Optimal. Leaf size=152 \[ \frac{8 \sqrt [4]{-1} a^3 d^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac{8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac{8 a^3 d \sqrt{d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{5/2}}{7 d f} \]
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Rubi [A] time = 0.266388, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3556, 3592, 3528, 3533, 205} \[ \frac{8 \sqrt [4]{-1} a^3 d^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac{8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac{8 a^3 d \sqrt{d \tan (e+f x)}}{f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{5/2}}{7 d f} \]
Antiderivative was successfully verified.
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Rule 3556
Rule 3592
Rule 3528
Rule 3533
Rule 205
Rubi steps
\begin{align*} \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3 \, dx &=-\frac{2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac{(2 a) \int (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x)) (6 a d+8 i a d \tan (e+f x)) \, dx}{7 d}\\ &=-\frac{32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac{2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac{(2 a) \int (d \tan (e+f x))^{3/2} \left (14 a^2 d+14 i a^2 d \tan (e+f x)\right ) \, dx}{7 d}\\ &=\frac{8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}-\frac{32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac{2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac{(2 a) \int \sqrt{d \tan (e+f x)} \left (-14 i a^2 d^2+14 a^2 d^2 \tan (e+f x)\right ) \, dx}{7 d}\\ &=\frac{8 a^3 d \sqrt{d \tan (e+f x)}}{f}+\frac{8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}-\frac{32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac{2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac{(2 a) \int \frac{-14 a^2 d^3-14 i a^2 d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{7 d}\\ &=\frac{8 a^3 d \sqrt{d \tan (e+f x)}}{f}+\frac{8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}-\frac{32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac{2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}+\frac{\left (112 a^5 d^5\right ) \operatorname{Subst}\left (\int \frac{1}{-14 a^2 d^4+14 i a^2 d^3 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=\frac{8 \sqrt [4]{-1} a^3 d^{3/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}+\frac{8 a^3 d \sqrt{d \tan (e+f x)}}{f}+\frac{8 i a^3 (d \tan (e+f x))^{3/2}}{3 f}-\frac{32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}-\frac{2 (d \tan (e+f x))^{5/2} \left (a^3+i a^3 \tan (e+f x)\right )}{7 d f}\\ \end{align*}
Mathematica [A] time = 3.02798, size = 138, normalized size = 0.91 \[ \frac{a^3 d \sqrt{d \tan (e+f x)} \left (\sqrt{i \tan (e+f x)} \sec ^3(e+f x) (95 i \sin (e+f x)+155 i \sin (3 (e+f x))+1197 \cos (e+f x)+483 \cos (3 (e+f x)))-1680 \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{210 f \sqrt{i \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.023, size = 426, normalized size = 2.8 \begin{align*}{\frac{-{\frac{2\,i}{7}}{a}^{3}}{f{d}^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{6\,{a}^{3}}{5\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{{\frac{8\,i}{3}}{a}^{3}}{f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+8\,{\frac{{a}^{3}d\sqrt{d\tan \left ( fx+e \right ) }}{f}}-{\frac{{a}^{3}d\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }-2\,{\frac{{a}^{3}d\sqrt [4]{{d}^{2}}\sqrt{2}}{f}\arctan \left ({\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }+2\,{\frac{{a}^{3}d\sqrt [4]{{d}^{2}}\sqrt{2}}{f}\arctan \left ( -{\frac{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt [4]{{d}^{2}}}}+1 \right ) }-{\frac{i{a}^{3}{d}^{2}\sqrt{2}}{f}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{2\,i{a}^{3}{d}^{2}\sqrt{2}}{f}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{2\,i{a}^{3}{d}^{2}\sqrt{2}}{f}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \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 = 2.22293, size = 1175, normalized size = 7.73 \begin{align*} -\frac{105 \, \sqrt{-\frac{64 i \, a^{6} d^{3}}{f^{2}}}{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-8 i \, a^{3} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{-\frac{64 i \, a^{6} d^{3}}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} d}\right ) - 105 \, \sqrt{-\frac{64 i \, a^{6} d^{3}}{f^{2}}}{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-8 i \, a^{3} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - \sqrt{-\frac{64 i \, a^{6} d^{3}}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} d}\right ) - 16 \,{\left (319 \, a^{3} d e^{\left (6 i \, f x + 6 i \, e\right )} + 646 \, a^{3} d e^{\left (4 i \, f x + 4 i \, e\right )} + 551 \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} + 164 \, a^{3} d\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{420 \,{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 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 \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}\, dx + \int - 3 \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx + \int 3 i \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan{\left (e + f x \right )}\, dx + \int - i \left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan ^{3}{\left (e + f x \right )}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.26121, size = 262, normalized size = 1.72 \begin{align*} -\frac{1}{105} \,{\left (\frac{840 i \, \sqrt{2} a^{3} \sqrt{d} \arctan \left (\frac{16 \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{30 i \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{21} f^{6} \tan \left (f x + e\right )^{3} + 126 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{21} f^{6} \tan \left (f x + e\right )^{2} - 280 i \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{21} f^{6} \tan \left (f x + e\right ) - 840 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{21} f^{6}}{d^{21} f^{7}}\right )} d \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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