Optimal. Leaf size=110 \[ \frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}+\frac{2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 (-1)^{3/4} a \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.17045, antiderivative size = 110, 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 = {3529, 3533, 208} \[ \frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}+\frac{2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac{2 (-1)^{3/4} a \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3529
Rule 3533
Rule 208
Rubi steps
\begin{align*} \int \frac{a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx &=-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac{\int \frac{-i a d-a d \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{d^2}\\ &=-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac{2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{\int \frac{-a d^2+i a d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{d^4}\\ &=-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac{2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}+\frac{\int \frac{i a d^3+a d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{d^6}\\ &=-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac{2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{\left (2 a^2\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 \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{d^{7/2} f}-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac{2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.169054, size = 41, normalized size = 0.37 \[ -\frac{2 a \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};-i \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.022, size = 397, normalized size = 3.6 \begin{align*} 2\,{\frac{a}{{d}^{3}f\sqrt{d\tan \left ( fx+e \right ) }}}+{\frac{{\frac{2\,i}{3}}a}{{d}^{2}f} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,a}{5\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}+{\frac{{\frac{i}{4}}a\sqrt{2}}{f{d}^{4}}\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\sqrt{2}}{f{d}^{4}}\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\sqrt{2}}{f{d}^{4}}\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\sqrt{2}}{4\,{d}^{3}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\sqrt{2}}{2\,{d}^{3}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\sqrt{2}}{2\,{d}^{3}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 = 2.49246, size = 1156, normalized size = 10.51 \begin{align*} -\frac{15 \,{\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt{-\frac{4 i \, a^{2}}{d^{7} f^{2}}} \log \left (-\frac{{\left ({\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} 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}} \sqrt{-\frac{4 i \, a^{2}}{d^{7} f^{2}}} + 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3} f}\right ) - 15 \,{\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt{-\frac{4 i \, a^{2}}{d^{7} f^{2}}} \log \left (\frac{{\left ({\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} 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}} \sqrt{-\frac{4 i \, a^{2}}{d^{7} f^{2}}} - 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3} f}\right ) -{\left (104 i \, a e^{\left (6 i \, f x + 6 i \, e\right )} - 88 i \, a e^{\left (4 i \, f x + 4 i \, e\right )} - 8 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} + 184 i \, a\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 (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} 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 = 1.20622, size = 176, normalized size = 1.6 \begin{align*} -\frac{2}{15} \, a{\left (-\frac{15 i \, \sqrt{2} \arctan \left (\frac{16 i \, \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 )}{d^{\frac{7}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{15 \, d^{2} \tan \left (f x + e\right )^{2} + 5 i \, d^{2} \tan \left (f x + e\right ) - 3 \, d^{2}}{\sqrt{d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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