Optimal. Leaf size=118 \[ \frac{i 2^{\frac{n+1}{2}} (1+i \tan (c+d x))^{\frac{1}{2} (-n-1)} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{1-n} \text{Hypergeometric2F1}\left (\frac{1-n}{2},\frac{1-n}{2},\frac{3-n}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d (1-n)} \]
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Rubi [A] time = 0.217809, antiderivative size = 118, 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, 70, 69} \[ \frac{i 2^{\frac{n+1}{2}} (1+i \tan (c+d x))^{\frac{1}{2} (-n-1)} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{1-n} \text{Hypergeometric2F1}\left (\frac{1-n}{2},\frac{1-n}{2},\frac{3-n}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d (1-n)} \]
Antiderivative was successfully verified.
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Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx &=\left ((e \sec (c+d x))^{1-n} (a-i a \tan (c+d x))^{\frac{1}{2} (-1+n)} (a+i a \tan (c+d x))^{\frac{1}{2} (-1+n)}\right ) \int (a-i a \tan (c+d x))^{\frac{1-n}{2}} (a+i a \tan (c+d x))^{\frac{1-n}{2}+n} \, dx\\ &=\frac{\left (a^2 (e \sec (c+d x))^{1-n} (a-i a \tan (c+d x))^{\frac{1}{2} (-1+n)} (a+i a \tan (c+d x))^{\frac{1}{2} (-1+n)}\right ) \operatorname{Subst}\left (\int (a-i a x)^{-1+\frac{1-n}{2}} (a+i a x)^{-1+\frac{1-n}{2}+n} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{-\frac{1}{2}+\frac{n}{2}} a (e \sec (c+d x))^{1-n} (a-i a \tan (c+d x))^{\frac{1}{2} (-1+n)} (a+i a \tan (c+d x))^{\frac{1}{2}+\frac{1}{2} (-1+n)+\frac{n}{2}} \left (\frac{a+i a \tan (c+d x)}{a}\right )^{-\frac{1}{2}-\frac{n}{2}}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{-1+\frac{1-n}{2}+n} (a-i a x)^{-1+\frac{1-n}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i 2^{\frac{1+n}{2}} \, _2F_1\left (\frac{1-n}{2},\frac{1-n}{2};\frac{3-n}{2};\frac{1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{1-n} (1+i \tan (c+d x))^{\frac{1}{2} (-1-n)} (a+i a \tan (c+d x))^n}{d (1-n)}\\ \end{align*}
Mathematica [A] time = 4.4395, size = 87, normalized size = 0.74 \[ -\frac{e (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n} (\text{Hypergeometric2F1}(1,n,n+1,-\sin (c+d x)+i \cos (c+d x))-\text{Hypergeometric2F1}(1,n,n+1,\sin (c+d x)-i \cos (c+d x)))}{d n} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.971, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{1-n} \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(-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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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 )^{-n + 1}, 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 )^{-n + 1}{\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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