Optimal. Leaf size=195 \[ \frac{-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}-\frac{3 (3 B+5 i A)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{3 B+5 i A}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{3 (3 B+5 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{32 \sqrt{2} a^2 \sqrt{c} f} \]
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Rubi [A] time = 0.247676, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3588, 78, 51, 63, 208} \[ \frac{-B+i A}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}-\frac{3 (3 B+5 i A)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{3 B+5 i A}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{3 (3 B+5 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{32 \sqrt{2} a^2 \sqrt{c} f} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 78
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^3 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{((5 A-3 i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^2 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(3 (5 A-3 i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=-\frac{3 (5 i A+3 B)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(3 (5 A-3 i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{64 a f}\\ &=-\frac{3 (5 i A+3 B)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(3 (5 i A+3 B)) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{32 a c f}\\ &=\frac{3 (5 i A+3 B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{32 \sqrt{2} a^2 \sqrt{c} f}-\frac{3 (5 i A+3 B)}{32 a^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{i A-B}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 i A+3 B}{16 a^2 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 4.74789, size = 160, normalized size = 0.82 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\sin (e+f x)+i \cos (e+f x)) \left (3 (5 A-3 i B) e^{i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )-2 \cos (e+f x) (2 (3 B+5 i A) \sin (2 (e+f x))+2 (3 A-5 i B) \cos (2 (e+f x))-9 A-i B)\right )}{64 a^2 c f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.161, size = 152, normalized size = 0.8 \begin{align*}{\frac{-2\,i{c}^{2}}{f{a}^{2}} \left ({\frac{1}{8\,{c}^{2}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ( \left ( -{\frac{i}{8}}B+{\frac{7\,A}{8}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ( -{\frac{9\,Ac}{4}}-{\frac{i}{4}}Bc \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( -9\,iB+15\,A \right ) \sqrt{2}}{16}{\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 ) }-{\frac{-A+iB}{8\,{c}^{2}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}} \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 = 1.54952, size = 1031, normalized size = 5.29 \begin{align*} \frac{{\left (\sqrt{\frac{1}{2}} a^{2} c f \sqrt{-\frac{225 \, A^{2} - 270 i \, A B - 81 \, B^{2}}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left (\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{225 \, A^{2} - 270 i \, A B - 81 \, B^{2}}{a^{4} c f^{2}}} + 15 i \, A + 9 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) - \sqrt{\frac{1}{2}} a^{2} c f \sqrt{-\frac{225 \, A^{2} - 270 i \, A B - 81 \, B^{2}}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac{{\left (\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{225 \, A^{2} - 270 i \, A B - 81 \, B^{2}}{a^{4} c f^{2}}} - 15 i \, A - 9 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) + \sqrt{2}{\left ({\left (-8 i \, A - 8 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (i \, A - 9 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (11 i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, A - 2 \, B\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 )}}{64 \, a^{2} c 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{B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{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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