Optimal. Leaf size=160 \[ -\frac{a^{3/2} \sqrt{c} (B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{a (B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{B (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f} \]
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Rubi [A] time = 0.261031, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3588, 80, 50, 63, 217, 203} \[ -\frac{a^{3/2} \sqrt{c} (B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{a (B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{B (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 80
Rule 50
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt{c-i c \tan (e+f x)} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x} (A+B x)}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{B (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{(a (2 A-i B) c) \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{a (2 i A+B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{B (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{\left (a^2 (2 A-i B) 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{a (2 i A+B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{B (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{(a (2 i A+B) c) \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{a (2 i A+B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{B (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{(a (2 i A+B) c) \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{a^{3/2} (2 i A+B) \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{a (2 i A+B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{B (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}\\ \end{align*}
Mathematica [A] time = 6.4519, size = 220, normalized size = 1.38 \[ \frac{(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \left (\frac{\cos (e) (\tan (e)+i) \sqrt{\sec (e+f x)} \sqrt{c-i c \tan (e+f x)} (2 A+B \tan (e+f x)-2 i B)}{2 \cos (f x)+2 i \sin (f x)}-\frac{i c (2 A-i B) e^{-2 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt{\frac{c}{1+e^{2 i (e+f x)}}}}\right )}{f \sec ^{\frac{5}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.098, size = 223, normalized size = 1.4 \begin{align*}{\frac{a}{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 ( iB\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) -iB\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+2\,iA\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }+2\,A\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 ) ac+2\,B\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \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 = 2.2869, size = 1038, normalized size = 6.49 \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 = 1.56229, size = 1165, normalized size = 7.28 \begin{align*} \frac{2 \,{\left ({\left (4 i \, A + 6 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (4 i \, A + 2 \, B\right )} a\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{{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{2 \,{\left ({\left ({\left (8 i \, A + 4 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (8 i \, A + 4 \, B\right )} a\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 )} + 2 \, \sqrt{\frac{{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A + B\right )} a}\right ) - \sqrt{\frac{{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{2 \,{\left ({\left ({\left (8 i \, A + 4 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (8 i \, A + 4 \, B\right )} a\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 )} - 2 \, \sqrt{\frac{{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A + B\right )} a}\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 (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{3}{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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