Optimal. Leaf size=126 \[ -\frac{(3 c+5 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}} \]
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Rubi [A] time = 0.0979192, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2750, 2650, 2649, 206} \[ -\frac{(3 c+5 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(c-d) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2750
Rule 2650
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{c+d \sin (e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac{(c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{(3 c+5 d) \int \frac{1}{(a+a \sin (e+f x))^{3/2}} \, dx}{8 a}\\ &=-\frac{(c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac{(3 c+5 d) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{32 a^2}\\ &=-\frac{(c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(3 c+5 d) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{16 a^2 f}\\ &=-\frac{(3 c+5 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(c-d) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+5 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 0.352523, size = 227, normalized size = 1.8 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (8 (c-d) \sin \left (\frac{1}{2} (e+f x)\right )-(3 c+5 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+2 (3 c+5 d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+4 (d-c) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+(1+i) (-1)^{3/4} (3 c+5 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{16 f (a (\sin (e+f x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.065, size = 279, normalized size = 2.2 \begin{align*} -{\frac{1}{ \left ( 32+32\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 2\,\sin \left ( fx+e \right ){\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \sqrt{2}{a}^{3} \left ( 3\,c+5\,d \right ) -{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) \sqrt{2}{a}^{3} \left ( 3\,c+5\,d \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+20\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{5/2}c+12\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{5/2}d-6\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}{a}^{3/2}c-10\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}{a}^{3/2}d+6\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3}c+10\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{3}d \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{a}^{-{\frac{11}{2}}}{\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{d \sin \left (f x + e\right ) + c}{{\left (a \sin \left (f x + e\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 = 2.00946, size = 1015, normalized size = 8.06 \begin{align*} \frac{\sqrt{2}{\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right ) +{\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right ) - 12 \, c - 20 \, d\right )} \sin \left (f x + e\right ) - 12 \, c - 20 \, d\right )} \sqrt{a} \log \left (-\frac{a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\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 ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right )^{2} +{\left (7 \, c + d\right )} \cos \left (f x + e\right ) +{\left ({\left (3 \, c + 5 \, d\right )} \cos \left (f x + e\right ) - 4 \, c + 4 \, d\right )} \sin \left (f x + e\right ) + 4 \, c - 4 \, d\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{64 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f +{\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{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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