Optimal. Leaf size=171 \[ -\frac {a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{8 \sqrt {2} d}+\frac {a^3 \sec (c+d x) \sqrt {a \sin (c+d x)+a}}{8 d}+\frac {a^2 \sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{12 d}+\frac {\sec ^7(c+d x) (a \sin (c+d x)+a)^{7/2}}{7 d}+\frac {a \sec ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{10 d} \]
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Rubi [A] time = 0.27, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 206
Rule 2649
Rule 2675
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*}
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Mathematica [C] time = 5.50, 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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fricas [B] time = 0.65, size = 312, normalized size = 1.82 \[ \frac {105 \, {\left (3 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{3} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right ) - {\left (\sqrt {2} a^{3} \cos \left (d x + c\right )^{3} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (385 \, a^{3} \cos \left (d x + c\right )^{2} - 764 \, a^{3} - 7 \, {\left (15 \, a^{3} \cos \left (d x + c\right )^{2} - 92 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3360 \, {\left (3 \, d \cos \left (d x + c\right )^{3} - 4 \, d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{3} - 4 \, d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 139, normalized size = 0.81 \[ \frac {\left (1+\sin \left (d x +c \right )\right ) \left (-210 a^{\frac {15}{2}} \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+770 a^{\frac {15}{2}} \left (\cos ^{2}\left (d x +c \right )\right )+1288 a^{\frac {15}{2}} \sin \left (d x +c \right )-1528 a^{\frac {15}{2}}+105 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4} \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}}\right )}{1680 a^{\frac {7}{2}} \left (\sin \left (d x +c \right )-1\right )^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^8} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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