Optimal. Leaf size=153 \[ \frac{6 i c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}} \]
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Rubi [A] time = 0.159191, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3523, 47, 50, 63, 217, 203} \[ \frac{6 i c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 3523
Rule 47
Rule 50
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \frac{(c-i c \tan (e+f x))^{5/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (3 c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (6 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c-\frac{c x^2}{a}}} \, dx,x,\sqrt{a+i a \tan (e+f x)}\right )}{a f}\\ &=\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (6 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c x^2}{a}} \, dx,x,\frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c-i c \tan (e+f x)}}\right )}{a f}\\ &=\frac{6 i c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{a f}+\frac{2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 3.47265, size = 113, normalized size = 0.74 \[ \frac{c^3 (\cos (f x)+i \sin (f x)) (\sin (f x)+i \cos (f x)) \left (\tan ^2(e+f x)-4 i \tan (e+f x)+6 \sec (e+f x) \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))+5\right )}{f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.08, size = 300, normalized size = 2. \begin{align*}{\frac{i{c}^{2}}{fa \left ( -\tan \left ( fx+e \right ) +i \right ) ^{2}}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) } \left ( 3\,i\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{2}ac-3\,i\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) ac-6\,i\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) +6\,\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) \tan \left ( fx+e \right ) ac+\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( \tan \left ( fx+e \right ) \right ) ^{2}-5\,\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 2.01967, size = 250, normalized size = 1.63 \begin{align*} -\frac{{\left (6 i \, c^{2} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) \cos \left (f x + e\right ) + 6 i \, c^{2} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) \cos \left (f x + e\right ) - 8 i \, c^{2} \cos \left (f x + e\right )^{2} + 3 \, c^{2} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 3 \, c^{2} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 8 \, c^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 i \, c^{2}\right )} \sqrt{c}}{2 \, \sqrt{a} f \cos \left (f x + e\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.66494, size = 1010, normalized size = 6.6 \begin{align*} -\frac{{\left (3 \, \sqrt{\frac{c^{5}}{a f^{2}}} a f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac{2 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} -{\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a f\right )} \sqrt{\frac{c^{5}}{a f^{2}}}}{2 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}\right )}}\right ) - 3 \, \sqrt{\frac{c^{5}}{a f^{2}}} a f e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac{2 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} -{\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a f\right )} \sqrt{\frac{c^{5}}{a f^{2}}}}{2 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}\right )}}\right ) -{\left (-10 i \, c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 12 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 10 i \, c^{2} e^{\left (i \, f x + i \, e\right )} + 8 i \, c^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a f} \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 \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{\sqrt{i \, a \tan \left (f x + e\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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