Optimal. Leaf size=186 \[ \frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}+\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{2 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{11 d e^8}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]
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Rubi [A] time = 0.173392, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3497, 3496, 3769, 3771, 2641} \[ \frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}+\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{2 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{11 d e^8}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]
Antiderivative was successfully verified.
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Rule 3497
Rule 3496
Rule 3769
Rule 3771
Rule 2641
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx &=-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}+\frac{(3 a) \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{11/2}} \, dx}{5 e^2}\\ &=-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (21 a^3\right ) \int \frac{1}{(e \sec (c+d x))^{7/2}} \, dx}{55 e^4}\\ &=\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (3 a^3\right ) \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{11 e^6}\\ &=\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{a^3 \int \sqrt{e \sec (c+d x)} \, dx}{11 e^8}\\ &=\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (a^3 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{11 e^8}\\ &=\frac{2 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{11 d e^8}+\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}\\ \end{align*}
Mathematica [A] time = 1.85125, size = 170, normalized size = 0.91 \[ \frac{a^3 \sqrt{e \sec (c+d x)} (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x))) \left (-114 \sin (c+d x)-81 \sin (3 (c+d x))+33 \sin (5 (c+d x))-332 i \cos (c+d x)-154 i \cos (3 (c+d x))+22 i \cos (5 (c+d x))+240 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (3 (c+d x))-i \sin (3 (c+d x)))\right )}{1320 d e^8 (\cos (d x)+i \sin (d x))^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.378, size = 232, normalized size = 1.3 \begin{align*}{\frac{2\,{a}^{3}}{165\,d \left ( \cos \left ( dx+c \right ) \right ) ^{8}} \left ( -44\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+44\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+15\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+7\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +15\,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 ) +15\,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 ) +9\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +15\,\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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (2640 \, d e^{8} e^{\left (2 i \, d x + 2 i \, c\right )}{\rm integral}\left (-\frac{i \, \sqrt{2} a^{3} \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 )}}{11 \, d e^{8}}, x\right ) + \sqrt{2}{\left (-11 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 73 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 218 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 446 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 235 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 55 i \, a^{3}\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 )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2640 \, 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 )}^{3}}{\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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