Optimal. Leaf size=173 \[ \frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}-\frac {b (C (1-n)-A n) (b \cos (c+d x))^{-1+n} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-1+n);\frac {1+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-n) n \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}} \]
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Rubi [A]
time = 0.14, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 3102, 2827,
2722} \begin {gather*} -\frac {b (C (1-n)-A n) \sin (c+d x) (b \cos (c+d x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {n-1}{2};\frac {n+1}{2};\cos ^2(c+d x)\right )}{d (1-n) n \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}}+\frac {b C \sin (c+d x) (b \cos (c+d x))^{n-1}}{d n} \end {gather*}
Antiderivative was successfully verified.
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Rule 16
Rule 2722
Rule 2827
Rule 3102
Rubi steps
\begin {align*} \int (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=b^2 \int (b \cos (c+d x))^{-2+n} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}+\frac {b \int (b \cos (c+d x))^{-2+n} (-b (C (1-n)-A n)+b B n \cos (c+d x)) \, dx}{n}\\ &=\frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}+(b B) \int (b \cos (c+d x))^{-1+n} \, dx-\frac {\left (b^2 (C (1-n)-A n)\right ) \int (b \cos (c+d x))^{-2+n} \, dx}{n}\\ &=\frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}-\frac {b (C (1-n)-A n) (b \cos (c+d x))^{-1+n} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-1+n);\frac {1+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-n) n \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 0.32, size = 131, normalized size = 0.76 \begin {gather*} \frac {(b \cos (c+d x))^n \left ((C-A n-C n) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-1+n);\frac {1+n}{2};\cos ^2(c+d x)\right )-(-1+n) \left (B \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\cos ^2(c+d x)\right )-C \sqrt {\sin ^2(c+d x)}\right )\right ) \tan (c+d x)}{d (-1+n) n \sqrt {\sin ^2(c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.27, size = 0, normalized size = 0.00 \[\int \left (b \cos \left (d x +c \right )\right )^{n} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{2}\left (d x +c \right )\right )\, dx\]
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 \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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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