Optimal. Leaf size=122 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} c^{5/2} f}+\frac{a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac{3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}} \]
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Rubi [A] time = 0.240452, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2736, 2680, 2649, 206} \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} c^{5/2} f}+\frac{a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac{3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2736
Rule 2680
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac{a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac{1}{4} \left (3 a^2\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac{a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac{3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}}+\frac{\left (3 a^2\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{8 c^2}\\ &=\frac{a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac{3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,-\frac{c \cos (e+f x)}{\sqrt{c-c \sin (e+f x)}}\right )}{4 c^2 f}\\ &=\frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} c^{5/2} f}+\frac{a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac{3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 0.960644, size = 163, normalized size = 1.34 \[ \frac{a^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (3 \sin \left (\frac{1}{2} (e+f x)\right )+5 \sin \left (\frac{3}{2} (e+f x)\right )+3 \cos \left (\frac{1}{2} (e+f x)\right )-5 \cos \left (\frac{3}{2} (e+f x)\right )+(3+3 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) (4 \sin (e+f x)+\cos (2 (e+f x))-3)\right )}{8 c^2 f (\sin (e+f x)-1)^2 \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.735, size = 191, normalized size = 1.6 \begin{align*} -{\frac{{a}^{2}}{ \left ( -8+8\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{2}-6\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ){c}^{2}+10\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{c}+3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2}-12\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{3/2} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \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{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\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 = 1.13205, size = 915, normalized size = 7.5 \begin{align*} \frac{3 \, \sqrt{2}{\left (a^{2} \cos \left (f x + e\right )^{3} + 3 \, a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} -{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{c} \log \left (-\frac{c \cos \left (f x + e\right )^{2} + 2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c} \sqrt{c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) +{\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \,{\left (5 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} -{\left (5 \, a^{2} \cos \left (f x + e\right ) + 4 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{16 \,{\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f -{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\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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