Optimal. Leaf size=86 \[ \frac {a (2 A+3 B) \tan (c+d x)}{3 d}+\frac {a (A+B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d} \]
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Rubi [A] time = 0.15, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {a (2 A+3 B) \tan (c+d x)}{3 d}+\frac {a (A+B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2748
Rule 2968
Rule 3021
Rule 3767
Rule 3768
Rule 3770
Rubi steps
\begin {align*} \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx &=\int \left (a A+(a A+a B) \cos (c+d x)+a B \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 (3 a (A+B)+a (2 A+3 B) \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d}+(a (A+B)) \int \sec ^3(c+d x) \, dx+\frac {1}{3} (a (2 A+3 B)) \int \sec ^2(c+d x) \, dx\\ &=\frac {a (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} (a (A+B)) \int \sec (c+d x) \, dx-\frac {(a (2 A+3 B)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {a (A+B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (2 A+3 B) \tan (c+d x)}{3 d}+\frac {a (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 = 0.35, size = 56, normalized size = 0.65 \[ \frac {a \left (3 (A+B) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 (A+B) \sec (c+d x)+6 (A+B)+2 A \tan ^2(c+d x)\right )\right )}{6 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.73, size = 105, normalized size = 1.22 \[ \frac {3 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \, A a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.50, size = 154, normalized size = 1.79 \[ \frac {3 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 128, normalized size = 1.49 \[ \frac {a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a B \tan \left (d x +c \right )}{d}+\frac {2 a A \tan \left (d x +c \right )}{3 d}+\frac {a A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.37, size = 127, normalized size = 1.48 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a - 3 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a \tan \left (d x + c\right )}{12 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.07, size = 126, normalized size = 1.47 \[ \frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+B\right )}{d}-\frac {\left (A\,a+B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A\,a}{3}-4\,B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+3\,B\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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