Optimal. Leaf size=74 \[ -\frac{i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-2 (n+1)} \text{Hypergeometric2F1}\left (2,-n-1,-n,\frac{1}{2} (1-i \tan (c+d x))\right )}{4 a d (n+1)} \]
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Rubi [A] time = 0.148941, antiderivative size = 74, normalized size of antiderivative = 1., 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 3505
Rule 3523
Rule 7
Rule 68
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*}
Mathematica [B] time = 13.0512, size = 151, normalized size = 2.04 \[ -\frac{i 2^{-n-3} \left (1+e^{2 i (c+d x)}\right )^3 \left (e^{i d x}\right )^n \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-n} \sec ^n(c+d x) (\cos (d x)+i \sin (d x))^{-n} \text{Hypergeometric2F1}\left (2,n+3,n+4,1+e^{2 i (c+d x)}\right ) (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-2 n}}{d e^2 (n+3)} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.239, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{-2\,n-2} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \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}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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