Optimal. Leaf size=114 \[ \frac {c 2^{\frac {1}{2}-m} \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \, _2F_1\left (\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+3);\frac {1}{2} (2 m+5);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)} \]
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Rubi [A] time = 0.37, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {2841, 2745, 2689, 70, 69} \[ \frac {c 2^{\frac {1}{2}-m} \cos ^3(e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2} \, _2F_1\left (\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+3);\frac {1}{2} (2 m+5);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 2689
Rule 2745
Rule 2841
Rubi steps
\begin {align*} \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m} \, dx &=\frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-m} \, dx}{a c}\\ &=\left (\cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 (1+m)}(e+f x) (c-c \sin (e+f x))^{-1-2 m} \, dx\\ &=\frac {\left (c^2 \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{m+\frac {1}{2} (-1-2 (1+m))} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 (1+m))}\right ) \operatorname {Subst}\left (\int (c-c x)^{-1-2 m+\frac {1}{2} (-1+2 (1+m))} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\left (2^{-\frac {1}{2}-m} c^2 \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-2 (1+m))} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 (1+m))}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-1-2 m+\frac {1}{2} (-1+2 (1+m))} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {2^{\frac {1}{2}-m} c \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2} (1+2 m),\frac {1}{2} (3+2 m);\frac {1}{2} (5+2 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)}\\ \end {align*}
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Mathematica [C] time = 29.65, size = 1045, normalized size = 9.17 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 1} \cos \left (f x + e\right )^{2}, x\right ) \]
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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 1} \cos \left (f x + e\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.35, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-1-m}\, dx \]
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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 1} \cos \left (f x + e\right )^{2}\,{d x} \]
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 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{m+1}} \,d x \]
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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