Optimal. Leaf size=154 \[ -\frac{3 i a^{5/2} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f} \]
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Rubi [A] time = 0.159651, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{3 i a^{5/2} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f} \]
Antiderivative was successfully verified.
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Rule 3523
Rule 50
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{\left (3 a^2 c\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{\left (3 a^3 c\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{\left (3 i a^2 c\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{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{\left (3 i a^2 c\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{3 i a^{5/2} \sqrt{c} \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{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}\\ \end{align*}
Mathematica [A] time = 3.3508, size = 87, normalized size = 0.56 \[ \frac{a^2 c (\tan (e+f x)+i) \sqrt{a+i a \tan (e+f x)} \left (i \tan (e+f x)-6 \cos (e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right )+4\right )}{2 f \sqrt{c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.087, size = 154, normalized size = 1. \begin{align*}{\frac{{a}^{2}}{2\,f}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) } \left ( 4\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}-\tan \left ( fx+e \right ) \sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+3\,ac\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 ) \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 = 2.21158, size = 900, normalized size = 5.84 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.57612, size = 919, normalized size = 5.97 \begin{align*} \frac{2 \,{\left (10 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, a^{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 )} - 3 \, \sqrt{\frac{a^{5} c}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{8 \,{\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 )} + \sqrt{\frac{a^{5} c}{f^{2}}}{\left (4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, f\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right ) + 3 \, \sqrt{\frac{a^{5} c}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{8 \,{\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 )} + \sqrt{\frac{a^{5} c}{f^{2}}}{\left (-4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, f\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right )}{4 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\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{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sqrt{-i \, c \tan \left (f x + e\right ) + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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