Optimal. Leaf size=287 \[ -\frac{\left (\frac{3}{4}+\frac{5 i}{4}\right ) d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a f}+\frac{\left (\frac{3}{4}+\frac{5 i}{4}\right ) d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a f}-\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{2 a f}+\frac{\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a f}-\frac{\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a f}-\frac{d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))} \]
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Rubi [A] time = 0.278934, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {3550, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (\frac{3}{4}+\frac{5 i}{4}\right ) d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a f}+\frac{\left (\frac{3}{4}+\frac{5 i}{4}\right ) d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a f}-\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{2 a f}+\frac{\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a f}-\frac{\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a f}-\frac{d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))} \]
Antiderivative was successfully verified.
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Rule 3550
Rule 3528
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{(d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx &=-\frac{d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}+\frac{\int \sqrt{d \tan (e+f x)} \left (\frac{3 a d^2}{2}-\frac{5}{2} i a d^2 \tan (e+f x)\right ) \, dx}{2 a^2}\\ &=-\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{2 a f}-\frac{d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}+\frac{\int \frac{\frac{5}{2} i a d^3+\frac{3}{2} a d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{2 a^2}\\ &=-\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{2 a f}-\frac{d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}+\frac{\operatorname{Subst}\left (\int \frac{\frac{5}{2} i a d^4+\frac{3}{2} a d^3 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 f}\\ &=-\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{2 a f}-\frac{d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}+-\frac{\left (\left (\frac{3}{4}-\frac{5 i}{4}\right ) d^3\right ) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a f}+\frac{\left (\left (\frac{3}{4}+\frac{5 i}{4}\right ) d^3\right ) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a f}\\ &=-\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{2 a f}-\frac{d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}+\frac{\left (\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}+2 x}{-d-\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a f}+\frac{\left (\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}-2 x}{-d+\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a f}+\frac{\left (\left (\frac{3}{8}+\frac{5 i}{8}\right ) d^3\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a f}+\frac{\left (\left (\frac{3}{8}+\frac{5 i}{8}\right ) d^3\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a f}\\ &=\frac{\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a f}-\frac{\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a f}-\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{2 a f}-\frac{d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}+-\frac{\left (\left (\frac{3}{4}+\frac{5 i}{4}\right ) d^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a f}+\frac{\left (\left (\frac{3}{4}+\frac{5 i}{4}\right ) d^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a f}\\ &=-\frac{\left (\frac{3}{4}+\frac{5 i}{4}\right ) d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a f}+\frac{\left (\frac{3}{4}+\frac{5 i}{4}\right ) d^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a f}+\frac{\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a f}-\frac{\left (\frac{3}{8}-\frac{5 i}{8}\right ) d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a f}-\frac{5 i d^2 \sqrt{d \tan (e+f x)}}{2 a f}-\frac{d (d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.79021, size = 164, normalized size = 0.57 \[ \frac{\left (\frac{1}{8}+\frac{i}{8}\right ) d^2 \csc (e+f x) \sqrt{d \tan (e+f x)} \left (-(2+2 i) \sin (e+f x) (4 \tan (e+f x)-5 i)+(-4-i) \sqrt{\sin (2 (e+f x))} (\tan (e+f x)-i) \sin ^{-1}(\cos (e+f x)-\sin (e+f x))+(1+4 i) \sqrt{\sin (2 (e+f x))} (\tan (e+f x)-i) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{a f (\tan (e+f x)-i)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.047, size = 130, normalized size = 0.5 \begin{align*}{\frac{-2\,i{d}^{2}}{fa}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{{d}^{3}}{2\,fa \left ( -id+d\tan \left ( fx+e \right ) \right ) }\sqrt{d\tan \left ( fx+e \right ) }}+2\,{\frac{{d}^{3}}{fa\sqrt{-id}}\arctan \left ({\frac{\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt{-id}}} \right ) }-{\frac{{d}^{3}}{2\,fa}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id}}}} \right ){\frac{1}{\sqrt{id}}}} \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: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.16328, size = 1436, normalized size = 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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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 [A] time = 1.2196, size = 266, normalized size = 0.93 \begin{align*} -\frac{1}{2} \, d^{2}{\left (\frac{4 i \, \sqrt{2} \sqrt{d} \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 )}{a f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{i \, \sqrt{2} \sqrt{d} \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 )}{a f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{4 i \, \sqrt{d \tan \left (f x + e\right )}}{a f} + \frac{\sqrt{d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a f}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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