Optimal. Leaf size=187 \[ \frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}+\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{2 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 d e^8}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]
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Rubi [A] time = 0.166563, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3496, 3769, 3771, 2641} \[ \frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}+\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{2 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 d e^8}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]
Antiderivative was successfully verified.
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Rule 3496
Rule 3769
Rule 3771
Rule 2641
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx &=-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}+\frac{a^2 \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{11/2}} \, dx}{5 e^2}\\ &=-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (7 a^4\right ) \int \frac{1}{(e \sec (c+d x))^{7/2}} \, dx}{55 e^4}\\ &=\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{a^4 \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{11 e^6}\\ &=\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{a^4 \int \sqrt{e \sec (c+d x)} \, dx}{33 e^8}\\ &=\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (a^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{33 e^8}\\ &=\frac{2 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 d e^8}+\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}\\ \end{align*}
Mathematica [A] time = 2.04776, size = 155, normalized size = 0.83 \[ -\frac{i a^4 \sqrt{e \sec (c+d x)} (\cos (4 (c+2 d x))+i \sin (4 (c+2 d x))) \left (-54 i \sin (2 (c+d x))-37 i \sin (4 (c+d x))+112 \cos (2 (c+d x))+48 \cos (4 (c+d x))+40 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\sin (4 (c+d x))+i \cos (4 (c+d x)))+64\right )}{660 d e^8 (\cos (d x)+i \sin (d x))^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.364, size = 232, normalized size = 1.2 \begin{align*}{\frac{2\,{a}^{4}}{165\,d \left ( \cos \left ( dx+c \right ) \right ) ^{8}} \left ( -88\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+88\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+60\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}-16\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +5\,i\cos \left ( dx+c \right ) \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 ) +5\,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 ) +3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{15}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1320 \, d e^{8}{\rm integral}\left (-\frac{i \, \sqrt{2} a^{4} \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 )}}{33 \, d e^{8}}, x\right ) + \sqrt{2}{\left (-11 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 58 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 128 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 166 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 85 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 )}}{1320 \, d e^{8}} \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{15}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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