Optimal. Leaf size=199 \[ -\frac{3 c^2 (-9 B+i A) \sqrt{c-i c \tan (e+f x)}}{8 a^2 f}+\frac{3 c^{5/2} (-9 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{4 \sqrt{2} a^2 f}-\frac{c (-9 B+i A) (c-i c \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(-B+i A) (c-i c \tan (e+f x))^{5/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
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Rubi [A] time = 0.24501, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {3588, 78, 47, 50, 63, 208} \[ -\frac{3 c^2 (-9 B+i A) \sqrt{c-i c \tan (e+f x)}}{8 a^2 f}+\frac{3 c^{5/2} (-9 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{4 \sqrt{2} a^2 f}-\frac{c (-9 B+i A) (c-i c \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(-B+i A) (c-i c \tan (e+f x))^{5/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 78
Rule 47
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{3/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{4 a^2 f (1+i \tan (e+f x))^2}-\frac{((A+9 i B) c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{(i A-9 B) c (c-i c \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{\left (3 (A+9 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=-\frac{3 (i A-9 B) c^2 \sqrt{c-i c \tan (e+f x)}}{8 a^2 f}-\frac{(i A-9 B) c (c-i c \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{\left (3 (A+9 i B) c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{8 a f}\\ &=-\frac{3 (i A-9 B) c^2 \sqrt{c-i c \tan (e+f x)}}{8 a^2 f}-\frac{(i A-9 B) c (c-i c \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac{\left (3 (i A-9 B) c^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{4 a f}\\ &=\frac{3 (i A-9 B) c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{4 \sqrt{2} a^2 f}-\frac{3 (i A-9 B) c^2 \sqrt{c-i c \tan (e+f x)}}{8 a^2 f}-\frac{(i A-9 B) c (c-i c \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{4 a^2 f (1+i \tan (e+f x))^2}\\ \end{align*}
Mathematica [F] time = 180.004, size = 0, normalized size = 0. \[ \text{\$Aborted} \]
Verification is Not applicable to the result.
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Maple [A] time = 0.108, size = 138, normalized size = 0.7 \begin{align*}{\frac{-2\,i{c}^{2}}{f{a}^{2}} \left ( iB\sqrt{c-ic\tan \left ( fx+e \right ) }+c \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ( \left ( -{\frac{13\,i}{8}}B-{\frac{5\,A}{8}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ({\frac{11\,i}{4}}Bc+{\frac{3\,Ac}{4}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( 27\,iB+3\,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 ) \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 = 1.41553, size = 1011, normalized size = 5.08 \begin{align*} \frac{{\left (\sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{{\left (9 \, A^{2} + 162 i \, A B - 729 \, B^{2}\right )} c^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left ({\left (3 i \, A - 27 \, B\right )} c^{3} + \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{{\left (9 \, A^{2} + 162 i \, A B - 729 \, B^{2}\right )} c^{5}}{a^{4} f^{2}}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a^{2} f}\right ) - \sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{{\left (9 \, A^{2} + 162 i \, A B - 729 \, B^{2}\right )} c^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left ({\left (3 i \, A - 27 \, B\right )} c^{3} - \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{{\left (9 \, A^{2} + 162 i \, A B - 729 \, B^{2}\right )} c^{5}}{a^{4} f^{2}}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a^{2} f}\right ) + \sqrt{2}{\left ({\left (-3 i \, A + 27 \, B\right )} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-i \, A + 9 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (2 i \, A - 2 \, B\right )} c^{2}\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 )}}{8 \, a^{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{{\left (B \tan \left (f x + e\right ) + A\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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