Optimal. Leaf size=156 \[ \frac{a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{16 \sqrt{2} c^{7/2} f}+\frac{a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}} \]
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Rubi [A] time = 0.272922, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2736, 2680, 2650, 2649, 206} \[ \frac{a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{16 \sqrt{2} c^{7/2} f}+\frac{a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2736
Rule 2680
Rule 2650
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{7/2}} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{11/2}} \, dx\\ &=\frac{a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac{1}{2} a^2 \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=\frac{a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^2 \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 c^2}\\ &=\frac{a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^2 \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{32 c^3}\\ &=\frac{a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{a^2 \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 )}{16 c^3 f}\\ &=\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{16 \sqrt{2} c^{7/2} f}+\frac{a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^2 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 0.97242, size = 307, normalized size = 1.97 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (64 \sin \left (\frac{1}{2} (e+f x)\right )+3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5+6 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-28 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-56 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+32 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(-3-3 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6\right )}{48 f (c-c \sin (e+f x))^{7/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.802, size = 245, normalized size = 1.6 \begin{align*} -{\frac{{a}^{2}}{96\, \left ( -1+\sin \left ( fx+e \right ) \right ) ^{2}\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 ) ^{3}{c}^{4}+24\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{7/2}-32\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}{c}^{5/2}-6\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}{c}^{3/2}-9\,\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}^{4}+9\,\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}^{4}-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}^{4} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{15}{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{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.15374, size = 1118, normalized size = 7.17 \begin{align*} \frac{3 \, \sqrt{2}{\left (a^{2} \cos \left (f x + e\right )^{4} - 3 \, a^{2} \cos \left (f x + e\right )^{3} - 8 \, a^{2} \cos \left (f x + e\right )^{2} + 4 \, a^{2} \cos \left (f x + e\right ) + 8 \, a^{2} +{\left (a^{2} \cos \left (f x + e\right )^{3} + 4 \, a^{2} \cos \left (f x + e\right )^{2} - 4 \, a^{2} \cos \left (f x + e\right ) - 8 \, 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 (3 \, a^{2} \cos \left (f x + e\right )^{3} + 25 \, a^{2} \cos \left (f x + e\right )^{2} - 10 \, a^{2} \cos \left (f x + e\right ) - 32 \, a^{2} +{\left (3 \, a^{2} \cos \left (f x + e\right )^{2} - 22 \, a^{2} \cos \left (f x + e\right ) - 32 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{192 \,{\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f +{\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} 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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