Optimal. Leaf size=132 \[ -\frac {b^3 (A (1-n)+C (2-n)) \, _2F_1\left (\frac {1}{2},\frac {3-n}{2};\frac {5-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-3+n} \sin (c+d x)}{d (1-n) (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)} \]
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Rubi [A]
time = 0.10, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {16, 4131, 3857,
2722} \begin {gather*} -\frac {b^3 (A (1-n)+C (2-n)) \sin (c+d x) (b \sec (c+d x))^{n-3} \, _2F_1\left (\frac {1}{2},\frac {3-n}{2};\frac {5-n}{2};\cos ^2(c+d x)\right )}{d (1-n) (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C \tan (c+d x) (b \sec (c+d x))^{n-2}}{d (1-n)} \end {gather*}
Antiderivative was successfully verified.
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Rule 16
Rule 2722
Rule 3857
Rule 4131
Rubi steps
\begin {align*} \int \cos ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx &=b^2 \int (b \sec (c+d x))^{-2+n} \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=-\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)}+\left (b^2 \left (A+\frac {C (2-n)}{1-n}\right )\right ) \int (b \sec (c+d x))^{-2+n} \, dx\\ &=-\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)}+\left (b^2 \left (A+\frac {C (2-n)}{1-n}\right ) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{2-n} \, dx\\ &=-\frac {\left (A+\frac {C (2-n)}{1-n}\right ) \cos ^3(c+d x) \, _2F_1\left (\frac {1}{2},\frac {3-n}{2};\frac {5-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^2 C (b \sec (c+d x))^{-2+n} \tan (c+d x)}{d (1-n)}\\ \end {align*}
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Mathematica [A]
time = 0.13, size = 107, normalized size = 0.81 \begin {gather*} \frac {\cot (c+d x) \left (A n \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-2+n);\frac {n}{2};\sec ^2(c+d x)\right )+C (-2+n) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (-2+n) n} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.61, size = 0, normalized size = 0.00 \[\int \left (\cos ^{2}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]
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 \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 (b \sec {\left (c + d x \right )}\right )^{n} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\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 )}^2\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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