Optimal. Leaf size=96 \[ \frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}} \]
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Rubi [A] time = 0.0652379, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3486, 3769, 3771, 2639} \[ \frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3486
Rule 3769
Rule 3771
Rule 2639
Rubi steps
\begin{align*} \int \frac{a+i a \tan (c+d x)}{(e \sec (c+d x))^{5/2}} \, dx &=-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+a \int \frac{1}{(e \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}+\frac{(3 a) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 e^2}\\ &=-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}+\frac{(3 a) \int \sqrt{\cos (c+d x)} \, dx}{5 e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 0.75465, size = 99, normalized size = 1.03 \[ -\frac{a (\tan (c+d x)-i) \left (-2 \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 i \sin (2 (c+d x))+2 \cos (2 (c+d x))+2\right )}{5 d e^2 \sqrt{e \sec (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.191, size = 339, normalized size = 3.5 \begin{align*} -{\frac{2\,a}{5\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}} \left ( 3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3\,i\sin \left ( dx+c \right ){\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-3\,i\sin \left ( dx+c \right ){\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-3\,\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{5}{2}}}} \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{i \, a \tan \left (d x + c\right ) + a}{\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*} \frac{\sqrt{2}{\left (-i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 7 i \, a e^{\left (i \, d x + i \, c\right )} - 5 i \, a\right )} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )} + 10 \,{\left (d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}{\rm integral}\left (\frac{\sqrt{2}{\left (-3 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, a e^{\left (i \, d x + i \, c\right )} - 3 i \, a\right )} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}}{5 \,{\left (d e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{10 \,{\left (d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - d e^{3} e^{\left (i \, d x + i \, c\right )}\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*} a \left (\int \frac{1}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx + \int \frac{i \tan{\left (c + d x \right )}}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx\right ) \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{i \, a \tan \left (d x + c\right ) + a}{\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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