Optimal. Leaf size=204 \[ \frac{2 i a^{7/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac{2 i a^3 \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.189474, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3523, 47, 63, 217, 203} \[ \frac{2 i a^{7/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac{2 i a^3 \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3523
Rule 47
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(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))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(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))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}+\frac{a^3 \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{c f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a^3 \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{a^4 \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))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a^3 \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{\left (2 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 )}{c^2 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a^3 \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{\left (2 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 )}{c^2 f}\\ &=\frac{2 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 )}{c^{5/2} f}-\frac{2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a^3 \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 14.4776, size = 205, normalized size = 1. \[ \frac{2 a^3 \cos ^2(e+f x) (\tan (e+f x)-i)^3 \sqrt{a+i a \tan (e+f x)} \left (\cos \left (\frac{1}{2} (e-4 f x)\right )-i \sin \left (\frac{1}{2} (e-4 f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+6 f x)\right )+i \cos \left (\frac{1}{2} (e+6 f x)\right )\right ) \left (6 \sin (e+f x)+6 \sin (3 (e+f x))+4 i \cos (e+f x)+9 i \cos (3 (e+f x))-15 i \cos (e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right ) (\cos (3 (e+f x))-i \sin (3 (e+f x)))\right )}{15 c^2 f \sqrt{c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.039, size = 429, normalized size = 2.1 \begin{align*} -{\frac{{a}^{3}}{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 ( 60\,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\,\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-60\,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-94\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \left ( \tan \left ( fx+e \right ) \right ) ^{2}-90\,\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-46\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }+26\,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 ) +74\,\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 = 1.91279, size = 594, normalized size = 2.91 \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 = 1.5624, size = 968, normalized size = 4.75 \begin{align*} \frac{15 \, c^{3} f \sqrt{\frac{a^{7}}{c^{5} f^{2}}} \log \left (\frac{2 \,{\left (4 \,{\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 )} +{\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^{7}}{c^{5} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 15 \, c^{3} f \sqrt{\frac{a^{7}}{c^{5} f^{2}}} \log \left (\frac{2 \,{\left (4 \,{\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 )} +{\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^{7}}{c^{5} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) +{\left (-12 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 8 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 40 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 60 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 )}}{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{7}{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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