Optimal. Leaf size=287 \[ \frac{\left (\frac{5}{4}+\frac{3 i}{4}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a d^{3/2} f}-\frac{\left (\frac{5}{4}+\frac{3 i}{4}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a d^{3/2} f}-\frac{\left (\frac{5}{8}-\frac{3 i}{8}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a d^{3/2} f}+\frac{\left (\frac{5}{8}-\frac{3 i}{8}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a d^{3/2} f}-\frac{5}{2 a d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f (a+i a \tan (e+f x)) \sqrt{d \tan (e+f x)}} \]
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Rubi [A] time = 0.309243, 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 = {3552, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\frac{5}{4}+\frac{3 i}{4}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a d^{3/2} f}-\frac{\left (\frac{5}{4}+\frac{3 i}{4}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a d^{3/2} f}-\frac{\left (\frac{5}{8}-\frac{3 i}{8}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a d^{3/2} f}+\frac{\left (\frac{5}{8}-\frac{3 i}{8}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a d^{3/2} f}-\frac{5}{2 a d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f (a+i a \tan (e+f x)) \sqrt{d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 3552
Rule 3529
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx &=\frac{1}{2 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))}-\frac{\int \frac{-\frac{5 a d}{2}+\frac{3}{2} i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{2 a^2 d}\\ &=-\frac{5}{2 a d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))}-\frac{\int \frac{\frac{3}{2} i a d^2+\frac{5}{2} a d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{2 a^2 d^3}\\ &=-\frac{5}{2 a d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))}-\frac{\operatorname{Subst}\left (\int \frac{\frac{3}{2} i a d^3+\frac{5}{2} a d^2 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 d^3 f}\\ &=-\frac{5}{2 a d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))}--\frac{\left (\frac{5}{4}-\frac{3 i}{4}\right ) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a d f}-\frac{\left (\frac{5}{4}+\frac{3 i}{4}\right ) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a d f}\\ &=-\frac{5}{2 a d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))}-\frac{\left (\frac{5}{8}-\frac{3 i}{8}\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 d^{3/2} f}-\frac{\left (\frac{5}{8}-\frac{3 i}{8}\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 d^{3/2} f}-\frac{\left (\frac{5}{8}+\frac{3 i}{8}\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 d f}-\frac{\left (\frac{5}{8}+\frac{3 i}{8}\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 d f}\\ &=-\frac{\left (\frac{5}{8}-\frac{3 i}{8}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a d^{3/2} f}+\frac{\left (\frac{5}{8}-\frac{3 i}{8}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a d^{3/2} f}-\frac{5}{2 a d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))}--\frac{\left (\frac{5}{4}+\frac{3 i}{4}\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 d^{3/2} f}-\frac{\left (\frac{5}{4}+\frac{3 i}{4}\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 d^{3/2} f}\\ &=\frac{\left (\frac{5}{4}+\frac{3 i}{4}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a d^{3/2} f}-\frac{\left (\frac{5}{4}+\frac{3 i}{4}\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a d^{3/2} f}-\frac{\left (\frac{5}{8}-\frac{3 i}{8}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a d^{3/2} f}+\frac{\left (\frac{5}{8}-\frac{3 i}{8}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a d^{3/2} f}-\frac{5}{2 a d f \sqrt{d \tan (e+f x)}}+\frac{1}{2 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.22812, size = 155, normalized size = 0.54 \[ \frac{-20 \tan (e+f x)+(5+3 i) \sqrt{\sin (2 (e+f x))} (\tan (e+f x)-i) \sec (e+f x) \sin ^{-1}(\cos (e+f x)-\sin (e+f x))+(5-3 i) \sqrt{\sin (2 (e+f x))} (\tan (e+f x)-i) \sec (e+f x) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+16 i}{8 a d f (\tan (e+f x)-i) \sqrt{d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.056, size = 129, normalized size = 0.5 \begin{align*} -{\frac{1}{2\,fad \left ( -id+d\tan \left ( fx+e \right ) \right ) }\sqrt{d\tan \left ( fx+e \right ) }}-2\,{\frac{1}{fad\sqrt{-id}}\arctan \left ({\frac{\sqrt{d\tan \left ( fx+e \right ) }}{\sqrt{-id}}} \right ) }-2\,{\frac{1}{fad\sqrt{d\tan \left ( fx+e \right ) }}}-{\frac{1}{2\,fad}\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 = 2.11409, size = 1706, normalized size = 5.94 \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 = 1.1759, size = 277, normalized size = 0.97 \begin{align*} -\frac{1}{2} \, d^{2}{\left (\frac{5 i \, d \tan \left (f x + e\right ) + 4 \, d}{{\left (i \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + \sqrt{d \tan \left (f x + e\right )} d\right )} a d^{3} f} + \frac{4 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 )}{a d^{\frac{7}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{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 )}{a d^{\frac{7}{2}} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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