Optimal. Leaf size=106 \[ -\frac{a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{8 \sqrt{2} d}-\frac{a^2 \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{8 d}+\frac{a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d} \]
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Rubi [A] time = 0.171282, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2676, 2675, 2667, 63, 206} \[ -\frac{a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{8 \sqrt{2} d}-\frac{a^2 \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{8 d}+\frac{a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d} \]
Antiderivative was successfully verified.
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Rule 2676
Rule 2675
Rule 2667
Rule 63
Rule 206
Rubi steps
\begin{align*} \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac{a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d}-\frac{1}{4} a^2 \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac{a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d}-\frac{1}{16} a^3 \int \sec (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac{a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d}-\frac{a^4 \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{16 d}\\ &=-\frac{a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac{a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d}-\frac{a^4 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{8 d}\\ &=-\frac{a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{8 \sqrt{2} d}-\frac{a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac{a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d}\\ \end{align*}
Mathematica [A] time = 0.291257, size = 108, normalized size = 1.02 \[ \frac{2 a^3 (\sin (c+d x)+3) \sqrt{a (\sin (c+d x)+1)}-\sqrt{2} a^{7/2} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4 \tanh ^{-1}\left (\frac{\sqrt{a (\sin (c+d x)+1)}}{\sqrt{2} \sqrt{a}}\right )}{16 d (\sin (c+d x)-1)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.145, size = 75, normalized size = 0.7 \begin{align*} -2\,{\frac{{a}^{5}}{d} \left ( -1/16\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) } \left ( 3+\sin \left ( dx+c \right ) \right ) }{ \left ( a\sin \left ( dx+c \right ) -a \right ) ^{2}}}+1/32\,{\frac{\sqrt{2}}{{a}^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.77725, size = 382, normalized size = 3.6 \begin{align*} \frac{{\left (\sqrt{2} a^{3} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{2} a^{3} \sin \left (d x + c\right ) - 2 \, \sqrt{2} a^{3}\right )} \sqrt{a} \log \left (-\frac{a \sin \left (d x + c\right ) - 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \,{\left (a^{3} \sin \left (d x + c\right ) + 3 \, a^{3}\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{32 \,{\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, 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 [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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