Optimal. Leaf size=343 \[ \frac{\left (\frac{5}{16}-\frac{7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 f}-\frac{\left (\frac{5}{16}-\frac{7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^3 f}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\left (\frac{5}{32}+\frac{7 i}{32}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 f}-\frac{\left (\frac{5}{32}+\frac{7 i}{32}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 f}+\frac{i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3} \]
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Rubi [A] time = 0.610754, 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 = {3558, 3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\frac{5}{16}-\frac{7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^3 f}-\frac{\left (\frac{5}{16}-\frac{7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^3 f}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\left (\frac{5}{32}+\frac{7 i}{32}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 f}-\frac{\left (\frac{5}{32}+\frac{7 i}{32}\right ) d^{7/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^3 f}+\frac{i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3} \]
Antiderivative was successfully verified.
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Rule 3558
Rule 3595
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))^3} \, dx &=-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}-\frac{\int \frac{(d \tan (e+f x))^{3/2} \left (-\frac{5 a d^2}{2}+\frac{11}{2} i a d^2 \tan (e+f x)\right )}{(a+i a \tan (e+f x))^2} \, dx}{6 a^2}\\ &=-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac{\int \frac{\sqrt{d \tan (e+f x)} \left (-12 i a^2 d^3-18 a^2 d^3 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{24 a^4}\\ &=-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\int \frac{15 a^3 d^4-21 i a^3 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{48 a^6}\\ &=-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{15 a^3 d^5-21 i a^3 d^4 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{24 a^6 f}\\ &=-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{\left (\left (\frac{5}{16}-\frac{7 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^3 f}-\frac{\left (\left (\frac{5}{16}+\frac{7 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^3 f}\\ &=-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac{\left (\left (\frac{5}{32}+\frac{7 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^3 f}--\frac{\left (\left (\frac{5}{32}+\frac{7 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^3 f}-\frac{\left (\left (\frac{5}{32}-\frac{7 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^3 f}-\frac{\left (\left (\frac{5}{32}-\frac{7 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^3 f}\\ &=\frac{\left (\frac{5}{32}+\frac{7 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^3 f}-\frac{\left (\frac{5}{32}+\frac{7 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^3 f}-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac{\left (\left (\frac{5}{16}-\frac{7 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^3 f}-\frac{\left (\left (\frac{5}{16}-\frac{7 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^3 f}\\ &=\frac{\left (\frac{5}{16}-\frac{7 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^3 f}-\frac{\left (\frac{5}{16}-\frac{7 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^3 f}+\frac{\left (\frac{5}{32}+\frac{7 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^3 f}-\frac{\left (\frac{5}{32}+\frac{7 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^3 f}-\frac{d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac{5 d^3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 1.25322, size = 234, normalized size = 0.68 \[ \frac{d^4 \sec ^4(e+f x) \left (-12 i \sin (2 (e+f x))+21 i \sin (4 (e+f x))+19 \cos (4 (e+f x))+(21+15 i) \sqrt{\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+(15+21 i) \sqrt{\sin (2 (e+f x))} \sin (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )+(21-15 i) \sqrt{\sin (2 (e+f x))} \cos (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )-19\right )}{96 a^3 f (\tan (e+f x)-i)^3 \sqrt{d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.058, size = 184, normalized size = 0.5 \begin{align*}{\frac{-{\frac{9\,i}{8}}{d}^{4}}{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}^{5}}{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{5\,i}{8}}{d}^{6}}{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}}{d}^{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}}{d}^{4}}{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 = 2.54349, size = 1593, normalized size = 4.64 \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 = 1.21084, size = 312, normalized size = 0.91 \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} \sqrt{d} 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} \sqrt{d} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{27 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 ) - 15 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} f}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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