Optimal. Leaf size=353 \[ \frac {a \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} F_1\left (\frac {1}{2};\frac {1-m}{2},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac {1}{2};-\frac {m}{2},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b d \left (a^2-b^2\right )}+\frac {a C \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{b^2 d (m+1) \sqrt {\sin ^2(c+d x)}}-\frac {C \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{b d (m+2) \sqrt {\sin ^2(c+d x)}} \]
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Rubi [A] time = 0.46, antiderivative size = 353, normalized size of antiderivative = 1.00, 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 = {3064, 2643, 2823, 3189, 429} \[ \frac {a \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{m-1}(c+d x) \cos ^2(c+d x)^{\frac {1-m}{2}} F_1\left (\frac {1}{2};\frac {1-m}{2},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b^2 d \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^m(c+d x) \cos ^2(c+d x)^{-m/2} F_1\left (\frac {1}{2};-\frac {m}{2},1;\frac {3}{2};\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{b d \left (a^2-b^2\right )}+\frac {a C \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{b^2 d (m+1) \sqrt {\sin ^2(c+d x)}}-\frac {C \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{b d (m+2) \sqrt {\sin ^2(c+d x)}} \]
Antiderivative was successfully verified.
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Rule 429
Rule 2643
Rule 2823
Rule 3064
Rule 3189
Rubi steps
\begin {align*} \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=-\frac {(a C) \int \cos ^m(c+d x) \, dx}{b^2}+\frac {C \int \cos ^{1+m}(c+d x) \, dx}{b}+\left (A+\frac {a^2 C}{b^2}\right ) \int \frac {\cos ^m(c+d x)}{a+b \cos (c+d x)} \, dx\\ &=\frac {a C \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 d (1+m) \sqrt {\sin ^2(c+d x)}}-\frac {C \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)}{b d (2+m) \sqrt {\sin ^2(c+d x)}}+\left (b \left (-A-\frac {a^2 C}{b^2}\right )\right ) \int \frac {\cos ^{1+m}(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx+\left (a \left (A+\frac {a^2 C}{b^2}\right )\right ) \int \frac {\cos ^m(c+d x)}{a^2-b^2 \cos ^2(c+d x)} \, dx\\ &=\frac {a C \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 d (1+m) \sqrt {\sin ^2(c+d x)}}-\frac {C \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)}{b d (2+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (a \left (A+\frac {a^2 C}{b^2}\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 )}{d}+\frac {\left (b \left (-A-\frac {a^2 C}{b^2}\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 )}{d}\\ &=\frac {a \left (A+\frac {a^2 C}{b^2}\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)}{\left (a^2-b^2\right ) d}-\frac {\left (A b^2+a^2 C\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)}{b \left (a^2-b^2\right ) d}+\frac {a C \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 d (1+m) \sqrt {\sin ^2(c+d x)}}-\frac {C \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)}{b d (2+m) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}
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Mathematica [B] time = 28.15, size = 10459, normalized size = 29.63 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{b \cos \left (d x + c\right ) + a}, x\right ) \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{b \cos \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 3.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{m}\left (d x +c \right )\right ) \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{a +b \cos \left (d x +c \right )}\, dx \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{m}}{b \cos \left (d x + c\right ) + a}\,{d x} \]
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 \[ \int \frac {{\cos \left (c+d\,x\right )}^m\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \]
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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