Optimal. Leaf size=196 \[ \frac{a^2 B \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^3 B \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac{a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}} \]
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Rubi [A] time = 0.48772, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2972, 2739, 2737, 2667, 31} \[ \frac{a^2 B \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^3 B \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac{a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2972
Rule 2739
Rule 2737
Rule 2667
Rule 31
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac{B \int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{(a B) \int \frac{(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{c^2}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^2 B \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{\left (a^2 B\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c-c \sin (e+f x)}} \, dx}{c^3}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^2 B \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{\left (a^3 B \cos (e+f x)\right ) \int \frac{\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^2 B \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (a^3 B \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^3 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac{a^2 B \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{a^3 B \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 1.23685, size = 204, normalized size = 1.04 \[ \frac{(a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (3 (A+5 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-6 (A+2 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+4 (A+B)+6 B \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{3 f (c-c \sin (e+f x))^{7/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.299, size = 832, normalized size = 4.2 \begin{align*} \text{result too large to display} \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 (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{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 [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left ({\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (A + B\right )} a^{2} +{\left (B a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{c^{4} \cos \left (f x + e\right )^{4} - 8 \, c^{4} \cos \left (f x + e\right )^{2} + 8 \, c^{4} + 4 \,{\left (c^{4} \cos \left (f x + e\right )^{2} - 2 \, c^{4}\right )} \sin \left (f x + e\right )}, x\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*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{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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