Optimal. Leaf size=303 \[ \frac{\sqrt{2} \sin (e+f x) \left (a^2 C-a b B (m+2)+A b^2 (m+2)+b^2 C (m+1)\right ) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\cos (e+f x)+1}}-\frac{\sqrt{2} (a+b) \sin (e+f x) (a C-b B (m+2)) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\cos (e+f x)+1}}+\frac{C \sin (e+f x) (a+b \cos (e+f x))^{m+1}}{b f (m+2)} \]
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Rubi [A] time = 0.375155, antiderivative size = 303, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {3023, 2756, 2665, 139, 138} \[ \frac{\sqrt{2} \sin (e+f x) \left (a^2 C-a b B (m+2)+A b^2 (m+2)+b^2 C (m+1)\right ) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\cos (e+f x)+1}}-\frac{\sqrt{2} (a+b) \sin (e+f x) (a C-b B (m+2)) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\cos (e+f x)+1}}+\frac{C \sin (e+f x) (a+b \cos (e+f x))^{m+1}}{b f (m+2)} \]
Antiderivative was successfully verified.
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Rule 3023
Rule 2756
Rule 2665
Rule 139
Rule 138
Rubi steps
\begin{align*} \int (a+b \cos (e+f x))^m \left (A+B \cos (e+f x)+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{\int (a+b \cos (e+f x))^m (b (C (1+m)+A (2+m))-(a C-b B (2+m)) \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+b B (2+m)) \int (a+b \cos (e+f x))^{1+m} \, dx}{b^2 (2+m)}+\frac{\left (b^2 (C (1+m)+A (2+m))-a (-a C+b B (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+b B (2+m)) \sin (e+f x)) \operatorname{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 (b^2 (C (1+m)+A (2+m))-a (-a C+b B (2+m))\right ) \sin (e+f x)\right ) \operatorname{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-b) (-a C+b B (2+m)) (a+b \cos (e+f x))^m \left (-\frac{a+b \cos (e+f x)}{-a-b}\right )^{-m} \sin (e+f x)\right ) \operatorname{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 (b^2 (C (1+m)+A (2+m))-a (-a C+b B (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 ) \operatorname{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+b) (a C-b B (2+m)) F_1\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 b^2 (2+m)-a b B (2+m)\right ) F_1\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*}
Mathematica [B] time = 26.8395, size = 16189, normalized size = 53.43 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 1.438, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\cos \left ( fx+e \right ) \right ) ^{m} \left ( A+B\cos \left ( fx+e \right ) +C \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (f x + e\right )^{2} + B \cos \left (f x + e\right ) + A\right )}{\left (b \cos \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (f x + e\right )^{2} + B \cos \left (f x + e\right ) + A\right )}{\left (b \cos \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (f x + e\right )^{2} + B \cos \left (f x + e\right ) + A\right )}{\left (b \cos \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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