Optimal. Leaf size=204 \[ \frac{15 i a^{7/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{c} f}-\frac{15 i a^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}} \]
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Rubi [A] time = 0.181473, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{15 i a^{7/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{c} f}-\frac{15 i a^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}} \]
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))^{7/2}}{\sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{\left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{\left (15 a^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{15 i a^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{\left (15 a^4\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 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{15 i a^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}+\frac{\left (15 i a^3\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 )}{f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{15 i a^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}+\frac{\left (15 i a^3\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 )}{f}\\ &=\frac{15 i a^{7/2} \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 )}{\sqrt{c} f}-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{15 i a^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}\\ \end{align*}
Mathematica [A] time = 10.0837, size = 340, normalized size = 1.67 \[ \frac{\cos ^3(e+f x) (a+i a \tan (e+f x))^{7/2} \left (\cos (2 f x) \left (-\frac{4 \sin (e)}{c}-\frac{4 i \cos (e)}{c}\right )+\sin (2 f x) \left (\frac{4 \cos (e)}{c}-\frac{4 i \sin (e)}{c}\right )+\sec (e) \sin (f x) \left (\frac{\cos (3 e)}{2 c}-\frac{i \sin (3 e)}{2 c}\right ) \sec (e+f x)+\sec (e) (16 \cos (e)+i \sin (e)) \left (-\frac{\sin (3 e)}{2 c}-\frac{i \cos (3 e)}{2 c}\right )\right ) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))}}{f (\cos (f x)+i \sin (f x))^3}+\frac{15 i \sqrt{e^{i f x}} e^{-i (4 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2}}{f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{7}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.088, size = 328, normalized size = 1.6 \begin{align*} -{\frac{{a}^{3}}{2\,cf \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( 30\,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+6\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \left ( \tan \left ( fx+e \right ) \right ) ^{2}+15\,\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- \left ( \tan \left ( fx+e \right ) \right ) ^{3}\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }-24\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}-15\,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 ) -31\,\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 = 2.35614, size = 1067, normalized size = 5.23 \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 = 1.55976, size = 1004, normalized size = 4.92 \begin{align*} \frac{2 \,{\left (-16 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 50 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 30 i \, a^{3}\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 )} + 15 \, \sqrt{\frac{a^{7}}{c f^{2}}}{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac{8 \,{\left (a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}\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^{7}}{c f^{2}}}{\left (4 i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, c f\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 15 \, \sqrt{\frac{a^{7}}{c f^{2}}}{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac{8 \,{\left (a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}\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^{7}}{c f^{2}}}{\left (-4 i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, c f\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right )}{4 \,{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c 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{7}{2}}}{\sqrt{-i \, c \tan \left (f x + e\right ) + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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