Optimal. Leaf size=253 \[ \frac{14 i a^{9/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac{7 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^3 f}-\frac{14 i a^3 (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.215672, antiderivative size = 253, 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} \[ \frac{14 i a^{9/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac{7 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^3 f}-\frac{14 i a^3 (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]
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))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{7/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{\left (7 a^2\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}+\frac{\left (7 a^3\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 c f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{14 i a^3 (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{\left (7 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 )}{c^2 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{14 i a^3 (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{7 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^3 f}-\frac{\left (7 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 )}{c^2 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{14 i a^3 (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{7 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^3 f}+\frac{\left (14 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^2 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{14 i a^3 (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{7 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^3 f}+\frac{\left (14 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^2 f}\\ &=\frac{14 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^{5/2} f}-\frac{2 i a (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{14 i a^2 (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{14 i a^3 (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{7 i a^4 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{c^3 f}\\ \end{align*}
Mathematica [A] time = 16.3115, size = 390, normalized size = 1.54 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{9/2} \left (\cos (2 f x) \left (-\frac{14 \sin (2 e)}{3 c^3}-\frac{14 i \cos (2 e)}{3 c^3}\right )+\cos (6 f x) \left (\frac{2 \sin (2 e)}{5 c^3}-\frac{2 i \cos (2 e)}{5 c^3}\right )+\sin (2 f x) \left (\frac{14 \cos (2 e)}{3 c^3}-\frac{14 i \sin (2 e)}{3 c^3}\right )+\sin (6 f x) \left (\frac{2 \cos (2 e)}{5 c^3}+\frac{2 i \sin (2 e)}{5 c^3}\right )-\frac{7 \sin (4 e)}{c^3}-\frac{7 i \cos (4 e)}{c^3}-\frac{14 \sin (4 f x)}{15 c^3}+\frac{14 i \cos (4 f x)}{15 c^3}\right ) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))}}{f (\cos (f x)+i \sin (f x))^4}+\frac{14 i \sqrt{e^{i f x}} e^{-i (5 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{9/2}}{c^2 f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{9/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.042, size = 460, normalized size = 1.8 \begin{align*} -{\frac{{a}^{4}}{15\,f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( 420\,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 ) ^{3}ac+15\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \left ( \tan \left ( fx+e \right ) \right ) ^{4}+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 ) ^{4}ac-420\,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 ) \tan \left ( fx+e \right ) ac-658\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \left ( \tan \left ( fx+e \right ) \right ) ^{2}-630\,\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 ) ^{2}ac-292\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }+167\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+105\,ac\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 ) +548\,\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 \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 = 1.9821, size = 1056, normalized size = 4.17 \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 [A] time = 1.61739, size = 975, normalized size = 3.85 \begin{align*} \frac{105 \, \sqrt{\frac{a^{9}}{c^{5} f^{2}}} c^{3} f \log \left (\frac{2 \,{\left (4 \,{\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 )} +{\left (2 i \, c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, c^{3} f\right )} \sqrt{\frac{a^{9}}{c^{5} f^{2}}}\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right ) - 105 \, \sqrt{\frac{a^{9}}{c^{5} f^{2}}} c^{3} f \log \left (\frac{2 \,{\left (4 \,{\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 )} +{\left (-2 i \, c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c^{3} f\right )} \sqrt{\frac{a^{9}}{c^{5} f^{2}}}\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right ) +{\left (-24 i \, a^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 56 i \, a^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 280 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 420 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 )}}{30 \, c^{3} f} \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{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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