Optimal. Leaf size=195 \[ -\frac{105 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{256 \sqrt{2} a^{3/2} d}-\frac{105 \cos (c+d x)}{256 d (a \sin (c+d x)+a)^{3/2}}+\frac{\sec ^3(c+d x)}{4 a d \sqrt{a \sin (c+d x)+a}}-\frac{\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}+\frac{35 \sec (c+d x)}{64 a d \sqrt{a \sin (c+d x)+a}}-\frac{7 \sec (c+d x)}{32 d (a \sin (c+d x)+a)^{3/2}} \]
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Rubi [A] time = 0.29053, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2681, 2687, 2650, 2649, 206} \[ -\frac{105 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{256 \sqrt{2} a^{3/2} d}-\frac{105 \cos (c+d x)}{256 d (a \sin (c+d x)+a)^{3/2}}+\frac{\sec ^3(c+d x)}{4 a d \sqrt{a \sin (c+d x)+a}}-\frac{\sec ^3(c+d x)}{6 d (a \sin (c+d x)+a)^{3/2}}+\frac{35 \sec (c+d x)}{64 a d \sqrt{a \sin (c+d x)+a}}-\frac{7 \sec (c+d x)}{32 d (a \sin (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2681
Rule 2687
Rule 2650
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=-\frac{\sec ^3(c+d x)}{6 d (a+a \sin (c+d x))^{3/2}}+\frac{3 \int \frac{\sec ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{4 a}\\ &=-\frac{\sec ^3(c+d x)}{6 d (a+a \sin (c+d x))^{3/2}}+\frac{\sec ^3(c+d x)}{4 a d \sqrt{a+a \sin (c+d x)}}+\frac{7}{8} \int \frac{\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{7 \sec (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac{\sec ^3(c+d x)}{6 d (a+a \sin (c+d x))^{3/2}}+\frac{\sec ^3(c+d x)}{4 a d \sqrt{a+a \sin (c+d x)}}+\frac{35 \int \frac{\sec ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{64 a}\\ &=-\frac{7 \sec (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac{\sec ^3(c+d x)}{6 d (a+a \sin (c+d x))^{3/2}}+\frac{35 \sec (c+d x)}{64 a d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^3(c+d x)}{4 a d \sqrt{a+a \sin (c+d x)}}+\frac{105}{128} \int \frac{1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{105 \cos (c+d x)}{256 d (a+a \sin (c+d x))^{3/2}}-\frac{7 \sec (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac{\sec ^3(c+d x)}{6 d (a+a \sin (c+d x))^{3/2}}+\frac{35 \sec (c+d x)}{64 a d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^3(c+d x)}{4 a d \sqrt{a+a \sin (c+d x)}}+\frac{105 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{512 a}\\ &=-\frac{105 \cos (c+d x)}{256 d (a+a \sin (c+d x))^{3/2}}-\frac{7 \sec (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac{\sec ^3(c+d x)}{6 d (a+a \sin (c+d x))^{3/2}}+\frac{35 \sec (c+d x)}{64 a d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^3(c+d x)}{4 a d \sqrt{a+a \sin (c+d x)}}-\frac{105 \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 )}{256 a d}\\ &=-\frac{105 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{256 \sqrt{2} a^{3/2} d}-\frac{105 \cos (c+d x)}{256 d (a+a \sin (c+d x))^{3/2}}-\frac{7 \sec (c+d x)}{32 d (a+a \sin (c+d x))^{3/2}}-\frac{\sec ^3(c+d x)}{6 d (a+a \sin (c+d x))^{3/2}}+\frac{35 \sec (c+d x)}{64 a d \sqrt{a+a \sin (c+d x)}}+\frac{\sec ^3(c+d x)}{4 a d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}
Mathematica [C] time = 0.325417, size = 334, normalized size = 1.71 \[ \frac{\frac{192 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{32 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}-123 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2+246 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{136 \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-\frac{32}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{64 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+(315+315 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )-68}{768 d (a (\sin (c+d x)+1))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.184, size = 289, normalized size = 1.5 \begin{align*}{\frac{1}{ \left ( 1536\,\sin \left ( dx+c \right ) -1536 \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) d} \left ( \left ( -840\,{a}^{9/2}-315\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3} \right ) \sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+ \left ( -384\,{a}^{9/2}+1260\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3} \right ) \sin \left ( dx+c \right ) +630\,{a}^{9/2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}+ \left ( -504\,{a}^{9/2}-945\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-128\,{a}^{9/2}+1260\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3} \right ){a}^{-{\frac{11}{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 [A] time = 2.48944, size = 729, normalized size = 3.74 \begin{align*} \frac{315 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{3}\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 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 (315 \, \cos \left (d x + c\right )^{4} - 252 \, \cos \left (d x + c\right )^{2} - 12 \,{\left (35 \, \cos \left (d x + c\right )^{2} + 16\right )} \sin \left (d x + c\right ) - 64\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{3072 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\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] time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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