Optimal. Leaf size=294 \[ \frac{b^2 \sin (c+d x) \cos ^{m+1}(c+d x) \cos ^2(c+d x)^{\frac{1}{2} (-m-1)} F_1\left (\frac{1}{2};\frac{1}{2} (-m-1),2;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}+\frac{a^2 \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} F_1\left (\frac{1}{2};\frac{1-m}{2},2;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}-\frac{2 a b \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac{1}{2};-\frac{m}{2},2;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.35351, antiderivative size = 294, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2824, 3189, 429} \[ \frac{b^2 \sin (c+d x) \cos ^{m+1}(c+d x) \cos ^2(c+d x)^{\frac{1}{2} (-m-1)} F_1\left (\frac{1}{2};\frac{1}{2} (-m-1),2;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}+\frac{a^2 \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} F_1\left (\frac{1}{2};\frac{1-m}{2},2;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2}-\frac{2 a b \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac{1}{2};-\frac{m}{2},2;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 2824
Rule 3189
Rule 429
Rubi steps
\begin{align*} \int \frac{\cos ^m(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\int \left (\frac{a^2 \cos ^m(c+d x)}{\left (a^2-b^2 \cos ^2(c+d x)\right )^2}-\frac{2 a b \cos ^{1+m}(c+d x)}{\left (a^2-b^2 \cos ^2(c+d x)\right )^2}+\frac{b^2 \cos ^{2+m}(c+d x)}{\left (-a^2+b^2 \cos ^2(c+d x)\right )^2}\right ) \, dx\\ &=a^2 \int \frac{\cos ^m(c+d x)}{\left (a^2-b^2 \cos ^2(c+d x)\right )^2} \, dx-(2 a b) \int \frac{\cos ^{1+m}(c+d x)}{\left (a^2-b^2 \cos ^2(c+d x)\right )^2} \, dx+b^2 \int \frac{\cos ^{2+m}(c+d x)}{\left (-a^2+b^2 \cos ^2(c+d x)\right )^2} \, dx\\ &=\frac{\left (b^2 \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+m}{2}}}{\left (-a^2+b^2-b^2 x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}+\frac{\left (a^2 \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)}}{\left (a^2-b^2+b^2 x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac{\left (2 a 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}}{\left (a^2-b^2+b^2 x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{b^2 F_1\left (\frac{1}{2};\frac{1}{2} (-1-m),2;\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}{2} (-1-m)} \sin (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac{a^2 F_1\left (\frac{1}{2};\frac{1-m}{2},2;\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 )^2 d}-\frac{2 a b F_1\left (\frac{1}{2};-\frac{m}{2},2;\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 )^2 d}\\ \end{align*}
Mathematica [B] time = 26.08, size = 7214, normalized size = 24.54 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.606, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{m}}{ \left ( a+b\cos \left ( dx+c \right ) \right ) ^{2}}}\, 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}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{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^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, 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}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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