Optimal. Leaf size=311 \[ -\frac{21 (B+11 i A)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{21 (B+11 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{512 \sqrt{2} a^3 c^{5/2} f}+\frac{-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}-\frac{7 (B+11 i A)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (B+11 i A)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{3 (B+11 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{B+11 i A}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.349697, antiderivative size = 311, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{21 (B+11 i A)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{21 (B+11 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{512 \sqrt{2} a^3 c^{5/2} f}+\frac{-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}-\frac{7 (B+11 i A)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (B+11 i A)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{3 (B+11 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{B+11 i A}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \]
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 (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^4 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{((11 A-i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^3 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{(3 (11 A-i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^2 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(21 (11 A-i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{128 a^2 f}\\ &=-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(21 (11 A-i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{256 a^2 f}\\ &=-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}+\frac{(21 (11 A-i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{512 a^2 c f}\\ &=-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (11 i A+B)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(21 (11 A-i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{1024 a^2 c^2 f}\\ &=-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (11 i A+B)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(21 (11 i A+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 )}{512 a^2 c^3 f}\\ &=\frac{21 (11 i A+B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{512 \sqrt{2} a^3 c^{5/2} f}-\frac{21 (11 i A+B)}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac{11 i A+B}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac{3 (11 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 (11 i A+B)}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{21 (11 i A+B)}{512 a^3 c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 12.2013, size = 256, normalized size = 0.82 \[ \frac{e^{-6 i (e+f x)} \sqrt{c-i c \tan (e+f x)} \left (315 (B+11 i A) e^{6 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )-i \left (1+e^{2 i (e+f x)}\right ) \left (A \left (-310 e^{2 i (e+f x)}-1335 e^{4 i (e+f x)}+2768 e^{6 i (e+f x)}+416 e^{8 i (e+f x)}+48 e^{10 i (e+f x)}-40\right )-i B \left (190 e^{2 i (e+f x)}+315 e^{4 i (e+f x)}+688 e^{6 i (e+f x)}+256 e^{8 i (e+f x)}+48 e^{10 i (e+f x)}+40\right )\right )\right )}{15360 a^3 c^3 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.117, size = 233, normalized size = 0.8 \begin{align*}{\frac{2\,i{c}^{3}}{f{a}^{3}} \left ( -{\frac{1}{32\,{c}^{5}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ( \left ({\frac{11\,i}{32}}B+{\frac{71\,A}{32}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}+ \left ( -{\frac{59\,Ac}{6}}-{\frac{11\,i}{6}}Bc \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ({\frac{21\,i}{8}}B{c}^{2}+{\frac{89\,A{c}^{2}}{8}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( -21\,iB+231\,A \right ) \sqrt{2}}{64}{\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{5\,A-iB}{32\,{c}^{5}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{2\,A-iB}{48\,{c}^{4}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{A-iB}{80\,{c}^{3}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}} \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 [A] time = 1.80242, size = 1319, normalized size = 4.24 \begin{align*} \frac{{\left (15 \, \sqrt{\frac{1}{2}} a^{3} c^{3} f \sqrt{-\frac{53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} 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} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} + 231 i \, A + 21 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) - 15 \, \sqrt{\frac{1}{2}} a^{3} c^{3} f \sqrt{-\frac{53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} 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} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{53361 \, A^{2} - 9702 i \, A B - 441 \, B^{2}}{a^{6} c^{5} f^{2}}} - 231 i \, A - 21 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) + \sqrt{2}{\left ({\left (-48 i \, A - 48 \, B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-464 i \, A - 304 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-3184 i \, A - 944 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-1433 i \, A - 1003 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (1645 i \, A - 505 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (350 i \, A - 230 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 40 i \, A - 40 \, 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 )}}{15360 \, a^{3} c^{3} 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}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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