\begin {align*} \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \left (4 A b+a (3 A+4 C) \cos (c+d x)+4 b C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {A b \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int (3 a (3 A+4 C)+4 b (2 A+3 C) \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\frac {A b \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{3} (b (2 A+3 C)) \int \sec ^2(c+d x) \, dx+\frac {1}{4} (a (3 A+4 C)) \int \sec ^3(c+d x) \, dx\\ &=\frac {a (3 A+4 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A b \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} (a (3 A+4 C)) \int \sec (c+d x) \, dx-\frac {(b (2 A+3 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {a (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b (2 A+3 C) \tan (c+d x)}{3 d}+\frac {a (3 A+4 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A b \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.46, size = 80, normalized size = 0.68 \[ \frac {\tan (c+d x) \left (3 a (3 A+4 C) \sec (c+d x)+6 a A \sec ^3(c+d x)+8 b \left (A \tan ^2(c+d x)+3 (A+C)\right )\right )+3 a (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{24 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.86, size = 129, normalized size = 1.10 \[ \frac {3 \, {\left (3 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, A + 3 \, C\right )} b \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, A b \cos \left (d x + c\right ) + 6 \, A a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.45, size = 304, normalized size = 2.60 \[ \frac {3 \, {\left (3 \, A a + 4 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, A a + 4 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.36, size = 149, normalized size = 1.27 \[ \frac {a A \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a C \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 A b \tan \left (d x +c \right )}{3 d}+\frac {A b \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {C b \tan \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.68, size = 152, normalized size = 1.30 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b - 3 \, A a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C b \tan \left (d x + c\right )}{48 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.74, size = 195, normalized size = 1.67 \[ \frac {\left (\frac {5\,A\,a}{4}-2\,A\,b+C\,a-2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,A\,a}{4}+\frac {10\,A\,b}{3}-C\,a+6\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,A\,a}{4}-\frac {10\,A\,b}{3}-C\,a-6\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,a}{4}+2\,A\,b+C\,a+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A+4\,C\right )}{4\,d} \]
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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