Optimal. Leaf size=155 \[ -\frac{(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{315 c^2 f (c-i c \tan (e+f x))^{5/2}}-\frac{(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]
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Rubi [A] time = 0.259379, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac{(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{315 c^2 f (c-i c \tan (e+f x))^{5/2}}-\frac{(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]
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))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}+\frac{(a (2 A+7 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{9 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac{(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}+\frac{(a (2 A+7 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{63 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac{(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac{(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{315 c^2 f (c-i c \tan (e+f x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 10.0313, size = 135, normalized size = 0.87 \[ \frac{a^2 \cos (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} (\cos (7 e+9 f x)+i \sin (7 e+9 f x)) (-7 (2 A+7 i B) \sin (2 (e+f x))+7 (2 B-7 i A) \cos (2 (e+f x))-45 i A)}{630 c^5 f (\cos (f x)+i \sin (f x))^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.119, size = 138, normalized size = 0.9 \begin{align*}{\frac{-{\frac{i}{315}}{a}^{2} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( -47\,A-33\,iA\tan \left ( fx+e \right ) -12\,A \left ( \tan \left ( fx+e \right ) \right ) ^{2}+2\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{3}-7\,iB-42\,B\tan \left ( fx+e \right ) -42\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{2}-7\,B \left ( \tan \left ( fx+e \right ) \right ) ^{3} \right ) }{f{c}^{5} \left ( \tan \left ( fx+e \right ) +i \right ) ^{6}}\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 = 2.16504, size = 270, normalized size = 1.74 \begin{align*} \frac{{\left (35 \,{\left (-i \, A - B\right )} a^{2} \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 ) - 90 i \, A a^{2} \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 ) + 63 \,{\left (-i \, A + B\right )} a^{2} \cos \left (\frac{5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (35 \, A - 35 i \, B\right )} a^{2} \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 ) + 90 \, A a^{2} \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 ) +{\left (63 \, A + 63 i \, B\right )} a^{2} \sin \left (\frac{5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt{a}}{1260 \, c^{\frac{9}{2}} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.41018, size = 373, normalized size = 2.41 \begin{align*} \frac{{\left ({\left (-35 i \, A - 35 \, B\right )} a^{2} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-125 i \, A - 35 \, B\right )} a^{2} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-153 i \, A + 63 \, B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-63 i \, A + 63 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 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 )}}{1260 \, c^{5} 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{9}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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