Optimal. Leaf size=190 \[ \frac{a \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}-\frac{b \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )} \]
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Rubi [A] time = 0.232686, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2823, 3189, 429} \[ \frac{a \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}-\frac{b \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 2823
Rule 3189
Rule 429
Rubi steps
\begin{align*} \int \frac{\cos ^m(c+d x)}{a+b \cos (c+d x)} \, dx &=a \int \frac{\cos ^m(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx-b \int \frac{\cos ^{1+m}(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx\\ &=\frac{\left (a \cos ^{2 \left (-\frac{1}{2}+\frac{m}{2}\right )}(c+d x) \cos ^2(c+d x)^{\frac{1}{2}-\frac{m}{2}}\right ) \operatorname{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 (c+d x)\right )}{d}-\frac{\left (b \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{m/2}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{a F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^{-1+m}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} \sin (c+d x)}{\left (a^2-b^2\right ) d}-\frac{b F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \sin (c+d x)}{\left (a^2-b^2\right ) d}\\ \end{align*}
Mathematica [B] time = 24.4947, size = 6703, normalized size = 35.28 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 1.111, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{m}}{a+b\cos \left ( dx+c \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 \frac{\cos \left (d x + c\right )^{m}}{b \cos \left (d x + c\right ) + a}\,{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 (\frac{\cos \left (d x + c\right )^{m}}{b \cos \left (d x + c\right ) + a}, 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 \frac{\cos \left (d x + c\right )^{m}}{b \cos \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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