Integrand size = 33, antiderivative size = 299 \[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=-\frac {(A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{m/2} (c \sec (e+f x))^{1+m} \sin (e+f x)}{\left (a^2-b^2\right ) c f}+\frac {a (A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1+m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos ^2(e+f x)^{\frac {1+m}{2}} (c \sec (e+f x))^{1+m} \sin (e+f x)}{b \left (a^2-b^2\right ) c f}-\frac {B c \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^{-1+m} \sin (e+f x)}{b f (1-m) \sqrt {\sin ^2(e+f x)}} \]
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Time = 0.80 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3039, 4123, 3857, 2722, 3954, 2902, 3268, 440} \[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=-\frac {(A b-a B) \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{m/2} (c \sec (e+f x))^{m+1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {m}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{c f \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (e+f x) \cos ^2(e+f x)^{\frac {m+1}{2}} (c \sec (e+f x))^{m+1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {m+1}{2},1,\frac {3}{2},\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{b c f \left (a^2-b^2\right )}-\frac {B c \sin (e+f x) (c \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{b f (1-m) \sqrt {\sin ^2(e+f x)}} \]
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Rule 440
Rule 2722
Rule 2902
Rule 3039
Rule 3268
Rule 3857
Rule 3954
Rule 4123
Rubi steps \begin{align*} \text {integral}& = \int \frac {(c \sec (e+f x))^m (B+A \sec (e+f x))}{b+a \sec (e+f x)} \, dx \\ & = \frac {B \int (c \sec (e+f x))^m \, dx}{b}+\frac {(A b-a B) \int \frac {(c \sec (e+f x))^{1+m}}{b+a \sec (e+f x)} \, dx}{b c} \\ & = \frac {\left (B \left (\frac {\cos (e+f x)}{c}\right )^m (c \sec (e+f x))^m\right ) \int \left (\frac {\cos (e+f x)}{c}\right )^{-m} \, dx}{b}+\frac {\left ((A b-a B) \cos ^{1+m}(e+f x) (c \sec (e+f x))^{1+m}\right ) \int \frac {\cos ^{-m}(e+f x)}{a+b \cos (e+f x)} \, dx}{b c} \\ & = -\frac {B \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{b f (1-m) \sqrt {\sin ^2(e+f x)}}-\frac {\left ((A b-a B) \cos ^{1+m}(e+f x) (c \sec (e+f x))^{1+m}\right ) \int \frac {\cos ^{1-m}(e+f x)}{a^2-b^2 \cos ^2(e+f x)} \, dx}{c}+\frac {\left (a (A b-a B) \cos ^{1+m}(e+f x) (c \sec (e+f x))^{1+m}\right ) \int \frac {\cos ^{-m}(e+f x)}{a^2-b^2 \cos ^2(e+f x)} \, dx}{b c} \\ & = -\frac {B \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{b f (1-m) \sqrt {\sin ^2(e+f x)}}+\frac {\left (a (A b-a B) \cos ^{1+2 \left (-\frac {1}{2}-\frac {m}{2}\right )+m}(e+f x) \cos ^2(e+f x)^{\frac {1}{2}+\frac {m}{2}} (c \sec (e+f x))^{1+m}\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 c f}-\frac {\left ((A b-a B) \cos (e+f x) \cos ^2(e+f x)^{m/2} (c \sec (e+f x))^{1+m}\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 )}{c 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 ) \cos (e+f x) \cos ^2(e+f x)^{m/2} (c \sec (e+f x))^{1+m} \sin (e+f x)}{\left (a^2-b^2\right ) c f}+\frac {a (A b-a B) \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 ) \cos ^2(e+f x)^{\frac {1+m}{2}} (c \sec (e+f x))^{1+m} \sin (e+f x)}{b \left (a^2-b^2\right ) c f}-\frac {B \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{b f (1-m) \sqrt {\sin ^2(e+f x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(10630\) vs. \(2(299)=598\).
Time = 32.02 (sec) , antiderivative size = 10630, normalized size of antiderivative = 35.55 \[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\text {Result too large to show} \]
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\[\int \frac {\left (A +\cos \left (f x +e \right ) B \right ) \left (c \sec \left (f x +e \right )\right )^{m}}{a +b \cos \left (f x +e \right )}d x\]
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\[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\int { \frac {{\left (B \cos \left (f x + e\right ) + A\right )} \left (c \sec \left (f x + e\right )\right )^{m}}{b \cos \left (f x + e\right ) + a} \,d x } \]
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\[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\int \frac {\left (c \sec {\left (e + f x \right )}\right )^{m} \left (A + B \cos {\left (e + f x \right )}\right )}{a + b \cos {\left (e + f x \right )}}\, dx \]
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\[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\int { \frac {{\left (B \cos \left (f x + e\right ) + A\right )} \left (c \sec \left (f x + e\right )\right )^{m}}{b \cos \left (f x + e\right ) + a} \,d x } \]
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Exception generated. \[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{a+b \cos (e+f x)} \, dx=\int \frac {{\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^m\,\left (A+B\,\cos \left (e+f\,x\right )\right )}{a+b\,\cos \left (e+f\,x\right )} \,d x \]
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