Optimal. Leaf size=326 \[ -\frac{\left (\frac{25}{16}-\frac{21 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{25}{16}-\frac{21 i}{16}\right ) d^{7/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 f}-\frac{25 d^3 \sqrt{d \tan (e+f x)}}{8 a^2 f}+\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{\left (\frac{25}{32}+\frac{21 i}{32}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{25}{32}+\frac{21 i}{32}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2} \]
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Rubi [A] time = 0.483863, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3558, 3595, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (\frac{25}{16}-\frac{21 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{25}{16}-\frac{21 i}{16}\right ) d^{7/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 f}-\frac{25 d^3 \sqrt{d \tan (e+f x)}}{8 a^2 f}+\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{\left (\frac{25}{32}+\frac{21 i}{32}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{25}{32}+\frac{21 i}{32}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 3558
Rule 3595
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))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx &=-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}-\frac{\int \frac{(d \tan (e+f x))^{3/2} \left (-\frac{5 a d^2}{2}+\frac{9}{2} i a d^2 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\int \sqrt{d \tan (e+f x)} \left (-\frac{21}{2} i a^2 d^3-\frac{25}{2} a^2 d^3 \tan (e+f x)\right ) \, dx}{8 a^4}\\ &=-\frac{25 d^3 \sqrt{d \tan (e+f x)}}{8 a^2 f}+\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\int \frac{\frac{25 a^2 d^4}{2}-\frac{21}{2} i a^2 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{8 a^4}\\ &=-\frac{25 d^3 \sqrt{d \tan (e+f x)}}{8 a^2 f}+\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\operatorname{Subst}\left (\int \frac{\frac{25 a^2 d^5}{2}-\frac{21}{2} i a^2 d^4 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^4 f}\\ &=-\frac{25 d^3 \sqrt{d \tan (e+f x)}}{8 a^2 f}+\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\left (\left (\frac{25}{16}-\frac{21 i}{16}\right ) d^4\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 (\left (\frac{25}{16}+\frac{21 i}{16}\right ) d^4\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{25 d^3 \sqrt{d \tan (e+f x)}}{8 a^2 f}+\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac{\left (\left (\frac{25}{32}+\frac{21 i}{32}\right ) d^{7/2}\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^2 f}+-\frac{\left (\left (\frac{25}{32}+\frac{21 i}{32}\right ) d^{7/2}\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^2 f}+\frac{\left (\left (\frac{25}{32}-\frac{21 i}{32}\right ) d^4\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 (\left (\frac{25}{32}-\frac{21 i}{32}\right ) d^4\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{25}{32}+\frac{21 i}{32}\right ) d^{7/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{25}{32}+\frac{21 i}{32}\right ) d^{7/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 f}-\frac{25 d^3 \sqrt{d \tan (e+f x)}}{8 a^2 f}+\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac{\left (\left (\frac{25}{16}-\frac{21 i}{16}\right ) d^{7/2}\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^2 f}+\frac{\left (\left (\frac{25}{16}-\frac{21 i}{16}\right ) d^{7/2}\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^2 f}\\ &=-\frac{\left (\frac{25}{16}-\frac{21 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{25}{16}-\frac{21 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}-\frac{\left (\frac{25}{32}+\frac{21 i}{32}\right ) d^{7/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{25}{32}+\frac{21 i}{32}\right ) d^{7/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 f}-\frac{25 d^3 \sqrt{d \tan (e+f x)}}{8 a^2 f}+\frac{7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\\ \end{align*}
Mathematica [A] time = 1.30362, size = 233, normalized size = 0.71 \[ -\frac{i d^4 \sec ^3(e+f x) \left (-23 i \sin (e+f x)+41 i \sin (3 (e+f x))-43 \cos (e+f x)+43 \cos (3 (e+f x))+(21+25 i) \sqrt{\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+(25+21 i) \sin ^{\frac{3}{2}}(2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(21-25 i) \sqrt{\sin (2 (e+f x))} \cos (2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{32 a^2 f (\tan (e+f x)-i)^2 \sqrt{d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.054, size = 168, normalized size = 0.5 \begin{align*} -2\,{\frac{{d}^{3}\sqrt{d\tan \left ( fx+e \right ) }}{{a}^{2}f}}+{\frac{{\frac{11\,i}{8}}{d}^{4}}{{a}^{2}f \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{9\,{d}^{5}}{8\,{a}^{2}f \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{{\frac{23\,i}{8}}{d}^{4}}{{a}^{2}f}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}+{\frac{{\frac{i}{4}}{d}^{4}}{{a}^{2}f}\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 = 2.52459, size = 1558, normalized size = 4.78 \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 = 1.21557, size = 304, normalized size = 0.93 \begin{align*} -\frac{1}{8} \, d^{3}{\left (\frac{23 \, \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^{2} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{2 \, \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^{2} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{16 \, \sqrt{d \tan \left (f x + e\right )}}{a^{2} f} + \frac{-11 i \, \sqrt{d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 9 \, \sqrt{d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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