Optimal. Leaf size=171 \[ \frac{a^2 \sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{12 d}+\frac{a^3 \sec (c+d x) \sqrt{a \sin (c+d x)+a}}{8 d}-\frac{a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{8 \sqrt{2} d}+\frac{\sec ^7(c+d x) (a \sin (c+d x)+a)^{7/2}}{7 d}+\frac{a \sec ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{10 d} \]
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Rubi [A] time = 0.267813, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2675, 2649, 206} \[ \frac{a^2 \sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{12 d}+\frac{a^3 \sec (c+d x) \sqrt{a \sin (c+d x)+a}}{8 d}-\frac{a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{8 \sqrt{2} d}+\frac{\sec ^7(c+d x) (a \sin (c+d x)+a)^{7/2}}{7 d}+\frac{a \sec ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{10 d} \]
Antiderivative was successfully verified.
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Rule 2675
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac{\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}+\frac{1}{2} a \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=\frac{a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac{\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}+\frac{1}{4} a^2 \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=\frac{a^2 \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{12 d}+\frac{a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac{\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}+\frac{1}{8} a^3 \int \sec ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=\frac{a^3 \sec (c+d x) \sqrt{a+a \sin (c+d x)}}{8 d}+\frac{a^2 \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{12 d}+\frac{a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac{\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}+\frac{1}{16} a^4 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{a^3 \sec (c+d x) \sqrt{a+a \sin (c+d x)}}{8 d}+\frac{a^2 \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{12 d}+\frac{a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac{\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}-\frac{a^4 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{8 d}\\ &=-\frac{a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{8 \sqrt{2} d}+\frac{a^3 \sec (c+d x) \sqrt{a+a \sin (c+d x)}}{8 d}+\frac{a^2 \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{12 d}+\frac{a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac{\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}\\ \end{align*}
Mathematica [C] time = 5.49601, size = 139, normalized size = 0.81 \[ \frac{(a (\sin (c+d x)+1))^{7/2} \left (\frac{-2471 \sin (c+d x)+105 \sin (3 (c+d x))-770 \cos (2 (c+d x))+2286}{4 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^7}+(105+105 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{840 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^7} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.143, size = 139, normalized size = 0.8 \begin{align*}{\frac{1+\sin \left ( dx+c \right ) }{1680\, \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}\cos \left ( dx+c \right ) d} \left ( -210\,{a}^{15/2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+770\,{a}^{15/2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1288\,{a}^{15/2}\sin \left ( dx+c \right ) -1528\,{a}^{15/2}+105\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{4} \left ( a-a\sin \left ( dx+c \right ) \right ) ^{7/2} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [B] time = 2.21181, size = 828, normalized size = 4.84 \begin{align*} \frac{105 \,{\left (3 \, \sqrt{2} a^{3} \cos \left (d x + c\right )^{3} - 4 \, \sqrt{2} a^{3} \cos \left (d x + c\right ) -{\left (\sqrt{2} a^{3} \cos \left (d x + c\right )^{3} - 4 \, \sqrt{2} a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a \sin \left (d x + c\right ) + a}{\left (\sqrt{2} \cos \left (d x + c\right ) - \sqrt{2} \sin \left (d x + c\right ) + \sqrt{2}\right )} \sqrt{a} + 3 \, a \cos \left (d x + c\right ) -{\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \,{\left (385 \, a^{3} \cos \left (d x + c\right )^{2} - 764 \, a^{3} - 7 \,{\left (15 \, a^{3} \cos \left (d x + c\right )^{2} - 92 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{3360 \,{\left (3 \, d \cos \left (d x + c\right )^{3} - 4 \, d \cos \left (d x + c\right ) -{\left (d \cos \left (d x + c\right )^{3} - 4 \, d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\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 [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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