Integrand size = 25, antiderivative size = 285 \[ \int (a+b \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=\frac {C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}-\frac {\sqrt {2} a (a+b) C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{b^2 f (2+m) \sqrt {1+\cos (e+f x)}}+\frac {\sqrt {2} \left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{b^2 f (2+m) \sqrt {1+\cos (e+f x)}} \]
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Time = 0.56 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3103, 2835, 2744, 144, 143} \[ \int (a+b \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=\frac {\sqrt {2} \sin (e+f x) \left (a^2 C+b^2 (A (m+2)+C (m+1))\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\cos (e+f x)+1}}-\frac {\sqrt {2} a C (a+b) \sin (e+f x) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m-1,\frac {3}{2},\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\cos (e+f x)+1}}+\frac {C \sin (e+f x) (a+b \cos (e+f x))^{m+1}}{b f (m+2)} \]
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Rule 143
Rule 144
Rule 2744
Rule 2835
Rule 3103
Rubi steps \begin{align*} \text {integral}& = \frac {C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}+\frac {\int (a+b \cos (e+f x))^m (b (C (1+m)+A (2+m))-a C \cos (e+f x)) \, dx}{b (2+m)} \\ & = \frac {C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}-\frac {(a C) \int (a+b \cos (e+f x))^{1+m} \, dx}{b^2 (2+m)}+\frac {\left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) \int (a+b \cos (e+f x))^m \, dx}{b^2 (2+m)} \\ & = \frac {C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}+\frac {(a C \sin (e+f x)) \text {Subst}\left (\int \frac {(a+b x)^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\cos (e+f x)} \sqrt {1+\cos (e+f x)}}-\frac {\left (\left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) \sin (e+f x)\right ) \text {Subst}\left (\int \frac {(a+b x)^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\cos (e+f x)} \sqrt {1+\cos (e+f x)}} \\ & = \frac {C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}-\frac {\left (a (-a-b) C (a+b \cos (e+f x))^m \left (-\frac {a+b \cos (e+f x)}{-a-b}\right )^{-m} \sin (e+f x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\cos (e+f x)} \sqrt {1+\cos (e+f x)}}-\frac {\left (\left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) (a+b \cos (e+f x))^m \left (-\frac {a+b \cos (e+f x)}{-a-b}\right )^{-m} \sin (e+f x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\cos (e+f x)} \sqrt {1+\cos (e+f x)}} \\ & = \frac {C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}-\frac {\sqrt {2} a (a+b) C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{b^2 f (2+m) \sqrt {1+\cos (e+f x)}}+\frac {\sqrt {2} \left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{b^2 f (2+m) \sqrt {1+\cos (e+f x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(10805\) vs. \(2(285)=570\).
Time = 25.94 (sec) , antiderivative size = 10805, normalized size of antiderivative = 37.91 \[ \int (a+b \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=\text {Result too large to show} \]
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\[\int \left (a +b \cos \left (f x +e \right )\right )^{m} \left (A +C \left (\cos ^{2}\left (f x +e \right )\right )\right )d x\]
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\[ \int (a+b \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=\int { {\left (C \cos \left (f x + e\right )^{2} + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (a+b \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=\text {Timed out} \]
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\[ \int (a+b \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=\int { {\left (C \cos \left (f x + e\right )^{2} + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (a+b \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=\int { {\left (C \cos \left (f x + e\right )^{2} + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (a+b \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=\int \left (C\,{\cos \left (e+f\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (e+f\,x\right )\right )}^m \,d x \]
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