Optimal. Leaf size=125 \[ \frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d e^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}} \]
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Rubi [A] time = 0.131385, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3496, 3771, 2639} \[ \frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d e^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}} \]
Antiderivative was successfully verified.
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Rule 3496
Rule 3771
Rule 2639
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{9/2}} \, dx &=-\frac{4 i a (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}-\frac{a^2 \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{5/2}} \, dx}{3 e^2}\\ &=-\frac{4 i a (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}+\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac{a^4 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{15 e^4}\\ &=-\frac{4 i a (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}+\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac{a^4 \int \sqrt{\cos (c+d x)} \, dx}{15 e^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=-\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d e^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{9 d (e \sec (c+d x))^{9/2}}+\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{15 d e^2 (e \sec (c+d x))^{5/2}}\\ \end{align*}
Mathematica [C] time = 3.667, size = 108, normalized size = 0.86 \[ -\frac{i a^4 e^{i (c+d x)} \left (-2 \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+7 e^{2 i (c+d x)}+5 e^{4 i (c+d x)}+2\right ) \sqrt{e \sec (c+d x)}}{45 d e^5} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.285, size = 370, normalized size = 3. \begin{align*}{\frac{2\,{a}^{4}}{45\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}} \left ( -40\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+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 ) -40\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}+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}}}+36\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+56\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-13\, \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{9}{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{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac{9}{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 (-5 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 7 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 7 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} + 4 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + 6 i \, a^{4}\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 )} + 45 \,{\left (d e^{5} e^{\left (i \, d x + i \, c\right )} - d e^{5}\right )}{\rm integral}\left (\frac{\sqrt{2}{\left (i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + i \, a^{4}\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 )}}{15 \,{\left (d e^{5} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{5} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{45 \,{\left (d e^{5} e^{\left (i \, d x + i \, c\right )} - d e^{5}\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 )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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