Optimal. Leaf size=88 \[ \frac {c^2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-2} (g \cos (e+f x))^{-2 m} \, _2F_1\left (3,-m+n-2;-m+n-1;\frac {1}{2} (1-\sin (e+f x))\right )}{8 f g^5 (m-n+2)} \]
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Rubi [A] time = 0.24, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2853, 12, 2667, 68} \[ \frac {c^2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-2} (g \cos (e+f x))^{-2 m} \, _2F_1\left (3,-m+n-2;-m+n-1;\frac {1}{2} (1-\sin (e+f x))\right )}{8 f g^5 (m-n+2)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 68
Rule 2667
Rule 2853
Rubi steps
\begin {align*} \int (g \cos (e+f x))^{-5-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx &=\left ((g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{-m+n}}{g^5} \, dx\\ &=\frac {\left ((g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \sec ^5(e+f x) (c-c \sin (e+f x))^{-m+n} \, dx}{g^5}\\ &=-\frac {\left (c^5 (g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \operatorname {Subst}\left (\int \frac {(c+x)^{-3-m+n}}{(c-x)^3} \, dx,x,-c \sin (e+f x)\right )}{f g^5}\\ &=\frac {c^2 (g \cos (e+f x))^{-2 m} \, _2F_1\left (3,-2-m+n;-1-m+n;\frac {1}{2} (1-\sin (e+f x))\right ) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2+n}}{8 f g^5 (2+m-n)}\\ \end {align*}
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Mathematica [A] time = 41.14, size = 136, normalized size = 1.55 \[ \frac {\cot ^4\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) (a (\sin (e+f x)+1))^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-2 m} \sec ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )^{n-m} \, _2F_1\left (-m+n-4,-m+n-2;-m+n-1;-\tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )}{32 f g^5 (m-n+2)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (g \cos \left (f x + e\right )\right )^{-2 \, m - 5} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \]
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 \[ \int \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 5} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 53.34, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{-5-2 m} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}\, dx \]
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 \[ \int \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 5} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
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 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{2\,m+5}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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