Optimal. Leaf size=107 \[ -\frac{19 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{13 \cos (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}-\frac{\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \]
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Rubi [A] time = 0.128479, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2758, 2750, 2649, 206} \[ -\frac{19 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{13 \cos (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}-\frac{\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2758
Rule 2750
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac{\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{\int \frac{-\frac{5 a}{2}+4 a \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac{19 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac{19 \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 )}{16 a^2 d}\\ &=-\frac{19 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 0.205651, size = 196, normalized size = 1.83 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (8 \sin \left (\frac{1}{2} (c+d x)\right )+13 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3-26 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2-4 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+(19+19 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{16 d (a (\sin (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.658, size = 193, normalized size = 1.8 \begin{align*} -{\frac{1}{ \left ( 32+32\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) d} \left ( -19\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+38\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \sin \left ( dx+c \right ){a}^{2}+38\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}+26\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}\sqrt{a}-44\,\sqrt{a-a\sin \left ( dx+c \right ) }{a}^{3/2} \right ) \sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{-{\frac{9}{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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.79337, size = 855, normalized size = 7.99 \begin{align*} \frac{19 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\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 (13 \, \cos \left (d x + c\right )^{2} +{\left (13 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) + 9 \, \cos \left (d x + c\right ) - 4\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\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] 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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