Optimal. Leaf size=242 \[ \frac{105 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{256 \sqrt{2} a^3 c^{3/2} f}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \]
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Rubi [A] time = 0.244592, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {3522, 3487, 51, 63, 206} \[ \frac{105 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{256 \sqrt{2} a^3 c^{3/2} f}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3522
Rule 3487
Rule 51
Rule 63
Rule 206
Rubi steps
\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{3/2} \, dx}{a^3 c^3}\\ &=\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^4 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{\left (3 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^3 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{4 a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{\left (21 i c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(105 i c) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(105 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^3 f}\\ &=-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(105 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{512 a^3 c f}\\ &=-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(105 i) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{256 a^3 c f}\\ &=\frac{105 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{256 \sqrt{2} a^3 c^{3/2} f}-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 3.70186, size = 160, normalized size = 0.66 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\cos (e+f x)-i \sin (e+f x)) \left (258 \sin (e+f x)+282 \sin (3 (e+f x))+24 \sin (5 (e+f x))+172 i \cos (e+f x)-166 i \cos (3 (e+f x))-8 i \cos (5 (e+f x))+315 i e^{i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )\right )}{1536 a^3 c^2 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.046, size = 159, normalized size = 0.7 \begin{align*}{\frac{2\,i{c}^{4}}{f{a}^{3}} \left ( -{\frac{1}{16\,{c}^{5}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{41}{32} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{35\,c}{6} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{55\,{c}^{2}}{8}\sqrt{c-ic\tan \left ( fx+e \right ) }} \right ) }-{\frac{105\,\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{1}{8\,{c}^{5}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{1}{48\,{c}^{4}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{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 = 1.51662, size = 994, normalized size = 4.11 \begin{align*} \frac{{\left (315 i \, \sqrt{\frac{1}{2}} a^{3} c^{2} f \sqrt{\frac{1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (13440 i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + 13440 i \, a^{3} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{6} c^{3} f^{2}}} + 13440 i\right )} e^{\left (-i \, f x - i \, e\right )}}{16384 \, a^{3} c f}\right ) - 315 i \, \sqrt{\frac{1}{2}} a^{3} c^{2} f \sqrt{\frac{1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (-13440 i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} - 13440 i \, a^{3} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{6} c^{3} f^{2}}} + 13440 i\right )} e^{\left (-i \, f x - i \, e\right )}}{16384 \, a^{3} c f}\right ) + \sqrt{2} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-16 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 224 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 43 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 215 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 58 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1536 \, a^{3} c^{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{1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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