Optimal. Leaf size=100 \[ \frac{a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{c f \sqrt{a \sin (e+f x)+a}}-\frac{a (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]
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Rubi [A] time = 0.337057, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2971, 2738, 2737, 2667, 31} \[ \frac{a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{c f \sqrt{a \sin (e+f x)+a}}-\frac{a (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2971
Rule 2738
Rule 2737
Rule 2667
Rule 31
Rubi steps
\begin{align*} \int \frac{\sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt{c-c \sin (e+f x)}} \, dx &=(A+B) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c-c \sin (e+f x)}} \, dx-\frac{B \int \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)} \, dx}{c}\\ &=\frac{a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{c f \sqrt{a+a \sin (e+f x)}}+\frac{(a (A+B) c \cos (e+f x)) \int \frac{\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{c f \sqrt{a+a \sin (e+f x)}}-\frac{(a (A+B) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{a (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}+\frac{a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{c f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}
Mathematica [C] time = 1.17749, size = 120, normalized size = 1.2 \[ -\frac{\sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (B \sin (e+f x)+(A+B) \left (2 \log \left (i-e^{i (e+f x)}\right )-i f x\right )\right )}{f \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.367, size = 394, normalized size = 3.9 \begin{align*} -{\frac{1}{f \left ( -1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) \right ) } \left ( A\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -2\,A\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +A\sin \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -2\,A\sin \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +B \left ( \cos \left ( fx+e \right ) \right ) ^{2}-B\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +B\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -2\,B\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +B\sin \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -2\,B\sin \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -A\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +2\,A\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +B\sin \left ( fx+e \right ) -B\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +2\,B\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -B \right ) \sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{-c \left ( -1+\sin \left ( fx+e \right ) \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.57322, size = 236, normalized size = 2.36 \begin{align*} \frac{B{\left (\frac{2 \, \sqrt{a} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt{c}} - \frac{\sqrt{a} \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt{c}} + \frac{2 \, \sqrt{a} \sqrt{c} \sin \left (f x + e\right )}{{\left (c + \frac{c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + A{\left (\frac{2 \, \sqrt{a} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt{c}} - \frac{\sqrt{a} \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt{c}}\right )}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{c \sin \left (f x + e\right ) - c}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )} \left (A + B \sin{\left (e + f x \right )}\right )}{\sqrt{- c \left (\sin{\left (e + f x \right )} - 1\right )}}\, dx \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*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{\sqrt{-c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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