Optimal. Leaf size=207 \[ \frac{\left (A \left (4 m^2-8 m+3\right )+C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{32 c^2 f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(A (5-2 m)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}} \]
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Rubi [A] time = 0.628137, antiderivative size = 207, 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 = {3036, 2972, 2745, 2667, 68} \[ \frac{\left (A \left (4 m^2-8 m+3\right )+C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{32 c^2 f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(A (5-2 m)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3036
Rule 2972
Rule 2745
Rule 2667
Rule 68
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{5/2}} \, dx &=\frac{(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}-\frac{\int \frac{(a+a \sin (e+f x))^m \left (-\frac{1}{2} a^2 (A (9-2 m)-C (7+2 m))-\frac{1}{2} a^2 (A (1-2 m)-C (15+2 m)) \sin (e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 a^2 c}\\ &=\frac{(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac{(A (5-2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac{\left (A \left (3-8 m+4 m^2\right )+C \left (19+24 m+4 m^2\right )\right ) \int \frac{(a+a \sin (e+f x))^m}{\sqrt{c-c \sin (e+f x)}} \, dx}{32 c^2}\\ &=\frac{(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac{(A (5-2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac{\left (\left (A \left (3-8 m+4 m^2\right )+C \left (19+24 m+4 m^2\right )\right ) \cos (e+f x)\right ) \int \sec (e+f x) (a+a \sin (e+f x))^{\frac{1}{2}+m} \, dx}{32 c^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac{(A (5-2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac{\left (a \left (A \left (3-8 m+4 m^2\right )+C \left (19+24 m+4 m^2\right )\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-\frac{1}{2}+m}}{a-x} \, dx,x,a \sin (e+f x)\right )}{32 c^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac{(A (5-2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac{\left (A \left (3-8 m+4 m^2\right )+C \left (19+24 m+4 m^2\right )\right ) \cos (e+f x) \, _2F_1\left (1,\frac{1}{2}+m;\frac{3}{2}+m;\frac{1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{32 c^2 f (1+2 m) \sqrt{c-c \sin (e+f x)}}\\ \end{align*}
Mathematica [B] time = 13.4105, size = 619, normalized size = 2.99 \[ -\frac{\sec ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )^{2 m} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5 (a \sin (e+f x)+a)^m \left (\frac{2^{1-2 m} (3 A+19 C) \, _2F_1\left (2 m,2 m;2 m+1;\frac{1}{2} \left (1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right )}{m}+\frac{2^{1-2 m} (3 A-13 C) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )-1\right ) \, _2F_1\left (2 m,2 m+1;2 (m+1);\frac{1}{2} \left (1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right )}{2 m+1}+\frac{4^{1-m} (A+C) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )-1\right ) \, _2F_1\left (2 m-1,2 m+1;2 (m+1);\frac{1}{2} \left (1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right )}{2 m+1}-\frac{2 (3 A+19 C) \sec ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )^{-2 m} \, _2F_1\left (1,2 m;2 m+1;\cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )}{m}-\frac{16 (A-3 C) \cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) \sec ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )^{-2 m} \, _2F_1\left (2,2 m+1;2 m+2;\cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )}{2 m+1}+\frac{8 m (A+C) \cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) \sec ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )^{-2 m} \, _2F_1\left (2,2 m+1;2 m+2;\cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )}{2 m+1}-(A+C) \cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) \csc ^4\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ) \sec ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )^{-2 m}\right )}{128 \sqrt{2} f (c-c \sin (e+f x))^{5/2}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.675, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \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 (C \sin \left (f x + e\right )^{2} + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{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 (C \cos \left (f x + e\right )^{2} - A - C\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} -{\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\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 (C \sin \left (f x + e\right )^{2} + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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