Optimal. Leaf size=148 \[ \frac{A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{x (-7 B+i A)}{8 a^3}-\frac{i B \log (\cos (c+d x))}{a^3 d}+\frac{(-B+i A) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.368236, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3595, 3589, 3475, 12, 3526, 8} \[ \frac{A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{x (-7 B+i A)}{8 a^3}-\frac{i B \log (\cos (c+d x))}{a^3 d}+\frac{(-B+i A) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3595
Rule 3589
Rule 3475
Rule 12
Rule 3526
Rule 8
Rubi steps
\begin{align*} \int \frac{\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx &=\frac{(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac{\int \frac{\tan ^2(c+d x) (3 a (i A-B)+6 i a B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\tan (c+d x) \left (-6 a^2 (A+3 i B)-24 a^2 B \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac{i \int -\frac{6 a^3 (i A-7 B) \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{24 a^5}+\frac{(i B) \int \tan (c+d x) \, dx}{a^3}\\ &=-\frac{i B \log (\cos (c+d x))}{a^3 d}+\frac{(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac{(A+7 i B) \int \frac{\tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac{i B \log (\cos (c+d x))}{a^3 d}+\frac{(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac{A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(i A-7 B) \int 1 \, dx}{8 a^3}\\ &=\frac{(i A-7 B) x}{8 a^3}-\frac{i B \log (\cos (c+d x))}{a^3 d}+\frac{(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac{A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 1.20566, size = 178, normalized size = 1.2 \[ \frac{\sec ^3(c+d x) ((-51 B+9 i A) \cos (c+d x)-2 \cos (3 (c+d x)) (6 A d x-i A-48 B \log (\cos (c+d x))+42 i B d x+B)-27 A \sin (c+d x)+2 A \sin (3 (c+d x))-12 i A d x \sin (3 (c+d x))-81 i B \sin (c+d x)+2 i B \sin (3 (c+d x))+84 B d x \sin (3 (c+d x))+96 i B \sin (3 (c+d x)) \log (\cos (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.034, size = 203, normalized size = 1.4 \begin{align*}{\frac{\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{16\,{a}^{3}d}}+{\frac{{\frac{15\,i}{16}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{{a}^{3}d}}+{\frac{17\,B}{8\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{7\,i}{8}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{5\,A}{8\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{{\frac{7\,i}{8}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{{\frac{i}{6}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{B}{6\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{16\,{a}^{3}d}}+{\frac{{\frac{i}{16}}B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{3}d}} \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 = 1.51358, size = 306, normalized size = 2.07 \begin{align*} \frac{{\left ({\left (12 i \, A - 180 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} - 96 i \, B e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 6 \,{\left (3 \, A + 11 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \,{\left (3 \, A + 5 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, A + 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 13.8499, size = 253, normalized size = 1.71 \begin{align*} - \frac{i B \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} + \frac{\left (\begin{cases} i A x e^{6 i c} + \frac{3 A e^{4 i c} e^{- 2 i d x}}{2 d} - \frac{3 A e^{2 i c} e^{- 4 i d x}}{4 d} + \frac{A e^{- 6 i d x}}{6 d} - 15 B x e^{6 i c} + \frac{11 i B e^{4 i c} e^{- 2 i d x}}{2 d} - \frac{5 i B e^{2 i c} e^{- 4 i d x}}{4 d} + \frac{i B e^{- 6 i d x}}{6 d} & \text{for}\: d \neq 0 \\x \left (i A e^{6 i c} - 3 i A e^{4 i c} + 3 i A e^{2 i c} - i A - 15 B e^{6 i c} + 11 B e^{4 i c} - 5 B e^{2 i c} + B\right ) & \text{otherwise} \end{cases}\right ) e^{- 6 i c}}{8 a^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.95469, size = 176, normalized size = 1.19 \begin{align*} \frac{\frac{6 \,{\left (A + 15 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac{6 \,{\left (A - i \, B\right )} \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}} - \frac{11 \, A \tan \left (d x + c\right )^{3} + 165 i \, B \tan \left (d x + c\right )^{3} + 51 i \, A \tan \left (d x + c\right )^{2} + 291 \, B \tan \left (d x + c\right )^{2} + 75 \, A \tan \left (d x + c\right ) - 171 i \, B \tan \left (d x + c\right ) - 29 i \, A - 29 \, B}{a^{3}{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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