Optimal. Leaf size=98 \[ -\frac{i 2^{n-\frac{1}{4}} (1+i \tan (c+d x))^{\frac{1}{4}-n} (a+i a \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (-\frac{5}{4},\frac{9}{4}-n,-\frac{1}{4},\frac{1}{2} (1-i \tan (c+d x))\right )}{5 a d (e \sec (c+d x))^{5/2}} \]
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Rubi [A] time = 0.210164, antiderivative size = 98, 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} \[ -\frac{i 2^{n-\frac{1}{4}} (1+i \tan (c+d x))^{\frac{1}{4}-n} (a+i a \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (-\frac{5}{4},\frac{9}{4}-n,-\frac{1}{4},\frac{1}{2} (1-i \tan (c+d x))\right )}{5 a d (e \sec (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx &=\frac{\left ((a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4}\right ) \int \frac{(a+i a \tan (c+d x))^{-\frac{5}{4}+n}}{(a-i a \tan (c+d x))^{5/4}} \, dx}{(e \sec (c+d x))^{5/2}}\\ &=\frac{\left (a^2 (a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4}\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-\frac{9}{4}+n}}{(a-i a x)^{9/4}} \, dx,x,\tan (c+d x)\right )}{d (e \sec (c+d x))^{5/2}}\\ &=\frac{\left (2^{-\frac{9}{4}+n} (a-i a \tan (c+d x))^{5/4} (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 \frac{\left (\frac{1}{2}+\frac{i x}{2}\right )^{-\frac{9}{4}+n}}{(a-i a x)^{9/4}} \, dx,x,\tan (c+d x)\right )}{d (e \sec (c+d x))^{5/2}}\\ &=-\frac{i 2^{-\frac{1}{4}+n} \, _2F_1\left (-\frac{5}{4},\frac{9}{4}-n;-\frac{1}{4};\frac{1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac{1}{4}-n} (a+i a \tan (c+d x))^{1+n}}{5 a d (e \sec (c+d x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 10.4131, size = 147, normalized size = 1.5 \[ -\frac{i 2^{n-\frac{3}{2}} e^{-3 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^4 \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n+\frac{1}{2}} \sec ^{\frac{1}{2}-n}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{9}{4},n-\frac{1}{4},-e^{2 i (c+d x)}\right ) (a+i a \tan (c+d x))^n}{d e^2 (4 n-5) \sqrt{e \sec (c+d x)}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.181, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n} \left ( e\sec \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}\, 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 \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}\,{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{\sqrt{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}}{\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-\frac{5}{2} i \, d x - \frac{5}{2} i \, c\right )}}{8 \, e^{3}}, 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 \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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