Optimal. Leaf size=282 \[ -\frac {b \left (3 a^2 (3-m)+b^2 (2-m)\right ) \sin (c+d x) \sec ^{m-4}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {4-m}{2};\frac {6-m}{2};\cos ^2(c+d x)\right )}{d (2-m) (4-m) \sqrt {\sin ^2(c+d x)}}-\frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {3-m}{2};\frac {5-m}{2};\cos ^2(c+d x)\right )}{d (1-m) (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}-\frac {a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (1-m) (2-m)} \]
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Rubi [A] time = 0.42, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3238, 3842, 4047, 3772, 2643, 4046} \[ -\frac {b \left (3 a^2 (3-m)+b^2 (2-m)\right ) \sin (c+d x) \sec ^{m-4}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {4-m}{2};\frac {6-m}{2};\cos ^2(c+d x)\right )}{d (2-m) (4-m) \sqrt {\sin ^2(c+d x)}}-\frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {3-m}{2};\frac {5-m}{2};\cos ^2(c+d x)\right )}{d (1-m) (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}-\frac {a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (1-m) (2-m)} \]
Antiderivative was successfully verified.
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Rule 2643
Rule 3238
Rule 3772
Rule 3842
Rule 4046
Rule 4047
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx &=\int \sec ^{-3+m}(c+d x) (b+a \sec (c+d x))^3 \, dx\\ &=-\frac {a^2 \sec ^{-2+m}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{d (1-m)}+\frac {\int \sec ^{-3+m}(c+d x) \left (-b \left (b^2 (1-m)+a^2 (3-m)\right )-a \left (3 b^2 (1-m)+a^2 (2-m)\right ) \sec (c+d x)-a^2 b (1-2 m) \sec ^2(c+d x)\right ) \, dx}{-1+m}\\ &=-\frac {a^2 \sec ^{-2+m}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{d (1-m)}+\left (a \left (3 b^2+\frac {a^2 (2-m)}{1-m}\right )\right ) \int \sec ^{-2+m}(c+d x) \, dx+\frac {\int \sec ^{-3+m}(c+d x) \left (-b \left (b^2 (1-m)+a^2 (3-m)\right )-a^2 b (1-2 m) \sec ^2(c+d x)\right ) \, dx}{-1+m}\\ &=-\frac {a^2 b (1-2 m) \sec ^{-2+m}(c+d x) \sin (c+d x)}{d (1-m) (2-m)}-\frac {a^2 \sec ^{-2+m}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{d (1-m)}+\left (b \left (b^2+\frac {3 a^2 (3-m)}{2-m}\right )\right ) \int \sec ^{-3+m}(c+d x) \, dx+\left (a \left (3 b^2+\frac {a^2 (2-m)}{1-m}\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{2-m}(c+d x) \, dx\\ &=-\frac {a^2 b (1-2 m) \sec ^{-2+m}(c+d x) \sin (c+d x)}{d (1-m) (2-m)}-\frac {a^2 \sec ^{-2+m}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{d (1-m)}-\frac {a \left (3 b^2+\frac {a^2 (2-m)}{1-m}\right ) \, _2F_1\left (\frac {1}{2},\frac {3-m}{2};\frac {5-m}{2};\cos ^2(c+d x)\right ) \sec ^{-3+m}(c+d x) \sin (c+d x)}{d (3-m) \sqrt {\sin ^2(c+d x)}}+\left (b \left (b^2+\frac {3 a^2 (3-m)}{2-m}\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{3-m}(c+d x) \, dx\\ &=-\frac {a^2 b (1-2 m) \sec ^{-2+m}(c+d x) \sin (c+d x)}{d (1-m) (2-m)}-\frac {a^2 \sec ^{-2+m}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{d (1-m)}-\frac {b \left (b^2+\frac {3 a^2 (3-m)}{2-m}\right ) \, _2F_1\left (\frac {1}{2},\frac {4-m}{2};\frac {6-m}{2};\cos ^2(c+d x)\right ) \sec ^{-4+m}(c+d x) \sin (c+d x)}{d (4-m) \sqrt {\sin ^2(c+d x)}}-\frac {a \left (3 b^2+\frac {a^2 (2-m)}{1-m}\right ) \, _2F_1\left (\frac {1}{2},\frac {3-m}{2};\frac {5-m}{2};\cos ^2(c+d x)\right ) \sec ^{-3+m}(c+d x) \sin (c+d x)}{d (3-m) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.88, size = 222, normalized size = 0.79 \[ \frac {\sqrt {-\tan ^2(c+d x)} \csc (c+d x) \sec ^{m-4}(c+d x) \left (\frac {1}{2} a (m-3) \sec ^3(c+d x) \left (2 a (m-2) \left (a (m-1) \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};\sec ^2(c+d x)\right )+3 b m \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m-1}{2};\frac {m+1}{2};\sec ^2(c+d x)\right )\right )+6 b^2 (m-1) m \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m-2}{2};\frac {m}{2};\sec ^2(c+d x)\right )\right )+b^3 m \left (m^2-3 m+2\right ) \, _2F_1\left (\frac {1}{2},\frac {m-3}{2};\frac {m-1}{2};\sec ^2(c+d x)\right )\right )}{d (m-3) (m-2) (m-1) m} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}\right )} \sec \left (d x + c\right )^{m}, x\right ) \]
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 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.51, size = 0, normalized size = 0.00 \[ \int \left (a +b \cos \left (d x +c \right )\right )^{3} \left (\sec ^{m}\left (d x +c \right )\right )\, dx \]
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 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m}\,{d x} \]
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 \[ \int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \]
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 \[ \int \left (a + b \cos {\left (c + d x \right )}\right )^{3} \sec ^{m}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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