Optimal. Leaf size=165 \[ \frac{4 a \left (2 a^2-b^2\right ) \tan ^3(c+d x)}{35 d}+\frac{12 a \left (2 a^2-b^2\right ) \tan (c+d x)}{35 d}+\frac{2 b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{35 d}+\frac{2 \sec ^5(c+d x) (a+b \sin (c+d x)) \left (\left (3 a^2-b^2\right ) \sin (c+d x)+2 a b\right )}{35 d}+\frac{\sec ^7(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d} \]
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Rubi [A] time = 0.207002, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2691, 2861, 2669, 3767} \[ \frac{4 a \left (2 a^2-b^2\right ) \tan ^3(c+d x)}{35 d}+\frac{12 a \left (2 a^2-b^2\right ) \tan (c+d x)}{35 d}+\frac{2 b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{35 d}+\frac{2 \sec ^5(c+d x) (a+b \sin (c+d x)) \left (\left (3 a^2-b^2\right ) \sin (c+d x)+2 a b\right )}{35 d}+\frac{\sec ^7(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d} \]
Antiderivative was successfully verified.
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Rule 2691
Rule 2861
Rule 2669
Rule 3767
Rubi steps
\begin{align*} \int \sec ^8(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{7 d}-\frac{1}{7} \int \sec ^6(c+d x) (a+b \sin (c+d x)) \left (-6 a^2+2 b^2-4 a b \sin (c+d x)\right ) \, dx\\ &=\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{7 d}+\frac{2 \sec ^5(c+d x) (a+b \sin (c+d x)) \left (2 a b+\left (3 a^2-b^2\right ) \sin (c+d x)\right )}{35 d}+\frac{1}{35} \int \sec ^4(c+d x) \left (12 a \left (2 a^2-b^2\right )+6 b \left (3 a^2-b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac{2 b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{35 d}+\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{7 d}+\frac{2 \sec ^5(c+d x) (a+b \sin (c+d x)) \left (2 a b+\left (3 a^2-b^2\right ) \sin (c+d x)\right )}{35 d}+\frac{1}{35} \left (12 a \left (2 a^2-b^2\right )\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{2 b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{35 d}+\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{7 d}+\frac{2 \sec ^5(c+d x) (a+b \sin (c+d x)) \left (2 a b+\left (3 a^2-b^2\right ) \sin (c+d x)\right )}{35 d}-\frac{\left (12 a \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{35 d}\\ &=\frac{2 b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{35 d}+\frac{\sec ^7(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{7 d}+\frac{2 \sec ^5(c+d x) (a+b \sin (c+d x)) \left (2 a b+\left (3 a^2-b^2\right ) \sin (c+d x)\right )}{35 d}+\frac{12 a \left (2 a^2-b^2\right ) \tan (c+d x)}{35 d}+\frac{4 a \left (2 a^2-b^2\right ) \tan ^3(c+d x)}{35 d}\\ \end{align*}
Mathematica [A] time = 0.90584, size = 245, normalized size = 1.48 \[ \frac{\sec ^7(c+d x) \left (35 b \left (17 b^2-75 a^2\right ) \cos (c+d x)-1575 a^2 b \cos (3 (c+d x))-525 a^2 b \cos (5 (c+d x))-75 a^2 b \cos (7 (c+d x))+15360 a^2 b+8960 a^3 \sin (c+d x)+5376 a^3 \sin (3 (c+d x))+1792 a^3 \sin (5 (c+d x))+256 a^3 \sin (7 (c+d x))+13440 a b^2 \sin (c+d x)-2688 a b^2 \sin (3 (c+d x))-896 a b^2 \sin (5 (c+d x))-128 a b^2 \sin (7 (c+d x))-3584 b^3 \cos (2 (c+d x))+357 b^3 \cos (3 (c+d x))+119 b^3 \cos (5 (c+d x))+17 b^3 \cos (7 (c+d x))+1536 b^3\right )}{35840 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.082, size = 219, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -{a}^{3} \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{35}} \right ) \tan \left ( dx+c \right ) +{\frac{3\,{a}^{2}b}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+3\,a{b}^{2} \left ( 1/7\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{b}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{35}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.968928, size = 167, normalized size = 1.01 \begin{align*} \frac{{\left (5 \, \tan \left (d x + c\right )^{7} + 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} a^{3} +{\left (15 \, \tan \left (d x + c\right )^{7} + 42 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3}\right )} a b^{2} - \frac{{\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} b^{3}}{\cos \left (d x + c\right )^{7}} + \frac{15 \, a^{2} b}{\cos \left (d x + c\right )^{7}}}{35 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.36043, size = 279, normalized size = 1.69 \begin{align*} -\frac{7 \, b^{3} \cos \left (d x + c\right )^{2} - 15 \, a^{2} b - 5 \, b^{3} -{\left (8 \,{\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{6} + 4 \,{\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{4} + 5 \, a^{3} + 15 \, a b^{2} + 3 \,{\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{35 \, d \cos \left (d x + c\right )^{7}} \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 [B] time = 1.13117, size = 483, normalized size = 2.93 \begin{align*} -\frac{2 \,{\left (35 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} + 105 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{12} - 70 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 140 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 70 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 301 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 112 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 525 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 70 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 212 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 456 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 140 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 301 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 112 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 315 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 28 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 70 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 140 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 14 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 35 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, a^{2} b - 2 \, b^{3}\right )}}{35 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{7} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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