Optimal. Leaf size=255 \[ -\frac{35 i a^{9/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac{35 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]
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Rubi [A] time = 0.214565, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3523, 47, 50, 63, 217, 203} \[ -\frac{35 i a^{9/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}+\frac{35 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3523
Rule 47
Rule 50
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{7/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac{\left (7 a^2\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{\left (35 a^3\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{3 c f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}+\frac{\left (35 a^4\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{35 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}+\frac{\left (35 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{35 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}-\frac{\left (35 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c-\frac{c x^2}{a}}} \, dx,x,\sqrt{a+i a \tan (e+f x)}\right )}{c f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{35 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}-\frac{\left (35 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c x^2}{a}} \, dx,x,\frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c-i c \tan (e+f x)}}\right )}{c f}\\ &=-\frac{35 i a^{9/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{3/2} f}-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{35 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^2 f}+\frac{35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{6 c^2 f}\\ \end{align*}
Mathematica [A] time = 12.4928, size = 386, normalized size = 1.51 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{9/2} \left (\cos (2 f x) \left (\frac{32 \sin (2 e)}{3 c^2}+\frac{32 i \cos (2 e)}{3 c^2}\right )+\sin (2 f x) \left (-\frac{32 \cos (2 e)}{3 c^2}+\frac{32 i \sin (2 e)}{3 c^2}\right )-\sec (e) \sin (f x) \left (\frac{\cos (4 e)}{2 c^2}-\frac{i \sin (4 e)}{2 c^2}\right ) \sec (e+f x)+\sec (e) (36 \cos (e)+i \sin (e)) \left (\frac{\sin (4 e)}{2 c^2}+\frac{i \cos (4 e)}{2 c^2}\right )+\frac{4 \sin (4 f x)}{3 c^2}-\frac{4 i \cos (4 f x)}{3 c^2}\right ) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))}}{f (\cos (f x)+i \sin (f x))^4}-\frac{35 i \sqrt{e^{i f x}} e^{-i (5 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{9/2}}{c f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{9/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.047, size = 409, normalized size = 1.6 \begin{align*}{\frac{{a}^{4}}{6\,f{c}^{2} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( 315\,i\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{2}ac+27\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \left ( \tan \left ( fx+e \right ) \right ) ^{3}+105\,\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{3}ac-3\,\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \left ( \tan \left ( fx+e \right ) \right ) ^{4}-105\,i\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) ac-393\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}\tan \left ( fx+e \right ) -315\,\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) \tan \left ( fx+e \right ) ac-259\,\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( \tan \left ( fx+e \right ) \right ) ^{2}+164\,\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.0541, size = 1241, normalized size = 4.87 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.64048, size = 1092, normalized size = 4.28 \begin{align*} \frac{2 \,{\left (-16 i \, a^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 112 i \, a^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 350 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 210 i \, a^{4}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} - 105 \, \sqrt{\frac{a^{9}}{c^{3} f^{2}}}{\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \log \left (\frac{8 \,{\left (a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} + \sqrt{\frac{a^{9}}{c^{3} f^{2}}}{\left (4 i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, c^{2} f\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right ) + 105 \, \sqrt{\frac{a^{9}}{c^{3} f^{2}}}{\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \log \left (\frac{8 \,{\left (a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} + \sqrt{\frac{a^{9}}{c^{3} f^{2}}}{\left (-4 i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, c^{2} f\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right )}{12 \,{\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{9}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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