Optimal. Leaf size=343 \[ -\frac{\left (\frac{7}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\left (\frac{7}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{5 \sqrt{d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\left (\frac{7}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\left (\frac{7}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3} \]
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Rubi [A] time = 0.589638, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {3559, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (\frac{7}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\left (\frac{7}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{5 \sqrt{d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\left (\frac{7}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\left (\frac{7}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3} \]
Antiderivative was successfully verified.
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Rule 3559
Rule 3596
Rule 3534
Rule 1168
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{1}{\sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^3} \, dx &=\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac{\int \frac{\frac{11 a d}{2}-\frac{5}{2} i a d \tan (e+f x)}{\sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2} \, dx}{6 a^2 d}\\ &=\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac{\sqrt{d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac{\int \frac{18 a^2 d^2-12 i a^2 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{24 a^4 d^2}\\ &=\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac{\sqrt{d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac{5 \sqrt{d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\int \frac{21 a^3 d^3-15 i a^3 d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{48 a^6 d^3}\\ &=\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac{\sqrt{d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac{5 \sqrt{d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{21 a^3 d^4-15 i a^3 d^3 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{24 a^6 d^3 f}\\ &=\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac{\sqrt{d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac{5 \sqrt{d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\left (\frac{7}{16}-\frac{5 i}{16}\right ) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^3 f}+\frac{\left (\frac{7}{16}+\frac{5 i}{16}\right ) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^3 f}\\ &=\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac{\sqrt{d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac{5 \sqrt{d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\left (\frac{7}{32}-\frac{5 i}{32}\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^3 f}+\frac{\left (\frac{7}{32}-\frac{5 i}{32}\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^3 f}+-\frac{\left (\frac{7}{32}+\frac{5 i}{32}\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^3 \sqrt{d} f}+-\frac{\left (\frac{7}{32}+\frac{5 i}{32}\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^3 \sqrt{d} f}\\ &=-\frac{\left (\frac{7}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\left (\frac{7}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac{\sqrt{d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac{5 \sqrt{d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+-\frac{\left (\frac{7}{16}-\frac{5 i}{16}\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^3 \sqrt{d} f}+\frac{\left (\frac{7}{16}-\frac{5 i}{16}\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^3 \sqrt{d} f}\\ &=-\frac{\left (\frac{7}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\left (\frac{7}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 \sqrt{d} f}-\frac{\left (\frac{7}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\left (\frac{7}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^3 \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac{\sqrt{d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac{5 \sqrt{d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 1.1363, size = 234, normalized size = 0.68 \[ \frac{\sec ^4(e+f x) \left (19 \cos (4 (e+f x))-(15+21 i) \sqrt{\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+i \left (12 \sin (2 (e+f x))+21 \sin (4 (e+f x))+(-15+21 i) \sqrt{\sin (2 (e+f x))} \sin (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(21+15 i) \sqrt{\sin (2 (e+f x))} \cos (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+19 i\right )\right )}{96 a^3 f (\tan (e+f x)-i)^3 \sqrt{d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.07, size = 173, normalized size = 0.5 \begin{align*}{\frac{-{\frac{5\,i}{8}}}{f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{19\,d}{12\,f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{9\,i}{8}}{d}^{2}}{f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{{\frac{3\,i}{4}}}{f{a}^{3}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}+{\frac{{\frac{i}{8}}}{f{a}^{3}}\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 = 2.49382, size = 1548, normalized size = 4.51 \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 = 1.17141, size = 316, normalized size = 0.92 \begin{align*} -\frac{1}{24} \, d^{4}{\left (-\frac{3 i \, \sqrt{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 )}{a^{3} d^{\frac{9}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{18 i \, \sqrt{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 )}{a^{3} d^{\frac{9}{2}} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{15 i \, \sqrt{d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} + 38 \, \sqrt{d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 27 i \, \sqrt{d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} d^{4} f}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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