Optimal. Leaf size=205 \[ -\frac{2 B c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{5/2} f}-\frac{2 B c^2 \sqrt{c-i c \tan (e+f x)}}{a^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{2 B c (c-i c \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^{3/2}} \]
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Rubi [A] time = 0.289492, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3588, 78, 47, 63, 217, 203} \[ -\frac{2 B c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{a^{5/2} f}-\frac{2 B c^2 \sqrt{c-i c \tan (e+f x)}}{a^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{2 B c (c-i c \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 78
Rule 47
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{3/2}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{(i B c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 B c (c-i c \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{\left (i B c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=-\frac{2 B c^2 \sqrt{c-i c \tan (e+f x)}}{a^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{2 B c (c-i c \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{\left (i B c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{a^2 f}\\ &=-\frac{2 B c^2 \sqrt{c-i c \tan (e+f x)}}{a^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{2 B c (c-i c \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{\left (2 B c^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 )}{a^3 f}\\ &=-\frac{2 B c^2 \sqrt{c-i c \tan (e+f x)}}{a^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{2 B c (c-i c \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}-\frac{\left (2 B c^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 )}{a^3 f}\\ &=-\frac{2 B c^{5/2} \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 )}{a^{5/2} f}-\frac{2 B c^2 \sqrt{c-i c \tan (e+f x)}}{a^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{2 B c (c-i c \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 12.7061, size = 129, normalized size = 0.63 \[ -\frac{\sqrt{2} c^2 e^{-4 i (e+f x)} \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \left (-3 i A+B \left (-10 e^{2 i (e+f x)}+30 e^{4 i (e+f x)}+3\right )+30 B e^{5 i (e+f x)} \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{15 a^2 f \sqrt{a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.119, size = 557, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 3.22251, size = 294, normalized size = 1.43 \begin{align*} -\frac{{\left (30 \, B c^{2} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 30 \, B c^{2} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 6 \,{\left (i \, A - B\right )} c^{2} \cos \left (5 \, f x + 5 \, e\right ) - 20 \, B c^{2} \cos \left (3 \, f x + 3 \, e\right ) + 60 \, B c^{2} \cos \left (f x + e\right ) + 15 i \, B c^{2} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 15 i \, B c^{2} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) -{\left (6 \, A + 6 i \, B\right )} c^{2} \sin \left (5 \, f x + 5 \, e\right ) + 20 i \, B c^{2} \sin \left (3 \, f x + 3 \, e\right ) - 60 i \, B c^{2} \sin \left (f x + e\right )\right )} \sqrt{c}}{30 \, a^{\frac{5}{2}} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.7066, size = 1223, normalized size = 5.97 \begin{align*} \frac{{\left (15 \, a^{3} f \sqrt{-\frac{B^{2} c^{5}}{a^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac{2 \,{\left (B c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B c^{2}\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 (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt{-\frac{B^{2} c^{5}}{a^{5} f^{2}}}}{2 \,{\left (B c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B c^{2}\right )}}\right ) - 15 \, a^{3} f \sqrt{-\frac{B^{2} c^{5}}{a^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac{2 \,{\left (B c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B c^{2}\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 (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt{-\frac{B^{2} c^{5}}{a^{5} f^{2}}}}{2 \,{\left (B c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B c^{2}\right )}}\right ) +{\left ({\left (-6 i \, A + 46 \, B\right )} c^{2} e^{\left (7 i \, f x + 7 i \, e\right )} - 60 \, B c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-6 i \, A + 46 \, B\right )} c^{2} e^{\left (5 i \, f x + 5 i \, e\right )} - 40 \, B c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (6 i \, A + 14 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (6 i \, A - 6 \, B\right )} c^{2}\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 )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{30 \, a^{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 (B \tan \left (f x + e\right ) + A\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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