Optimal. Leaf size=166 \[ -\frac{a^3 (3 c+d) \cos (e+f x)}{d^2 f (c+d) \sqrt{a \sin (e+f x)+a}}+\frac{a^{5/2} (c-d) (3 c+5 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{5/2} f (c+d)^{3/2}}+\frac{a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))} \]
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Rubi [A] time = 0.388863, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2762, 2981, 2773, 208} \[ -\frac{a^3 (3 c+d) \cos (e+f x)}{d^2 f (c+d) \sqrt{a \sin (e+f x)+a}}+\frac{a^{5/2} (c-d) (3 c+5 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{d^{5/2} f (c+d)^{3/2}}+\frac{a^2 (c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))} \]
Antiderivative was successfully verified.
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Rule 2762
Rule 2981
Rule 2773
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^2} \, dx &=\frac{a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}-\frac{a \int \frac{\sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a (c-5 d)-\frac{1}{2} a (3 c+d) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{d (c+d)}\\ &=-\frac{a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt{a+a \sin (e+f x)}}+\frac{a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}-\frac{\left (a^2 (c-d) (3 c+5 d)\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{2 d^2 (c+d)}\\ &=-\frac{a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt{a+a \sin (e+f x)}}+\frac{a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}+\frac{\left (a^3 (c-d) (3 c+5 d)\right ) \operatorname{Subst}\left (\int \frac{1}{a c+a d-d x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{d^2 (c+d) f}\\ &=\frac{a^{5/2} (c-d) (3 c+5 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a+a \sin (e+f x)}}\right )}{d^{5/2} (c+d)^{3/2} f}-\frac{a^3 (3 c+d) \cos (e+f x)}{d^2 (c+d) f \sqrt{a+a \sin (e+f x)}}+\frac{a^2 (c-d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [B] time = 4.11335, size = 350, normalized size = 2.11 \[ \frac{(a (\sin (e+f x)+1))^{5/2} \left (\frac{\left (-3 c^2-2 c d+5 d^2\right ) \left (2 \log \left (\sqrt{d} \sqrt{c+d} \left (\tan ^2\left (\frac{1}{4} (e+f x)\right )+2 \tan \left (\frac{1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac{1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}+\frac{\left (3 c^2+2 c d-5 d^2\right ) \left (2 \log \left (-\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (-\sqrt{d} \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{3/2}}-\frac{4 \sqrt{d} (c-d)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}+8 \sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )-8 \sqrt{d} \cos \left (\frac{1}{2} (e+f x)\right )\right )}{4 d^{5/2} f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.224, size = 392, normalized size = 2.4 \begin{align*} -{\frac{{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) }{ \left ( c+d \right ){d}^{2} \left ( c+d\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( \sin \left ( fx+e \right ) d \left ( -3\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) a{c}^{2}-2\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) acd+5\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) a{d}^{2}+2\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}c+2\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}d \right ) -3\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) a{c}^{3}-2\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) a{c}^{2}d+5\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }d}{\sqrt{acd+a{d}^{2}}}} \right ) ac{d}^{2}+3\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}{c}^{2}+\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a \left ( c+d \right ) d}{d}^{2} \right ){\frac{1}{\sqrt{a \left ( c+d \right ) d}}}{\frac{1}{\sqrt{a+a\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 )}^{\frac{5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.70633, size = 2916, normalized size = 17.57 \begin{align*} \text{result too large to display} \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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