Optimal. Leaf size=96 \[ \frac{i a 2^{n+\frac{9}{4}} (e \sec (c+d x))^{5/2} (1+i \tan (c+d x))^{-n-\frac{1}{4}} (a+i a \tan (c+d x))^{n-1} \text{Hypergeometric2F1}\left (\frac{5}{4},-n-\frac{1}{4},\frac{9}{4},\frac{1}{2} (1-i \tan (c+d x))\right )}{5 d} \]
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Rubi [A] time = 0.201624, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3505, 3523, 70, 69} \[ \frac{i a 2^{n+\frac{9}{4}} (e \sec (c+d x))^{5/2} (1+i \tan (c+d x))^{-n-\frac{1}{4}} (a+i a \tan (c+d x))^{n-1} \text{Hypergeometric2F1}\left (\frac{5}{4},-n-\frac{1}{4},\frac{9}{4},\frac{1}{2} (1-i \tan (c+d x))\right )}{5 d} \]
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))^{5/2} (a+i a \tan (c+d x))^n \, dx &=\frac{(e \sec (c+d x))^{5/2} \int (a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{\frac{5}{4}+n} \, dx}{(a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4}}\\ &=\frac{\left (a^2 (e \sec (c+d x))^{5/2}\right ) \operatorname{Subst}\left (\int \sqrt [4]{a-i a x} (a+i a x)^{\frac{1}{4}+n} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4}}\\ &=\frac{\left (2^{\frac{1}{4}+n} a^2 (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{-1+n} \left (\frac{a+i a \tan (c+d x)}{a}\right )^{-\frac{1}{4}-n}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{\frac{1}{4}+n} \sqrt [4]{a-i a x} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/4}}\\ &=\frac{i 2^{\frac{9}{4}+n} a \, _2F_1\left (\frac{5}{4},-\frac{1}{4}-n;\frac{9}{4};\frac{1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{5/2} (1+i \tan (c+d x))^{-\frac{1}{4}-n} (a+i a \tan (c+d x))^{-1+n}}{5 d}\\ \end{align*}
Mathematica [A] time = 8.86914, size = 156, normalized size = 1.62 \[ -\frac{i 2^{n+\frac{7}{2}} e^{i (c+d x)} \left (e^{i d x}\right )^n \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n+\frac{3}{2}} (e \sec (c+d x))^{5/2} \sec ^{-n-\frac{5}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{-n} \text{Hypergeometric2F1}\left (-\frac{1}{4},1,n+\frac{9}{4},-e^{2 i (c+d x)}\right ) (a+i a \tan (c+d x))^n}{d (4 n+5)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.198, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{{\frac{5}{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 )^{\frac{5}{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 (\frac{4 \, \sqrt{2} e^{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} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{5}{2} i \, d x + \frac{5}{2} i \, c\right )}}{e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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(-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 )^{\frac{5}{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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