Integrand size = 33, antiderivative size = 286 \[ \int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx=\frac {a (A b-a B) c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) (c \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} \sin (e+f x)}{b \left (a^2-b^2\right ) f}-\frac {(A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) (c \cos (e+f x))^m \cos ^2(e+f x)^{-m/2} \sin (e+f x)}{\left (a^2-b^2\right ) f}-\frac {B (c \cos (e+f x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{b c f (1+m) \sqrt {\sin ^2(e+f x)}} \]
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Time = 0.48 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3081, 2722, 2902, 3268, 440} \[ \int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx=\frac {a c (A b-a B) \sin (e+f x) \cos ^2(e+f x)^{\frac {1-m}{2}} (c \cos (e+f x))^{m-1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{b f \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (e+f x) \cos ^2(e+f x)^{-m/2} (c \cos (e+f x))^m \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {B \sin (e+f x) (c \cos (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(e+f x)\right )}{b c f (m+1) \sqrt {\sin ^2(e+f x)}} \]
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Rule 440
Rule 2722
Rule 2902
Rule 3081
Rule 3268
Rubi steps \begin{align*} \text {integral}& = \frac {B \int (c \cos (e+f x))^m \, dx}{b}-\frac {(-A b+a B) \int \frac {(c \cos (e+f x))^m}{a+b \cos (e+f x)} \, dx}{b} \\ & = -\frac {B (c \cos (e+f x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{b c f (1+m) \sqrt {\sin ^2(e+f x)}}+\frac {(a (A b-a B)) \int \frac {(c \cos (e+f x))^m}{a^2-b^2 \cos ^2(e+f x)} \, dx}{b}-\frac {(A b-a B) \int \frac {(c \cos (e+f x))^{1+m}}{a^2-b^2 \cos ^2(e+f x)} \, dx}{c} \\ & = -\frac {B (c \cos (e+f x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{b c f (1+m) \sqrt {\sin ^2(e+f x)}}+\frac {\left (a (A b-a B) c (c \cos (e+f x))^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )} \cos ^2(e+f x)^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1}{2} (-1+m)}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (e+f x)\right )}{b f}-\frac {\left ((A b-a B) (c \cos (e+f x))^m \cos ^2(e+f x)^{-m/2}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{m/2}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {a (A b-a B) c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) (c \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} \sin (e+f x)}{b \left (a^2-b^2\right ) f}-\frac {(A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) (c \cos (e+f x))^m \cos ^2(e+f x)^{-m/2} \sin (e+f x)}{\left (a^2-b^2\right ) f}-\frac {B (c \cos (e+f x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{b c f (1+m) \sqrt {\sin ^2(e+f x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(10482\) vs. \(2(286)=572\).
Time = 30.99 (sec) , antiderivative size = 10482, normalized size of antiderivative = 36.65 \[ \int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx=\text {Result too large to show} \]
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\[\int \frac {\left (c \cos \left (f x +e \right )\right )^{m} \left (A +\cos \left (f x +e \right ) B \right )}{a +b \cos \left (f x +e \right )}d x\]
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\[ \int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx=\int { \frac {{\left (B \cos \left (f x + e\right ) + A\right )} \left (c \cos \left (f x + e\right )\right )^{m}}{b \cos \left (f x + e\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx=\int { \frac {{\left (B \cos \left (f x + e\right ) + A\right )} \left (c \cos \left (f x + e\right )\right )^{m}}{b \cos \left (f x + e\right ) + a} \,d x } \]
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Exception generated. \[ \int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx=\int \frac {{\left (c\,\cos \left (e+f\,x\right )\right )}^m\,\left (A+B\,\cos \left (e+f\,x\right )\right )}{a+b\,\cos \left (e+f\,x\right )} \,d x \]
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