Optimal. Leaf size=150 \[ \frac{4 a^3 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{8 i a^3 \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac{8 i a^3 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]
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Rubi [A] time = 0.423102, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3556, 3592, 3528, 3537, 63, 208} \[ \frac{4 a^3 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{8 i a^3 \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac{8 i a^3 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]
Antiderivative was successfully verified.
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Rule 3556
Rule 3592
Rule 3528
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int (a+i a \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)} \, dx &=-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac{(2 a) \int (a+i a \tan (e+f x)) (a (i c+4 d)+a (c+6 i d) \tan (e+f x)) \sqrt{c+d \tan (e+f x)} \, dx}{5 d}\\ &=\frac{4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac{(2 a) \int \sqrt{c+d \tan (e+f x)} \left (10 a^2 d+10 i a^2 d \tan (e+f x)\right ) \, dx}{5 d}\\ &=\frac{8 i a^3 \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac{(2 a) \int \frac{10 a^2 (c-i d) d+10 a^2 d (i c+d) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{5 d}\\ &=\frac{8 i a^3 \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac{\left (40 i a^5 (c-i d)^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{\left (100 a^4 d^2 (i c+d)^2+10 a^2 (c-i d) d x\right ) \sqrt{c+\frac{x}{10 a^2 (i c+d)}}} \, dx,x,10 a^2 d (i c+d) \tan (e+f x)\right )}{f}\\ &=\frac{8 i a^3 \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac{\left (800 a^7 (c-i d)^3 d\right ) \operatorname{Subst}\left (\int \frac{1}{-100 a^4 c (c-i d) d (i c+d)+100 a^4 d^2 (i c+d)^2+100 a^4 (c-i d) d (i c+d) x^2} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{f}\\ &=-\frac{8 i a^3 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{8 i a^3 \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\\ \end{align*}
Mathematica [A] time = 5.31257, size = 219, normalized size = 1.46 \[ \frac{a^3 (\cos (e+f x)+i \sin (e+f x))^3 \left (\frac{(\sin (3 e)+i \cos (3 e)) \sec ^2(e+f x) \sqrt{c+d \tan (e+f x)} \left (\left (2 c^2+15 i c d+63 d^2\right ) \cos (2 (e+f x))+2 c^2-d (c-15 i d) \sin (2 (e+f x))+15 i c d+57 d^2\right )}{15 d^2}-8 i e^{-3 i e} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )\right )}{f (\cos (f x)+i \sin (f x))^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.079, size = 1272, normalized size = 8.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} \sqrt{d \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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Fricas [B] time = 2.63262, size = 1353, normalized size = 9.02 \begin{align*} \frac{15 \,{\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt{-\frac{64 \, a^{6} c - 64 i \, a^{6} d}{f^{2}}} \log \left (\frac{{\left (8 \, a^{3} c +{\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{64 \, a^{6} c - 64 i \, a^{6} d}{f^{2}}} +{\left (8 \, a^{3} c - 8 i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 15 \,{\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt{-\frac{64 \, a^{6} c - 64 i \, a^{6} d}{f^{2}}} \log \left (\frac{{\left (8 \, a^{3} c +{\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{64 \, a^{6} c - 64 i \, a^{6} d}{f^{2}}} +{\left (8 \, a^{3} c - 8 i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) +{\left (16 i \, a^{3} c^{2} - 112 \, a^{3} c d + 384 i \, a^{3} d^{2} +{\left (16 i \, a^{3} c^{2} - 128 \, a^{3} c d + 624 i \, a^{3} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (32 i \, a^{3} c^{2} - 240 \, a^{3} c d + 912 i \, a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \,{\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int - 3 \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int 3 i \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx + \int - i \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \sqrt{c + d \tan{\left (e + f x \right )}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.54773, size = 374, normalized size = 2.49 \begin{align*} \frac{2 \,{\left (16 i \, a^{3} c + 16 \, a^{3} d\right )} \arctan \left (\frac{4 \,{\left (\sqrt{d \tan \left (f x + e\right ) + c} c - \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} - i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{\sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} f{\left (-\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} - \frac{6 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a^{3} d^{8} f^{4} - 10 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{3} c d^{8} f^{4} + 30 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{3} d^{9} f^{4} - 120 i \, \sqrt{d \tan \left (f x + e\right ) + c} a^{3} d^{10} f^{4}}{15 \, d^{10} f^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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