Optimal. Leaf size=181 \[ \frac{5 i \sqrt{c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac{5 i \sqrt{c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac{i \sqrt{c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{5 i \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{64 \sqrt{2} a^3 f} \]
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Rubi [A] time = 0.208587, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{5 i \sqrt{c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac{5 i \sqrt{c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac{i \sqrt{c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{5 i \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{64 \sqrt{2} a^3 f} \]
Antiderivative was successfully verified.
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Rule 3522
Rule 3487
Rule 51
Rule 63
Rule 206
Rubi steps
\begin{align*} \int \frac{\sqrt{c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx &=\frac{\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{7/2} \, dx}{a^3 c^3}\\ &=\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^4 \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac{i \sqrt{c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{\left (5 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^3 \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{12 a^3 f}\\ &=\frac{i \sqrt{c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{5 i \sqrt{c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac{\left (5 i c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^3 f}\\ &=\frac{i \sqrt{c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{5 i \sqrt{c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac{5 i \sqrt{c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac{(5 i c) \operatorname{Subst}\left (\int \frac{1}{(c-x) \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=\frac{i \sqrt{c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{5 i \sqrt{c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac{5 i \sqrt{c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac{(5 i c) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{64 a^3 f}\\ &=\frac{5 i \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{64 \sqrt{2} a^3 f}+\frac{i \sqrt{c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{5 i \sqrt{c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac{5 i \sqrt{c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.90396, size = 149, normalized size = 0.82 \[ \frac{(\sin (3 (e+f x))+i \cos (3 (e+f x))) \left (\sqrt{c-i c \tan (e+f x)} \left (93 \cos (e+f x)+41 \cos (3 (e+f x))+100 i \sin (e+f x) \cos ^2(e+f x)\right )+15 \sqrt{2} \sqrt{c} (\cos (3 (e+f x))+i \sin (3 (e+f x))) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )\right )}{384 a^3 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.074, size = 162, normalized size = 0.9 \begin{align*}{\frac{2\,i{c}^{4}}{f{a}^{3}} \left ( -{\frac{1}{12\,c \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}}\sqrt{c-ic\tan \left ( fx+e \right ) }}-{\frac{5}{12\,c} \left ( -{\frac{1}{8\,c \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}}\sqrt{c-ic\tan \left ( fx+e \right ) }}-{\frac{3}{8\,c} \left ( -{\frac{1}{4\,c \left ( -c-ic\tan \left ( fx+e \right ) \right ) }\sqrt{c-ic\tan \left ( fx+e \right ) }}+{\frac{\sqrt{2}}{8}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}} \right ) } \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 = 1.39228, size = 794, normalized size = 4.39 \begin{align*} \frac{{\left (15 \, \sqrt{\frac{1}{2}} a^{3} f \sqrt{-\frac{c}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{5 \,{\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{c}{a^{6} f^{2}}} + i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{3} f}\right ) - 15 \, \sqrt{\frac{1}{2}} a^{3} f \sqrt{-\frac{c}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac{5 \,{\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{c}{a^{6} f^{2}}} - i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{3} f}\right ) + \sqrt{2} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (33 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 59 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 34 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 )}}{384 \, a^{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{\sqrt{-i \, c \tan \left (f x + e\right ) + c}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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