Optimal. Leaf size=66 \[ \frac {i (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n} \, _2F_1\left (2,n;n+1;\frac {1}{2} (1-i \tan (e+f x))\right )}{4 a f n} \]
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Rubi [A] time = 0.18, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3505, 3522, 3487, 68} \[ \frac {i (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n} \text {Hypergeometric2F1}\left (2,n,n+1,\frac {1}{2} (1-i \tan (e+f x))\right )}{4 a f n} \]
Antiderivative was successfully verified.
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Rule 68
Rule 3487
Rule 3505
Rule 3522
Rubi steps
\begin {align*} \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx &=\left ((d \sec (e+f x))^{2 n} (a-i a \tan (e+f x))^{-n} (a+i a \tan (e+f x))^{-n}\right ) \int \frac {(a-i a \tan (e+f x))^n}{a+i a \tan (e+f x)} \, dx\\ &=\frac {\left ((d \sec (e+f x))^{2 n} (a-i a \tan (e+f x))^{-n} (a+i a \tan (e+f x))^{-n}\right ) \int \cos ^2(e+f x) (a-i a \tan (e+f x))^{1+n} \, dx}{a^2}\\ &=\frac {\left (i a (d \sec (e+f x))^{2 n} (a-i a \tan (e+f x))^{-n} (a+i a \tan (e+f x))^{-n}\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{-1+n}}{(a-x)^2} \, dx,x,-i a \tan (e+f x)\right )}{f}\\ &=\frac {i \, _2F_1\left (2,n;1+n;\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-n}}{4 a f n}\\ \end {align*}
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Mathematica [B] time = 14.30, size = 165, normalized size = 2.50 \[ \frac {i e^{i e} 2^{n-2} \left (1+e^{2 i (e+f x)}\right )^2 \left (e^{i f x}\right )^{-n} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^n \, _2F_1\left (2,2-n;3-n;1+e^{2 i (e+f x)}\right ) \sec ^{1-n}(e+f x) (\cos (f x)+i \sin (f x))^{n+1} (a+i a \tan (e+f x))^{-n-1} (d \sec (e+f x))^{2 n}}{f (n-2)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\frac {2 \, d e^{\left (i \, f x + i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{2 \, n} e^{\left (-i \, e n + {\left (-i \, f n - i \, f\right )} x - {\left (n + 1\right )} \log \left (\frac {2 \, d e^{\left (i \, f x + i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right ) - {\left (n + 1\right )} \log \left (\frac {a}{d}\right ) - i \, e\right )}, x\right ) \]
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 \[ \int \left (d \sec \left (f x + e\right )\right )^{2 \, n} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{-n - 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 4.73, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{2 n} \left (a +i a \tan \left (f x +e \right )\right )^{-1-n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{2\,n}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{n+1}} \,d x \]
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 \[ \int \left (d \sec {\left (e + f x \right )}\right )^{2 n} \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{- n - 1}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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