Optimal. Leaf size=113 \[ \frac{i a 2^{\frac{n}{2}+1} (1+i \tan (c+d x))^{-n/2} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{2-n} \text{Hypergeometric2F1}\left (\frac{2-n}{2},-\frac{n}{2},\frac{4-n}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d (2-n)} \]
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Rubi [A] time = 0.183201, antiderivative size = 113, 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} \[ \frac{i a 2^{\frac{n}{2}+1} (1+i \tan (c+d x))^{-n/2} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{2-n} \text{Hypergeometric2F1}\left (\frac{2-n}{2},-\frac{n}{2},\frac{4-n}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d (2-n)} \]
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))^{2-n} (a+i a \tan (c+d x))^n \, dx &=\left ((e \sec (c+d x))^{2-n} (a-i a \tan (c+d x))^{\frac{1}{2} (-2+n)} (a+i a \tan (c+d x))^{\frac{1}{2} (-2+n)}\right ) \int (a-i a \tan (c+d x))^{\frac{2-n}{2}} (a+i a \tan (c+d x))^{\frac{2-n}{2}+n} \, dx\\ &=\frac{\left (a^2 (e \sec (c+d x))^{2-n} (a-i a \tan (c+d x))^{\frac{1}{2} (-2+n)} (a+i a \tan (c+d x))^{\frac{1}{2} (-2+n)}\right ) \operatorname{Subst}\left (\int (a-i a x)^{-1+\frac{2-n}{2}} (a+i a x)^{-1+\frac{2-n}{2}+n} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{n/2} a^2 (e \sec (c+d x))^{2-n} (a-i a \tan (c+d x))^{\frac{1}{2} (-2+n)} (a+i a \tan (c+d x))^{\frac{1}{2} (-2+n)+\frac{n}{2}} \left (\frac{a+i a \tan (c+d x)}{a}\right )^{-n/2}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{-1+\frac{2-n}{2}+n} (a-i a x)^{-1+\frac{2-n}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i 2^{1+\frac{n}{2}} a \, _2F_1\left (\frac{2-n}{2},-\frac{n}{2};\frac{4-n}{2};\frac{1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{2-n} (1+i \tan (c+d x))^{-n/2} (a+i a \tan (c+d x))^{-1+n}}{d (2-n)}\\ \end{align*}
Mathematica [A] time = 12.5509, size = 112, normalized size = 0.99 \[ \frac{4 e^2 (\cos (2 c)-i \sin (2 c)) (\tan (d x)+i) (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n} \text{Hypergeometric2F1}\left (2,1-\frac{n}{2},2-\frac{n}{2},-\cos (2 (c+d x))+i \sin (2 (c+d x))\right )}{d (n-2) (-1-i \tan (d x))} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.967, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{2-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 + 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 )^{-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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