Optimal. Leaf size=81 \[ \frac {2^{m+\frac {3}{2}} \sec (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac {1}{2},-m-\frac {1}{2};\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{a c^2 f} \]
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Rubi [A] time = 0.21, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2840, 2689, 70, 69} \[ \frac {2^{m+\frac {3}{2}} \sec (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac {1}{2},-m-\frac {1}{2};\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{a c^2 f} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 2689
Rule 2840
Rubi steps
\begin {align*} \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx &=\frac {\int \sec ^2(e+f x) (a+a \sin (e+f x))^{2+m} \, dx}{a^2 c^2}\\ &=\frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{c^2 f}\\ &=\frac {\left (2^{\frac {1}{2}+m} \sec (e+f x) \sqrt {a-a \sin (e+f x)} (a+a \sin (e+f x))^{1+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{-\frac {1}{2}-m}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{\frac {1}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{c^2 f}\\ &=\frac {2^{\frac {3}{2}+m} \, _2F_1\left (-\frac {1}{2},-\frac {1}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^{1+m}}{a c^2 f}\\ \end {align*}
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Mathematica [A] time = 0.21, size = 88, normalized size = 1.09 \[ \frac {2^{m+\frac {3}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a (\sin (e+f x)+1))^m \, _2F_1\left (-\frac {1}{2},-m-\frac {1}{2};\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{c^2 f (1-\sin (e+f x))} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 3.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (c -c \sin \left (f x +e \right )\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (a \sin {\left (e + f x \right )} + a\right )^{m} \cos ^{2}{\left (e + f x \right )}}{\sin ^{2}{\left (e + f x \right )} - 2 \sin {\left (e + f x \right )} + 1}\, dx}{c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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