Optimal. Leaf size=353 \[ -\frac{\left (\frac{49}{16}+\frac{45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{49}{16}+\frac{45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 f}-\frac{45 i d^4 \sqrt{d \tan (e+f x)}}{8 a^2 f}-\frac{49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{\left (\frac{49}{32}-\frac{45 i}{32}\right ) d^{9/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}-\frac{\left (\frac{49}{32}-\frac{45 i}{32}\right ) d^{9/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2} \]
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Rubi [A] time = 0.535632, antiderivative size = 353, normalized size of antiderivative = 1., number of steps used = 14, 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} \[ -\frac{\left (\frac{49}{16}+\frac{45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{49}{16}+\frac{45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 f}-\frac{45 i d^4 \sqrt{d \tan (e+f x)}}{8 a^2 f}-\frac{49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac{\left (\frac{49}{32}-\frac{45 i}{32}\right ) d^{9/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}-\frac{\left (\frac{49}{32}-\frac{45 i}{32}\right ) d^{9/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 f}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2} \]
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))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx &=-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}-\frac{\int \frac{(d \tan (e+f x))^{5/2} \left (-\frac{7 a d^2}{2}+\frac{11}{2} i a d^2 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\int (d \tan (e+f x))^{3/2} \left (-\frac{45}{2} i a^2 d^3-\frac{49}{2} a^2 d^3 \tan (e+f x)\right ) \, dx}{8 a^4}\\ &=-\frac{49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\int \sqrt{d \tan (e+f x)} \left (\frac{49 a^2 d^4}{2}-\frac{45}{2} i a^2 d^4 \tan (e+f x)\right ) \, dx}{8 a^4}\\ &=-\frac{45 i d^4 \sqrt{d \tan (e+f x)}}{8 a^2 f}-\frac{49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\int \frac{\frac{45}{2} i a^2 d^5+\frac{49}{2} a^2 d^5 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{8 a^4}\\ &=-\frac{45 i d^4 \sqrt{d \tan (e+f x)}}{8 a^2 f}-\frac{49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\operatorname{Subst}\left (\int \frac{\frac{45}{2} i a^2 d^6+\frac{49}{2} a^2 d^5 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^4 f}\\ &=-\frac{45 i d^4 \sqrt{d \tan (e+f x)}}{8 a^2 f}-\frac{49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac{\left (\left (\frac{49}{16}-\frac{45 i}{16}\right ) d^5\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{49}{16}+\frac{45 i}{16}\right ) d^5\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{45 i d^4 \sqrt{d \tan (e+f x)}}{8 a^2 f}-\frac{49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac{\left (\left (\frac{49}{32}-\frac{45 i}{32}\right ) d^{9/2}\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{49}{32}-\frac{45 i}{32}\right ) d^{9/2}\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{49}{32}+\frac{45 i}{32}\right ) d^5\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{49}{32}+\frac{45 i}{32}\right ) d^5\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{49}{32}-\frac{45 i}{32}\right ) d^{9/2} \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{49}{32}-\frac{45 i}{32}\right ) d^{9/2} \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{45 i d^4 \sqrt{d \tan (e+f x)}}{8 a^2 f}-\frac{49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac{\left (\left (\frac{49}{16}+\frac{45 i}{16}\right ) d^{9/2}\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{49}{16}+\frac{45 i}{16}\right ) d^{9/2}\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{49}{16}+\frac{45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{49}{16}+\frac{45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 f}+\frac{\left (\frac{49}{32}-\frac{45 i}{32}\right ) d^{9/2} \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{49}{32}-\frac{45 i}{32}\right ) d^{9/2} \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{45 i d^4 \sqrt{d \tan (e+f x)}}{8 a^2 f}-\frac{49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac{9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac{d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\\ \end{align*}
Mathematica [A] time = 2.14688, size = 346, normalized size = 0.98 \[ \frac{d^5 \sec ^4(e+f x) \left (142 i \sin (2 (e+f x))+199 i \sin (4 (e+f x))+64 \cos (2 (e+f x))+205 \cos (4 (e+f x))+(294+270 i) \sqrt{\sin (2 (e+f x))} \cos (e+f x) \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+(135+147 i) \sqrt{\sin (2 (e+f x))} \sin (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(147-135 i) \sqrt{\sin (2 (e+f x))} \cos (e+f x) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(147-135 i) \sqrt{\sin (2 (e+f x))} \cos (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(135+147 i) \sin (e+f x) \sqrt{\sin (2 (e+f x))} \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )-269\right )}{192 a^2 f (\tan (e+f x)-i)^2 \sqrt{d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.055, size = 188, normalized size = 0.5 \begin{align*} -{\frac{2\,{d}^{3}}{3\,{a}^{2}f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{4\,i{d}^{4}}{{a}^{2}f}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{15\,{d}^{5}}{8\,{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{{\frac{13\,i}{8}}{d}^{6}}{{a}^{2}f \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{47\,{d}^{5}}{8\,{a}^{2}f}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}+{\frac{{d}^{5}}{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 = 2.69814, size = 1840, normalized size = 5.21 \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 = 1.2359, size = 354, normalized size = 1. \begin{align*} -\frac{1}{24} \, d^{3}{\left (\frac{141 i \, \sqrt{2} d^{\frac{3}{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} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{6 i \, \sqrt{2} d^{\frac{3}{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} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{3 \,{\left (15 \, \sqrt{d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) - 13 i \, \sqrt{d \tan \left (f x + e\right )} d^{3}\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f} + \frac{16 \,{\left (\sqrt{d \tan \left (f x + e\right )} a^{4} d f^{2} \tan \left (f x + e\right ) + 6 i \, \sqrt{d \tan \left (f x + e\right )} a^{4} d f^{2}\right )}}{a^{6} f^{3}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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