Optimal. Leaf size=318 \[ -\frac {\sin (c+d x) \left (a^2 (m+4) (A (m+2)+C (m+1))+b^2 (m+1) (A (m+4)+C (m+3))\right ) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) (m+4) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \left (2 a^2 C+b^2 (A (m+4)+C (m+3))\right ) \cos ^{m+1}(c+d x)}{d (m+2) (m+4)}-\frac {2 a b (A (m+3)+C (m+2)) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {C \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}+\frac {2 a b C \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3) (m+4)} \]
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Rubi [A] time = 0.86, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3050, 3033, 3023, 2748, 2643} \[ -\frac {\sin (c+d x) \left (a^2 (m+4) (A (m+2)+C (m+1))+b^2 (m+1) (A (m+4)+C (m+3))\right ) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) (m+2) (m+4) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \left (2 a^2 C+b^2 (A (m+4)+C (m+3))\right ) \cos ^{m+1}(c+d x)}{d (m+2) (m+4)}-\frac {2 a b (A (m+3)+C (m+2)) \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {C \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}+\frac {2 a b C \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3) (m+4)} \]
Antiderivative was successfully verified.
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Rule 2643
Rule 2748
Rule 3023
Rule 3033
Rule 3050
Rubi steps
\begin {align*} \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}+\frac {\int \cos ^m(c+d x) (a+b \cos (c+d x)) \left (a (C (1+m)+A (4+m))+b (C (3+m)+A (4+m)) \cos (c+d x)+2 a C \cos ^2(c+d x)\right ) \, dx}{4+m}\\ &=\frac {2 a b C \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}+\frac {\int \cos ^m(c+d x) \left (a^2 (3+m) (C (1+m)+A (4+m))+2 a b (4+m) (C (2+m)+A (3+m)) \cos (c+d x)+(3+m) \left (2 a^2 C+b^2 (C (3+m)+A (4+m))\right ) \cos ^2(c+d x)\right ) \, dx}{12+7 m+m^2}\\ &=\frac {\left (2 a^2 C+b^2 C (3+m)+A b^2 (4+m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {2 a b C \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}+\frac {\int \cos ^m(c+d x) \left ((3+m) \left (a^2 (4+m) (C (1+m)+A (2+m))+b^2 (1+m) (C (3+m)+A (4+m))\right )+2 a b (2+m) (4+m) (C (2+m)+A (3+m)) \cos (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=\frac {\left (2 a^2 C+b^2 C (3+m)+A b^2 (4+m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {2 a b C \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}+\frac {(2 a b (2+m) (4+m) (C (2+m)+A (3+m))) \int \cos ^{1+m}(c+d x) \, dx}{24+26 m+9 m^2+m^3}+\frac {\left (a^2 (4+m) (C (1+m)+A (2+m))+b^2 (1+m) (C (3+m)+A (4+m))\right ) \int \cos ^m(c+d x) \, dx}{8+6 m+m^2}\\ &=\frac {\left (2 a^2 C+b^2 C (3+m)+A b^2 (4+m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {2 a b C \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {C \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}-\frac {\left (a^2 (4+m) (C (1+m)+A (2+m))+b^2 (1+m) (C (3+m)+A (4+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)}{d (1+m) \left (8+6 m+m^2\right ) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b (C (2+m)+A (3+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)}{d \left (6+5 m+m^2\right ) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}
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Mathematica [A] time = 2.29, size = 250, normalized size = 0.79 \[ \frac {\sin (c+d x) \cos ^{m+1}(c+d x) \left (\cos (c+d x) \left (\cos (c+d x) \left (b C \cos (c+d x) \left (-\frac {2 a \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};\cos ^2(c+d x)\right )}{m+4}-\frac {b \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+5}{2};\frac {m+7}{2};\cos ^2(c+d x)\right )}{m+5}\right )-\frac {\left (a^2 C+A b^2\right ) \, _2F_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};\cos ^2(c+d x)\right )}{m+3}\right )-\frac {2 a A b \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\cos ^2(c+d x)\right )}{m+2}\right )-\frac {a^2 A \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(c+d x)\right )}{m+1}\right )}{d \sqrt {\sin ^2(c+d x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + 2 \, C a b \cos \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) + A a^{2} + {\left (C a^{2} + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \cos \left (d x + c\right )^{m}, x\right ) \]
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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.21, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{m}\left (d x +c \right )\right ) \left (a +b \cos \left (d x +c \right )\right )^{2} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]
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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m}\,{d x} \]
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 \[ \int {\cos \left (c+d\,x\right )}^m\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \]
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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