Optimal. Leaf size=123 \[ -\frac{4 a^3 (A-2 i B)}{c^2 f (\tan (e+f x)+i)}+\frac{2 a^3 (B+i A)}{c^2 f (\tan (e+f x)+i)^2}-\frac{a^3 (5 B+i A) \log (\cos (e+f x))}{c^2 f}+\frac{a^3 x (A-5 i B)}{c^2}+\frac{i a^3 B \tan (e+f x)}{c^2 f} \]
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Rubi [A] time = 0.178271, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3588, 77} \[ -\frac{4 a^3 (A-2 i B)}{c^2 f (\tan (e+f x)+i)}+\frac{2 a^3 (B+i A)}{c^2 f (\tan (e+f x)+i)^2}-\frac{a^3 (5 B+i A) \log (\cos (e+f x))}{c^2 f}+\frac{a^3 x (A-5 i B)}{c^2}+\frac{i a^3 B \tan (e+f x)}{c^2 f} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 77
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^2 (A+B x)}{(c-i c x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (\frac{i a^2 B}{c^3}-\frac{4 i a^2 (A-i B)}{c^3 (i+x)^3}+\frac{4 a^2 (A-2 i B)}{c^3 (i+x)^2}+\frac{a^2 (i A+5 B)}{c^3 (i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a^3 (A-5 i B) x}{c^2}-\frac{a^3 (i A+5 B) \log (\cos (e+f x))}{c^2 f}+\frac{i a^3 B \tan (e+f x)}{c^2 f}+\frac{2 a^3 (i A+B)}{c^2 f (i+\tan (e+f x))^2}-\frac{4 a^3 (A-2 i B)}{c^2 f (i+\tan (e+f x))}\\ \end{align*}
Mathematica [B] time = 9.54447, size = 1063, normalized size = 8.64 \[ \frac{x \left (\frac{A \cos ^3(e)}{2 c^2}-\frac{5 i B \cos ^3(e)}{2 c^2}-\frac{2 i A \sin (e) \cos ^2(e)}{c^2}-\frac{10 B \sin (e) \cos ^2(e)}{c^2}-\frac{3 A \sin ^2(e) \cos (e)}{c^2}+\frac{15 i B \sin ^2(e) \cos (e)}{c^2}-\frac{A \cos (e)}{2 c^2}+\frac{5 i B \cos (e)}{2 c^2}+\frac{2 i A \sin ^3(e)}{c^2}+\frac{10 B \sin ^3(e)}{c^2}+\frac{i A \sin (e)}{c^2}+\frac{5 B \sin (e)}{c^2}+\frac{A \sin ^3(e) \tan (e)}{2 c^2}-\frac{5 i B \sin ^3(e) \tan (e)}{2 c^2}+\frac{A \sin (e) \tan (e)}{2 c^2}-\frac{5 i B \sin (e) \tan (e)}{2 c^2}+i (A-5 i B) \left (\frac{\cos (3 e)}{c^2}-\frac{i \sin (3 e)}{c^2}\right ) \tan (e)\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{(\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(A-3 i B) \left (\frac{i \sin (e)}{c^2}-\frac{\cos (e)}{c^2}\right ) \sin (2 f x) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(A-i B) \left (\frac{\cos (e)}{2 c^2}+\frac{i \sin (e)}{2 c^2}\right ) \sin (4 f x) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(i A+3 B) \cos (2 f x) \left (\frac{\cos (e)}{c^2}-\frac{i \sin (e)}{c^2}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(A-i B) \cos (4 f x) \left (\frac{\sin (e)}{2 c^2}-\frac{i \cos (e)}{2 c^2}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(A-5 i B) \left (\frac{f x \cos (3 e)}{c^2}-\frac{i f x \sin (3 e)}{c^2}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{(A-5 i B) \left (-\frac{i \cos (3 e) \log \left (\cos ^2(e+f x)\right )}{2 c^2}-\frac{\sin (3 e) \log \left (\cos ^2(e+f x)\right )}{2 c^2}\right ) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^4(e+f x)}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac{i B \left (\frac{\cos (3 e)}{c^2}-\frac{i \sin (3 e)}{c^2}\right ) \sin (f x) (i \tan (e+f x) a+a)^3 (A+B \tan (e+f x)) \cos ^3(e+f x)}{f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}\right )+\sin \left (\frac{e}{2}\right )\right ) (\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 = 0.043, size = 160, normalized size = 1.3 \begin{align*}{\frac{iB{a}^{3}\tan \left ( fx+e \right ) }{{c}^{2}f}}+{\frac{8\,i{a}^{3}B}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) }}-4\,{\frac{A{a}^{3}}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{iA{a}^{3}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{{c}^{2}f}}+5\,{\frac{B{a}^{3}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{{c}^{2}f}}+{\frac{2\,i{a}^{3}A}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+2\,{\frac{B{a}^{3}}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.42335, size = 352, normalized size = 2.86 \begin{align*} \frac{{\left (-i \, A - B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (i \, A + 5 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (2 i \, A + 6 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, B a^{3} +{\left ({\left (-2 i \, A - 10 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-2 i \, A - 10 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{2 \,{\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 4.13402, size = 228, normalized size = 1.85 \begin{align*} - \frac{2 B a^{3} e^{- 2 i e}}{c^{2} f \left (e^{2 i f x} + e^{- 2 i e}\right )} - \frac{a^{3} \left (i A + 5 B\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{2} f} + \frac{\begin{cases} - \frac{i A a^{3} e^{4 i e} e^{4 i f x}}{2 f} + \frac{i A a^{3} e^{2 i e} e^{2 i f x}}{f} - \frac{B a^{3} e^{4 i e} e^{4 i f x}}{2 f} + \frac{3 B a^{3} e^{2 i e} e^{2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (2 A a^{3} e^{4 i e} - 2 A a^{3} e^{2 i e} - 2 i B a^{3} e^{4 i e} + 6 i B a^{3} e^{2 i e}\right ) & \text{otherwise} \end{cases}}{c^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.5781, size = 485, normalized size = 3.94 \begin{align*} \frac{\frac{12 \,{\left (i \, A a^{3} + 5 \, B a^{3}\right )} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c^{2}} + \frac{6 \,{\left (-i \, A a^{3} - 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{2}} - \frac{6 \,{\left (i \, A a^{3} + 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{2}} - \frac{6 \,{\left (-i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 5 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i \, A a^{3} + 5 \, B a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} c^{2}} - \frac{25 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 125 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 100 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 548 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 198 i \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 894 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 100 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 548 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 25 i \, A a^{3} + 125 \, B a^{3}}{c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{4}}}{6 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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