Optimal. Leaf size=107 \[ \frac{a^5 (2 A-3 B) \cos (c+d x)}{15 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^3}{5 d} \]
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Rubi [A] time = 0.153776, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2855, 2670, 2650, 2648} \[ \frac{a^5 (2 A-3 B) \cos (c+d x)}{15 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^3}{5 d} \]
Antiderivative was successfully verified.
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Rule 2855
Rule 2670
Rule 2650
Rule 2648
Rubi steps
\begin{align*} \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac{1}{5} (a (2 A-3 B)) \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac{1}{5} \left (a^5 (2 A-3 B)\right ) \int \frac{1}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac{a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac{1}{15} \left (a^4 (2 A-3 B)\right ) \int \frac{1}{a-a \sin (c+d x)} \, dx\\ &=\frac{a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{a^4 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))}+\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}\\ \end{align*}
Mathematica [A] time = 0.164378, size = 94, normalized size = 0.88 \[ -\frac{a^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (6 (2 A-3 B) \sin (c+d x)+(2 A-3 B) \cos (2 (c+d x))-16 A+9 B)}{30 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.119, size = 333, normalized size = 3.1 \begin{align*}{\frac{1}{d} \left ({a}^{3}A \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{15}} \right ) +{\frac{B{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+3\,{a}^{3}A \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +3\,B{a}^{3} \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+1/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-1/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}-1/15\, \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{\frac{3\,{a}^{3}A}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+3\,B{a}^{3} \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) -{a}^{3}A \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) +{\frac{B{a}^{3}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.06842, size = 254, normalized size = 2.37 \begin{align*} \frac{3 \, B a^{3} \tan \left (d x + c\right )^{5} +{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 3 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} A a^{3} + 3 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} B a^{3} - \frac{{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} A a^{3}}{\cos \left (d x + c\right )^{5}} - \frac{3 \,{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} B a^{3}}{\cos \left (d x + c\right )^{5}} + \frac{9 \, A a^{3}}{\cos \left (d x + c\right )^{5}} + \frac{3 \, B a^{3}}{\cos \left (d x + c\right )^{5}}}{15 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.65749, size = 464, normalized size = 4.34 \begin{align*} \frac{{\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} - 2 \,{\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - 3 \,{\left (3 \, A - 2 \, B\right )} a^{3} \cos \left (d x + c\right ) - 3 \,{\left (A + B\right )} a^{3} +{\left ({\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right ) - 3 \,{\left (A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{3} + 3 \, d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) -{\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.37364, size = 197, normalized size = 1.84 \begin{align*} -\frac{2 \,{\left (15 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 20 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7 \, A a^{3} - 3 \, B a^{3}\right )}}{15 \, d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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