Optimal. Leaf size=203 \[ \frac{2 a^{5/2} B \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac{2 a^2 B \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}} \]
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Rubi [A] time = 0.289427, antiderivative size = 203, 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} \[ \frac{2 a^{5/2} B \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac{2 a^2 B \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}} \]
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+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{(i a B) \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{(i A+B) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{\left (i a^2 B\right ) \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{(i A+B) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a^2 B \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{\left (i a^3 B\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{(i A+B) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a^2 B \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{\left (2 a^2 B\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{(i A+B) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a^2 B \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{\left (2 a^2 B\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 a^{5/2} B \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{(i A+B) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 a^2 B \sqrt{a+i a \tan (e+f x)}}{c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 15.5644, size = 203, normalized size = 1. \[ \frac{a^2 \cos ^2(e+f x) (\tan (e+f x)-i)^2 \sqrt{a+i a \tan (e+f x)} \left (\cos \left (\frac{1}{2} (e-2 f x)\right )-i \sin \left (\frac{1}{2} (e-2 f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e-2 f x)\right )+i \sin \left (\frac{1}{2} (e-2 f x)\right )\right ) \left ((33 B+3 i A) \cos (2 (e+f x))-3 A \sin (2 (e+f x))-27 i B \sin (2 (e+f x))-30 B \tan ^{-1}\left (e^{i (e+f x)}\right ) (\cos (3 (e+f x))-i \sin (3 (e+f x)))-10 B\right )}{15 c^2 f \sqrt{c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.131, size = 555, 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 = 2.35966, size = 290, normalized size = 1.43 \begin{align*} \frac{{\left (30 \, B a^{2} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 30 \, B a^{2} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 6 \,{\left (i \, A + B\right )} a^{2} \cos \left (5 \, f x + 5 \, e\right ) + 20 \, B a^{2} \cos \left (3 \, f x + 3 \, e\right ) - 60 \, B a^{2} \cos \left (f x + e\right ) + 15 i \, B a^{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 a^{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 )} a^{2} \sin \left (5 \, f x + 5 \, e\right ) + 20 i \, B a^{2} \sin \left (3 \, f x + 3 \, e\right ) - 60 i \, B a^{2} \sin \left (f x + e\right )\right )} \sqrt{a}}{30 \, c^{\frac{5}{2}} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.68933, size = 1023, normalized size = 5.04 \begin{align*} -\frac{15 \, c^{3} f \sqrt{-\frac{B^{2} a^{5}}{c^{5} f^{2}}} \log \left (\frac{4 \,{\left (2 \,{\left (B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{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 (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt{-\frac{B^{2} a^{5}}{c^{5} f^{2}}}\right )}}{B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{2}}\right ) - 15 \, c^{3} f \sqrt{-\frac{B^{2} a^{5}}{c^{5} f^{2}}} \log \left (\frac{4 \,{\left (2 \,{\left (B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{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 (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt{-\frac{B^{2} a^{5}}{c^{5} f^{2}}}\right )}}{B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{2}}\right ) -{\left ({\left (-6 i \, A - 6 \, B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-6 i \, A + 14 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 40 \, B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 60 \, B a^{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 )}}{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 (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{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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