Optimal. Leaf size=245 \[ \frac{-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}-\frac{5 (5 B+7 i A)}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{5 (5 B+7 i A)}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{5 B+7 i A}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 (5 B+7 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^3 \sqrt{c} f} \]
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Rubi [A] time = 0.273547, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}-\frac{5 (5 B+7 i A)}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{5 (5 B+7 i A)}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{5 B+7 i A}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 (5 B+7 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^3 \sqrt{c} f} \]
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))^3 \sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^4 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{((7 A-5 i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^3 (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{(5 (7 A-5 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 )}{96 a f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(5 (7 A-5 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 )}{128 a^2 f}\\ &=-\frac{5 (7 i A+5 B)}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(5 (7 A-5 i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{256 a^2 f}\\ &=-\frac{5 (7 i A+5 B)}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(5 (7 i A+5 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 )}{128 a^2 c f}\\ &=\frac{5 (7 i A+5 B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^3 \sqrt{c} f}-\frac{5 (7 i A+5 B)}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 5.547, size = 181, normalized size = 0.74 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\cos (2 (e+f x))-i \sin (2 (e+f x))) \left (15 (5 B+7 i A) e^{2 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )+2 \cos (e+f x) ((7 A-5 i B) (8 \sin (3 (e+f x))-7 \sin (e+f x))+(7 B+125 i A) \cos (e+f x)+(-56 B-40 i A) \cos (3 (e+f x)))\right )}{768 a^3 c f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.165, size = 179, normalized size = 0.7 \begin{align*}{\frac{2\,i{c}^{3}}{f{a}^{3}} \left ( -{\frac{1}{16\,{c}^{3}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ( \left ( -{\frac{9\,i}{16}}B+{\frac{19\,A}{16}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}+ \left ({\frac{7\,i}{3}}Bc-{\frac{17\,Ac}{3}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ( -{\frac{7\,i}{4}}B{c}^{2}+{\frac{29\,A{c}^{2}}{4}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( -25\,iB+35\,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 ) }-{\frac{A-iB}{16\,{c}^{3}}{\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 = 1.59236, size = 1118, normalized size = 4.56 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} a^{3} c f \sqrt{-\frac{1225 \, A^{2} - 1750 i \, A B - 625 \, B^{2}}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{1225 \, A^{2} - 1750 i \, A B - 625 \, B^{2}}{a^{6} c f^{2}}} + 35 i \, A + 25 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{3} f}\right ) - 3 \, \sqrt{\frac{1}{2}} a^{3} c f \sqrt{-\frac{1225 \, A^{2} - 1750 i \, A B - 625 \, B^{2}}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{1225 \, A^{2} - 1750 i \, A B - 625 \, B^{2}}{a^{6} c f^{2}}} - 35 i \, A - 25 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{3} f}\right ) + \sqrt{2}{\left ({\left (-48 i \, A - 48 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (39 i \, A - 27 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (125 i \, A + 7 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (46 i \, A - 22 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, A - 8 \, B\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{768 \, a^{3} 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 )}^{3} \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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