Optimal. Leaf size=160 \[ -\frac{c^{3/2} (7 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a^2 f}+\frac{c (7 B+i A) \sqrt{c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
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Rubi [A] time = 0.226024, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3588, 78, 47, 63, 208} \[ -\frac{c^{3/2} (7 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a^2 f}+\frac{c (7 B+i A) \sqrt{c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 78
Rule 47
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) \sqrt{c-i c x}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{((A-7 i B) c) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac{(i A+7 B) c \sqrt{c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2}-\frac{\left ((A-7 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=\frac{(i A+7 B) c \sqrt{c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2}-\frac{((i A+7 B) c) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{8 a f}\\ &=-\frac{(i A+7 B) c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a^2 f}+\frac{(i A+7 B) c \sqrt{c-i c \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{3/2}}{4 a^2 f (1+i \tan (e+f x))^2}\\ \end{align*}
Mathematica [A] time = 3.86559, size = 205, normalized size = 1.28 \[ \frac{\sec (e+f x) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \left (\sqrt{2} c^{3/2} (A-7 i B) (\sin (2 e)-i \cos (2 e)) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )+2 c \cos (e+f x) (\cos (2 f x)-i \sin (2 f x)) \sqrt{c-i c \tan (e+f x)} ((A+9 i B) \sin (e+f x)+(5 B+3 i A) \cos (e+f x))\right )}{16 f (a+i a \tan (e+f x))^2 (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.099, size = 117, normalized size = 0.7 \begin{align*}{\frac{-2\,i{c}^{2}}{f{a}^{2}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ( \left ( -{\frac{9\,i}{16}}B-{\frac{A}{16}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ({\frac{7\,i}{8}}Bc-{\frac{Ac}{8}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }+{\frac{ \left ( -7\,iB+A \right ) \sqrt{2}}{32}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.44929, size = 973, normalized size = 6.08 \begin{align*} \frac{{\left (\sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{{\left (A^{2} - 14 i \, A B - 49 \, B^{2}\right )} c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left ({\left (-i \, A - 7 \, B\right )} c^{2} + \sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{{\left (A^{2} - 14 i \, A B - 49 \, B^{2}\right )} c^{3}}{a^{4} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) - \sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{{\left (A^{2} - 14 i \, A B - 49 \, B^{2}\right )} c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left ({\left (-i \, A - 7 \, B\right )} c^{2} - \sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{{\left (A^{2} - 14 i \, A B - 49 \, B^{2}\right )} c^{3}}{a^{4} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) + \sqrt{2}{\left ({\left (i \, A + 7 \, B\right )} c e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (3 i \, A + 5 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A - 2 \, B\right )} c\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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 (B \tan \left (f x + e\right ) + A\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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