Optimal. Leaf size=84 \[ \frac{2 a \left (a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac{2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d} \]
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Rubi [A] time = 0.088474, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2691, 12, 2669, 3767, 8} \[ \frac{2 a \left (a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac{2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d} \]
Antiderivative was successfully verified.
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Rule 2691
Rule 12
Rule 2669
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}-\frac{1}{3} \int \left (-2 a^2+2 b^2\right ) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx\\ &=\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}+\frac{1}{3} \left (2 \left (a^2-b^2\right )\right ) \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx\\ &=\frac{2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}+\frac{1}{3} \left (2 a \left (a^2-b^2\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}-\frac{\left (2 a \left (a^2-b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}+\frac{2 a \left (a^2-b^2\right ) \tan (c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.412579, size = 136, normalized size = 1.62 \[ \frac{\sec ^3(c+d x) \left (\left (15 b^3-9 a^2 b\right ) \cos (c+d x)-3 a^2 b \cos (3 (c+d x))+24 a^2 b+12 a^3 \sin (c+d x)+4 a^3 \sin (3 (c+d x))+18 a b^2 \sin (c+d x)-6 a b^2 \sin (3 (c+d x))-12 b^3 \cos (2 (c+d x))+5 b^3 \cos (3 (c+d x))-4 b^3\right )}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 122, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( -{a}^{3} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) +{\frac{{a}^{2}b}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{b}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.954564, size = 108, normalized size = 1.29 \begin{align*} \frac{3 \, a b^{2} \tan \left (d x + c\right )^{3} +{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - \frac{{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{3}}{\cos \left (d x + c\right )^{3}} + \frac{3 \, a^{2} b}{\cos \left (d x + c\right )^{3}}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.42204, size = 176, normalized size = 2.1 \begin{align*} -\frac{3 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3} -{\left (a^{3} + 3 \, a b^{2} +{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \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 = 1.11149, size = 173, normalized size = 2.06 \begin{align*} -\frac{2 \,{\left (3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, a^{3} \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 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2} b - 2 \, b^{3}\right )}}{3 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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