Optimal. Leaf size=179 \[ \frac {b^2 \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}-\frac {\left (b^2 (1+m)+a^2 (2+m)\right ) \cos ^{1+m}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+m) (2+m) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \cos ^{2+m}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) \sqrt {\sin ^2(c+d x)}} \]
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Rubi [A]
time = 0.08, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2868, 2722,
3093} \begin {gather*} -\frac {\left (a^2 (m+2)+b^2 (m+1)\right ) \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 )}{d (m+1) (m+2) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \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) \sqrt {\sin ^2(c+d x)}}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 2722
Rule 2868
Rule 3093
Rubi steps
\begin {align*} \int \cos ^m(c+d x) (a+b \cos (c+d x))^2 \, dx &=(2 a b) \int \cos ^{1+m}(c+d x) \, dx+\int \cos ^m(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac {b^2 \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}-\frac {2 a b \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 (2+m) \sqrt {\sin ^2(c+d x)}}+\left (a^2+\frac {b^2 (1+m)}{2+m}\right ) \int \cos ^m(c+d x) \, dx\\ &=\frac {b^2 \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}-\frac {\left (a^2+\frac {b^2 (1+m)}{2+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) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \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 (2+m) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 0.37, size = 168, normalized size = 0.94 \begin {gather*} -\frac {\cos ^{1+m}(c+d x) \csc (c+d x) \left (a^2 \left (6+5 m+m^2\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\cos ^2(c+d x)\right )+b (1+m) \cos (c+d x) \left (2 a (3+m) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};\cos ^2(c+d x)\right )+b (2+m) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {3+m}{2};\frac {5+m}{2};\cos ^2(c+d x)\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (1+m) (2+m) (3+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int \left (\cos ^{m}\left (d x +c \right )\right ) \left (a +b \cos \left (d x +c \right )\right )^{2}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \cos ^{m}{\left (c + d x \right )}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^m\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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