Optimal. Leaf size=74 \[ -\frac {i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-2 (n+1)} \, _2F_1\left (2,-n-1;-n;\frac {1}{2} (1-i \tan (c+d x))\right )}{4 a d (n+1)} \]
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Rubi [A] time = 0.15, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3505, 3523, 7, 68} \[ -\frac {i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-2 (n+1)} \text {Hypergeometric2F1}\left (2,-n-1,-n,\frac {1}{2} (1-i \tan (c+d x))\right )}{4 a d (n+1)} \]
Antiderivative was successfully verified.
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Rule 7
Rule 68
Rule 3505
Rule 3523
Rubi steps
\begin {align*} \int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx &=\left ((e \sec (c+d x))^{-2-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (2+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (2+2 n)}\right ) \int (a-i a \tan (c+d x))^{\frac {1}{2} (-2-2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-2-2 n)+n} \, dx\\ &=\frac {\left (a^2 (e \sec (c+d x))^{-2-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (2+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (2+2 n)}\right ) \operatorname {Subst}\left (\int (a-i a x)^{-1+\frac {1}{2} (-2-2 n)} (a+i a x)^{-1+\frac {1}{2} (-2-2 n)+n} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (a^2 (e \sec (c+d x))^{-2-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (2+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (2+2 n)}\right ) \operatorname {Subst}\left (\int \frac {(a-i a x)^{-1+\frac {1}{2} (-2-2 n)}}{(a+i a x)^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {i \, _2F_1\left (2,-1-n;-n;\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{-2 (1+n)} (a+i a \tan (c+d x))^{1+n}}{4 a d (1+n)}\\ \end {align*}
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Mathematica [B] time = 13.12, size = 151, normalized size = 2.04 \[ -\frac {i 2^{-n-3} \left (1+e^{2 i (c+d x)}\right )^3 \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-n} \, _2F_1\left (2,n+3;n+4;1+e^{2 i (c+d x)}\right ) \sec ^n(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-2 n}}{d e^2 (n+3)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-2 \, n - 2} e^{\left (i \, d n x + i \, c n + n \log \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right ) + n \log \left (\frac {a}{e}\right )\right )}, x\right ) \]
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 \[ \int \left (e \sec \left (d x + c\right )\right )^{-2 \, n - 2} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.67, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x +c \right )\right )^{-2-2 n} \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx \]
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 \[ \int \left (e \sec \left (d x + c\right )\right )^{-2 \, n - 2} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{2\,n+2}} \,d x \]
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 \[ \int \left (e \sec {\left (c + d x \right )}\right )^{- 2 n - 2} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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