Optimal. Leaf size=564 \[ \frac{\sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} \left (a^2 b^2 (A (-m)+A+C (m+2))+a^3 b B m+a^4 (-C) (m+1)-a b^3 B (m+1)+A b^4 m\right ) F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \left (a^2 b^2 (A (-m)+A+C (m+2))+a^3 b B m+a^4 (-C) (m+1)-a b^3 B (m+1)+A b^4 m\right ) F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{a b d \left (a^2-b^2\right )^2}+\frac{\sin (c+d x) \cos ^{m+1}(c+d x) \left (a^2 (-C) (m+1)+a b B m+b^2 (C-A m)\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{b^2 d (m+1) \left (a^2-b^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{(m+1) \sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{a b d (m+2) \left (a^2-b^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
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Rubi [A] time = 1.02937, antiderivative size = 564, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3055, 3063, 2643, 2823, 3189, 429} \[ \frac{\sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac{1-m}{2}} \left (a^2 b^2 (A (-m)+A+C (m+2))+a^3 b B m+a^4 (-C) (m+1)-a b^3 B (m+1)+A b^4 m\right ) F_1\left (\frac{1}{2};\frac{1-m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )^2}-\frac{\sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \left (a^2 b^2 (A (-m)+A+C (m+2))+a^3 b B m+a^4 (-C) (m+1)-a b^3 B (m+1)+A b^4 m\right ) F_1\left (\frac{1}{2};-\frac{m}{2},1;\frac{3}{2};\sin ^2(c+d x),-\frac{b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{a b d \left (a^2-b^2\right )^2}+\frac{\sin (c+d x) \cos ^{m+1}(c+d x) \left (a^2 (-C) (m+1)+a b B m+b^2 (C-A m)\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{b^2 d (m+1) \left (a^2-b^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{(m+1) \sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{a b d (m+2) \left (a^2-b^2\right ) \sqrt{\sin ^2(c+d x)}}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 3055
Rule 3063
Rule 2643
Rule 2823
Rule 3189
Rule 429
Rubi steps
\begin{align*} \int \frac{\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\cos ^m(c+d x) \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)-a (A b-a B+b C) \cos (c+d x)-\left (A b^2-a (b B-a C)\right ) (1+m) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (\left (A b^2-a (b B-a C)\right ) (1+m)\right ) \int \cos ^{1+m}(c+d x) \, dx}{a b \left (a^2-b^2\right )}-\frac{\left (a b B m-a^2 C (1+m)+b^2 (C-A m)\right ) \int \cos ^m(c+d x) \, dx}{b^2 \left (a^2-b^2\right )}+\frac{\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \int \frac{\cos ^m(c+d x)}{a+b \cos (c+d x)} \, dx}{a b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a b B m-a^2 C (1+m)+b^2 (C-A m)\right ) \cos ^{1+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d (1+m) \sqrt{\sin ^2(c+d x)}}+\frac{\left (A b^2-a (b B-a C)\right ) (1+m) \cos ^{2+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{a b \left (a^2-b^2\right ) d (2+m) \sqrt{\sin ^2(c+d x)}}+\frac{\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \int \frac{\cos ^m(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx}{b^2 \left (a^2-b^2\right )}-\frac{\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \int \frac{\cos ^{1+m}(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx}{a b \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a b B m-a^2 C (1+m)+b^2 (C-A m)\right ) \cos ^{1+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d (1+m) \sqrt{\sin ^2(c+d x)}}+\frac{\left (A b^2-a (b B-a C)\right ) (1+m) \cos ^{2+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{a b \left (a^2-b^2\right ) d (2+m) \sqrt{\sin ^2(c+d x)}}+\frac{\left (\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \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 )}{b^2 \left (a^2-b^2\right ) d}-\frac{\left (\left (a^2 b (A b-a B+b C)-a^2 \left (A b^2-a (b B-a C)\right ) (1+m)+b^2 \left (A b^2 m-a b B (1+m)+a^2 (A+C+C m)\right )\right ) \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 )}{a b \left (a^2-b^2\right ) d}\\ &=\frac{\left (A b^4 m+a^3 b B m-a b^3 B (1+m)-a^4 C (1+m)+a^2 b^2 (A-A m+C (2+m))\right ) 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)}{b^2 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^4 m+a^3 b B m-a b^3 B (1+m)-a^4 C (1+m)+a^2 b^2 (A (1-m)+C (2+m))\right ) 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)}{a b \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a b B m-a^2 C (1+m)+b^2 (C-A m)\right ) \cos ^{1+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d (1+m) \sqrt{\sin ^2(c+d x)}}+\frac{\left (A b^2-a (b B-a C)\right ) (1+m) \cos ^{2+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{a b \left (a^2-b^2\right ) d (2+m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}
Mathematica [B] time = 35.4635, size = 25789, normalized size = 45.73 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.608, 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 ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) }{ \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{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \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{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \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{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \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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