Optimal. Leaf size=107 \[ -\frac{2 (-1)^{3/4} a d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f} \]
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Rubi [A] time = 0.160267, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3528, 3533, 205} \[ -\frac{2 (-1)^{3/4} a d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f} \]
Antiderivative was successfully verified.
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Rule 3528
Rule 3533
Rule 205
Rubi steps
\begin{align*} \int (d \tan (e+f x))^{5/2} (a+i a \tan (e+f x)) \, dx &=\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}+\int (d \tan (e+f x))^{3/2} (-i a d+a d \tan (e+f x)) \, dx\\ &=\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}+\int \sqrt{d \tan (e+f x)} \left (-a d^2-i a d^2 \tan (e+f x)\right ) \, dx\\ &=-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}+\int \frac{i a d^3-a d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}-\frac{\left (2 a^2 d^6\right ) \operatorname{Subst}\left (\int \frac{1}{i a d^4+a d^3 x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=-\frac{2 (-1)^{3/4} a d^{5/2} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{f}-\frac{2 i a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 i a (d \tan (e+f x))^{5/2}}{5 f}\\ \end{align*}
Mathematica [A] time = 2.45962, size = 125, normalized size = 1.17 \[ \frac{a d^2 \sqrt{d \tan (e+f x)} \left (30 i \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+\sqrt{i \tan (e+f x)} \sec ^2(e+f x) (5 \sin (2 (e+f x))-18 i \cos (2 (e+f x))-12 i)\right )}{15 f \sqrt{i \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.026, size = 393, normalized size = 3.7 \begin{align*}{\frac{{\frac{2\,i}{5}}a}{f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,ad}{3\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{2\,ia{d}^{2}}{f}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{{\frac{i}{4}}a{d}^{2}\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 ) }+{\frac{{\frac{i}{2}}a{d}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{{\frac{i}{2}}a{d}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{a{d}^{3}\sqrt{2}}{4\,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{a{d}^{3}\sqrt{2}}{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{a{d}^{3}\sqrt{2}}{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 = 2.06221, size = 1013, normalized size = 9.47 \begin{align*} -\frac{15 \, \sqrt{\frac{4 i \, a^{2} d^{5}}{f^{2}}}{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-2 i \, a d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{4 i \, a^{2} d^{5}}{f^{2}}}{\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, 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 )}}{a d^{2}}\right ) - 15 \, \sqrt{\frac{4 i \, a^{2} d^{5}}{f^{2}}}{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{{\left (-2 i \, a d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt{\frac{4 i \, a^{2} d^{5}}{f^{2}}}{\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, 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 )}}{a d^{2}}\right ) -{\left (-184 i \, a d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 192 i \, a d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 104 i \, a d^{2}\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}}}{60 \,{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, 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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.18068, size = 205, normalized size = 1.92 \begin{align*} -\frac{2}{15} \, a d{\left (\frac{15 \, \sqrt{2} d^{\frac{3}{2}} \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{-3 i \, \sqrt{d \tan \left (f x + e\right )} d^{6} f^{4} \tan \left (f x + e\right )^{2} - 5 \, \sqrt{d \tan \left (f x + e\right )} d^{6} f^{4} \tan \left (f x + e\right ) + 15 i \, \sqrt{d \tan \left (f x + e\right )} d^{6} f^{4}}{d^{5} f^{5}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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