Optimal. Leaf size=208 \[ -\frac{2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{9009 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac{2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{(-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}} \]
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Rubi [A] time = 0.285863, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac{2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{9009 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac{2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{(-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 78
Rule 45
Rule 37
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{15/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}+\frac{(a (3 A+10 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{13/2}} \, dx,x,\tan (e+f x)\right )}{13 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}+\frac{(2 a (3 A+10 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{143 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{2 (3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}+\frac{(2 a (3 A+10 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{1287 c^2 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{2 (3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{2 (3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{9009 c^3 f (c-i c \tan (e+f x))^{7/2}}\\ \end{align*}
Mathematica [B] time = 17.0674, size = 495, normalized size = 2.38 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((B-i A) \cos (6 f x) \left (\frac{\cos (3 e)}{112 c^7}+\frac{i \sin (3 e)}{112 c^7}\right )+(A+i B) \sin (6 f x) \left (\frac{\cos (3 e)}{112 c^7}+\frac{i \sin (3 e)}{112 c^7}\right )+(8 B-15 i A) \cos (8 f x) \left (\frac{\cos (5 e)}{504 c^7}+\frac{i \sin (5 e)}{504 c^7}\right )+(B-30 i A) \cos (10 f x) \left (\frac{\cos (7 e)}{792 c^7}+\frac{i \sin (7 e)}{792 c^7}\right )+(25 A-12 i B) \cos (12 f x) \left (\frac{\sin (9 e)}{1144 c^7}-\frac{i \cos (9 e)}{1144 c^7}\right )+(A-i B) \cos (14 f x) \left (\frac{\sin (11 e)}{208 c^7}-\frac{i \cos (11 e)}{208 c^7}\right )+(15 A+8 i B) \sin (8 f x) \left (\frac{\cos (5 e)}{504 c^7}+\frac{i \sin (5 e)}{504 c^7}\right )+(30 A+i B) \sin (10 f x) \left (\frac{\cos (7 e)}{792 c^7}+\frac{i \sin (7 e)}{792 c^7}\right )+(25 A-12 i B) \sin (12 f x) \left (\frac{\cos (9 e)}{1144 c^7}+\frac{i \sin (9 e)}{1144 c^7}\right )+(A-i B) \sin (14 f x) \left (\frac{\cos (11 e)}{208 c^7}+\frac{i \sin (11 e)}{208 c^7}\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.114, size = 184, normalized size = 0.9 \begin{align*}{\frac{{\frac{i}{9009}}{a}^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 6\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{5}-160\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{4}-20\,B \left ( \tan \left ( fx+e \right ) \right ) ^{5}-177\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{3}-48\,A \left ( \tan \left ( fx+e \right ) \right ) ^{4}-1643\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{2}+590\,B \left ( \tan \left ( fx+e \right ) \right ) ^{3}-1569\,iA\tan \left ( fx+e \right ) +408\,A \left ( \tan \left ( fx+e \right ) \right ) ^{2}-97\,iB-776\,B\tan \left ( fx+e \right ) -930\,A \right ) }{f{c}^{7} \left ( \tan \left ( fx+e \right ) +i \right ) ^{8}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 3.11868, size = 373, normalized size = 1.79 \begin{align*} \frac{{\left (693 \,{\left (-i \, A - B\right )} a^{3} \cos \left (\frac{13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 819 \,{\left (-3 i \, A - B\right )} a^{3} \cos \left (\frac{11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1001 \,{\left (-3 i \, A + B\right )} a^{3} \cos \left (\frac{9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1287 \,{\left (-i \, A + B\right )} a^{3} \cos \left (\frac{7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (693 \, A - 693 i \, B\right )} a^{3} \sin \left (\frac{13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (2457 \, A - 819 i \, B\right )} a^{3} \sin \left (\frac{11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (3003 \, A + 1001 i \, B\right )} a^{3} \sin \left (\frac{9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (1287 \, A + 1287 i \, B\right )} a^{3} \sin \left (\frac{7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt{a}}{72072 \, c^{\frac{13}{2}} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.30005, size = 458, normalized size = 2.2 \begin{align*} \frac{{\left ({\left (-693 i \, A - 693 \, B\right )} a^{3} e^{\left (14 i \, f x + 14 i \, e\right )} +{\left (-3150 i \, A - 1512 \, B\right )} a^{3} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-5460 i \, A + 182 \, B\right )} a^{3} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-4290 i \, A + 2288 \, B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-1287 i \, A + 1287 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )}\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 )}}{72072 \, c^{7} 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{7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{13}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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