Optimal. Leaf size=106 \[ -\frac{2 i \sqrt{a} c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}-\frac{i c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{f} \]
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Rubi [A] time = 0.135949, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3523, 50, 63, 217, 203} \[ -\frac{2 i \sqrt{a} c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}-\frac{i c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{f} \]
Antiderivative was successfully verified.
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Rule 3523
Rule 50
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{i c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{f}+\frac{\left (a c^2\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{i c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{f}-\frac{\left (2 i c^2\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 )}{f}\\ &=-\frac{i c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{f}-\frac{\left (2 i c^2\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 )}{f}\\ &=-\frac{2 i \sqrt{a} c^{3/2} \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 )}{f}-\frac{i c \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{f}\\ \end{align*}
Mathematica [A] time = 2.21537, size = 100, normalized size = 0.94 \[ -\frac{i \sqrt{2} c e^{-i (e+f x)} \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \left (e^{i (e+f x)}+\left (1+e^{2 i (e+f x)}\right ) \tan ^{-1}\left (e^{i (e+f x)}\right )\right ) \sqrt{a+i a \tan (e+f x)}}{f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.074, size = 122, normalized size = 1.2 \begin{align*}{\frac{c}{f} \left ( -i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+ac\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 ) \right ) \sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \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 [B] time = 1.87914, size = 344, normalized size = 3.25 \begin{align*} -\frac{{\left (2 \,{\left (c \cos \left (2 \, f x + 2 \, e\right ) + i \, c \sin \left (2 \, f x + 2 \, e\right ) + c\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (c \cos \left (2 \, f x + 2 \, e\right ) + i \, c \sin \left (2 \, f x + 2 \, e\right ) + c\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + 4 \, c \cos \left (f x + e\right ) -{\left (-i \, c \cos \left (2 \, f x + 2 \, e\right ) + c \sin \left (2 \, f x + 2 \, e\right ) - i \, c\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 ) -{\left (i \, c \cos \left (2 \, f x + 2 \, e\right ) - c \sin \left (2 \, f x + 2 \, e\right ) + i \, c\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 ) + 4 i \, c \sin \left (f x + e\right )\right )} \sqrt{a} \sqrt{c}}{f{\left (-2 i \, \cos \left (2 \, f x + 2 \, e\right ) + 2 \, \sin \left (2 \, f x + 2 \, e\right ) - 2 i\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.54345, size = 749, normalized size = 7.07 \begin{align*} \frac{-8 i \, c \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 )} - 2 \, \sqrt{\frac{a c^{3}}{f^{2}}} f \log \left (\frac{2 \,{\left (4 \,{\left (c e^{\left (2 i \, f x + 2 i \, e\right )} + c\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 )} + \sqrt{\frac{a c^{3}}{f^{2}}}{\left (2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, f\right )}\right )}}{c e^{\left (2 i \, f x + 2 i \, e\right )} + c}\right ) + 2 \, \sqrt{\frac{a c^{3}}{f^{2}}} f \log \left (\frac{2 \,{\left (4 \,{\left (c e^{\left (2 i \, f x + 2 i \, e\right )} + c\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 )} + \sqrt{\frac{a c^{3}}{f^{2}}}{\left (-2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, f\right )}\right )}}{c e^{\left (2 i \, f x + 2 i \, e\right )} + c}\right )}{4 \, 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 \sqrt{i \, a \tan \left (f x + e\right ) + a}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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