Optimal. Leaf size=564 \[ \frac {\sin (c+d x) \cos ^{m+1}(c+d x) \left (a^2 (-C) (m+1)+a b B m+b^2 (C-A m)\right ) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{b^2 d (m+1) \left (a^2-b^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {(m+1) \sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{a b d (m+2) \left (a^2-b^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right ) \cos ^{m+1}(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} \left (a^4 (-C) (m+1)+a^3 b B m+a^2 b^2 (A (-m)+A+C (m+2))-a b^3 B (m+1)+A b^4 m\right ) F_1\left (\frac {1}{2};\frac {1-m}{2},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )^2}-\frac {\sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} \left (a^4 (-C) (m+1)+a^3 b B m+a^2 b^2 (A (-m)+A+C (m+2))-a b^3 B (m+1)+A b^4 m\right ) F_1\left (\frac {1}{2};-\frac {m}{2},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{a b d \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 1.03, antiderivative size = 564, normalized size of antiderivative = 1.00, 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 429
Rule 2643
Rule 2823
Rule 3055
Rule 3063
Rule 3189
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*}
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Mathematica [B] time = 45.63, size = 12349, normalized size = 21.90 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{m}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.38, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{m}\left (d x +c \right )\right ) \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (a +b \cos \left (d x +c \right )\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{m}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^m\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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