Optimal. Leaf size=301 \[ -\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 \sqrt{d} f}+\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 \sqrt{d} f}+\frac{5 \sqrt{d \tan (e+f x)}}{8 a^2 d f (1+i \tan (e+f x))}-\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 \sqrt{d} f}+\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{4 d f (a+i a \tan (e+f x))^2} \]
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Rubi [A] time = 0.405107, antiderivative size = 301, 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 = {3559, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 \sqrt{d} f}+\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 \sqrt{d} f}+\frac{5 \sqrt{d \tan (e+f x)}}{8 a^2 d f (1+i \tan (e+f x))}-\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 \sqrt{d} f}+\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{4 d f (a+i a \tan (e+f x))^2} \]
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))^2} \, dx &=\frac{\sqrt{d \tan (e+f x)}}{4 d f (a+i a \tan (e+f x))^2}+\frac{\int \frac{\frac{7 a d}{2}-\frac{3}{2} i a d \tan (e+f x)}{\sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{4 a^2 d}\\ &=\frac{5 \sqrt{d \tan (e+f x)}}{8 a^2 d f (1+i \tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 d f (a+i a \tan (e+f x))^2}+\frac{\int \frac{\frac{9 a^2 d^2}{2}-\frac{5}{2} i a^2 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{8 a^4 d^2}\\ &=\frac{5 \sqrt{d \tan (e+f x)}}{8 a^2 d f (1+i \tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 d f (a+i a \tan (e+f x))^2}+\frac{\operatorname{Subst}\left (\int \frac{\frac{9 a^2 d^3}{2}-\frac{5}{2} i a^2 d^2 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^4 d^2 f}\\ &=\frac{5 \sqrt{d \tan (e+f x)}}{8 a^2 d f (1+i \tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 d f (a+i a \tan (e+f x))^2}+\frac{\left (\frac{9}{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^2 f}+\frac{\left (\frac{9}{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^2 f}\\ &=\frac{5 \sqrt{d \tan (e+f x)}}{8 a^2 d f (1+i \tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 d f (a+i a \tan (e+f x))^2}+\frac{\left (\frac{9}{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^2 f}+\frac{\left (\frac{9}{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^2 f}+-\frac{\left (\frac{9}{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^2 \sqrt{d} f}+-\frac{\left (\frac{9}{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^2 \sqrt{d} f}\\ &=-\frac{\left (\frac{9}{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^2 \sqrt{d} f}+\frac{\left (\frac{9}{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^2 \sqrt{d} f}+\frac{5 \sqrt{d \tan (e+f x)}}{8 a^2 d f (1+i \tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 d f (a+i a \tan (e+f x))^2}+-\frac{\left (\frac{9}{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^2 \sqrt{d} f}+\frac{\left (\frac{9}{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^2 \sqrt{d} f}\\ &=-\frac{\left (\frac{9}{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^2 \sqrt{d} f}+\frac{\left (\frac{9}{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^2 \sqrt{d} f}-\frac{\left (\frac{9}{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^2 \sqrt{d} f}+\frac{\left (\frac{9}{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^2 \sqrt{d} f}+\frac{5 \sqrt{d \tan (e+f x)}}{8 a^2 d f (1+i \tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 d f (a+i a \tan (e+f x))^2}\\ \end{align*}
Mathematica [A] time = 0.961424, size = 228, normalized size = 0.76 \[ \frac{\sec ^3(e+f x) \left (-7 \sin (e+f x)-7 \sin (3 (e+f x))-5 i \cos (e+f x)+5 i \cos (3 (e+f x))+(5+9 i) \sqrt{\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\sin (2 (e+f x))-i \cos (2 (e+f x)))+(5-9 i) \sin ^{\frac{3}{2}}(2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )-(9+5 i) \sqrt{\sin (2 (e+f x))} \cos (2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{32 a^2 f (\tan (e+f x)-i)^2 \sqrt{d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.065, size = 136, normalized size = 0.5 \begin{align*}{\frac{-{\frac{5\,i}{8}}}{f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{7\,d}{8\,f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{{\frac{7\,i}{8}}}{f{a}^{2}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}+{\frac{{\frac{i}{4}}}{f{a}^{2}}\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 = 2.47761, size = 1511, normalized size = 5.02 \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 = 1.16063, size = 278, normalized size = 0.92 \begin{align*} \frac{1}{8} \, d^{3}{\left (\frac{7 \, \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^{2} d^{\frac{7}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{2 \, \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^{2} d^{\frac{7}{2}} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{5 i \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 7 \, \sqrt{d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} d^{3} f}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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