Optimal. Leaf size=88 \[ \frac{i a 2^{n+\frac{1}{2}} \sec (c+d x) (1+i \tan (c+d x))^{\frac{1}{2}-n} (a+i a \tan (c+d x))^{n-1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2}-n,\frac{3}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d} \]
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Rubi [A] time = 0.149262, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3505, 3523, 70, 69} \[ \frac{i a 2^{n+\frac{1}{2}} \sec (c+d x) (1+i \tan (c+d x))^{\frac{1}{2}-n} (a+i a \tan (c+d x))^{n-1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2}-n,\frac{3}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \sec (c+d x) (a+i a \tan (c+d x))^n \, dx &=\frac{\sec (c+d x) \int \sqrt{a-i a \tan (c+d x)} (a+i a \tan (c+d x))^{\frac{1}{2}+n} \, dx}{\sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{\left (a^2 \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-\frac{1}{2}+n}}{\sqrt{a-i a x}} \, dx,x,\tan (c+d x)\right )}{d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{\left (2^{-\frac{1}{2}+n} a^2 \sec (c+d x) (a+i a \tan (c+d x))^{-1+n} \left (\frac{a+i a \tan (c+d x)}{a}\right )^{\frac{1}{2}-n}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{i x}{2}\right )^{-\frac{1}{2}+n}}{\sqrt{a-i a x}} \, dx,x,\tan (c+d x)\right )}{d \sqrt{a-i a \tan (c+d x)}}\\ &=\frac{i 2^{\frac{1}{2}+n} a \, _2F_1\left (\frac{1}{2},\frac{1}{2}-n;\frac{3}{2};\frac{1}{2} (1-i \tan (c+d x))\right ) \sec (c+d x) (1+i \tan (c+d x))^{\frac{1}{2}-n} (a+i a \tan (c+d x))^{-1+n}}{d}\\ \end{align*}
Mathematica [A] time = 8.20648, size = 134, normalized size = 1.52 \[ -\frac{i 2^{n+1} e^{i (c+d x)} \left (e^{i d x}\right )^n \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} \text{Hypergeometric2F1}\left (\frac{1}{2},1,n+\frac{3}{2},-e^{2 i (c+d x)}\right ) (a+i a \tan (c+d x))^n}{d (2 n+1)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.444, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( dx+c \right ) \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 (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )\,{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 (\frac{2 \, \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} e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{n} \sec{\left (c + d x \right )}\, dx \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 (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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