Optimal. Leaf size=142 \[ \frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{3 (-3 B+i A) \tan (c+d x)}{4 a^2 d}+\frac{(A+2 i B) \log (\cos (c+d x))}{a^2 d}-\frac{3 x (-3 B+i A)}{4 a^2}+\frac{(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.280072, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3595, 3525, 3475} \[ \frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{3 (-3 B+i A) \tan (c+d x)}{4 a^2 d}+\frac{(A+2 i B) \log (\cos (c+d x))}{a^2 d}-\frac{3 x (-3 B+i A)}{4 a^2}+\frac{(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3595
Rule 3525
Rule 3475
Rubi steps
\begin{align*} \int \frac{\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\frac{(i A-B) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\tan ^2(c+d x) (3 a (i A-B)+a (A+5 i B) \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan (c+d x) \left (-8 a^2 (A+2 i B)+6 a^2 (i A-3 B) \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{3 (i A-3 B) x}{4 a^2}+\frac{3 (i A-3 B) \tan (c+d x)}{4 a^2 d}+\frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(A+2 i B) \int \tan (c+d x) \, dx}{a^2}\\ &=-\frac{3 (i A-3 B) x}{4 a^2}+\frac{(A+2 i B) \log (\cos (c+d x))}{a^2 d}+\frac{3 (i A-3 B) \tan (c+d x)}{4 a^2 d}+\frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}
Mathematica [B] time = 6.91042, size = 956, normalized size = 6.73 \[ \frac{i \sec (c) \sec ^2(c+d x) (-B \cos (2 c-d x)+B \cos (2 c+d x)-i B \sin (2 c-d x)+i B \sin (2 c+d x)) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{2 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{x \sec (c+d x) (-\tan (c) A+i A-2 B-2 i B \tan (c)+(A+2 i B) (-\cos (2 c)-i \sin (2 c)) \tan (c)) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{(A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}-\frac{(2 A+3 i B) \cos (2 d x) \sec (c+d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{4 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{\sec (c+d x) (A \cos (c)+2 i B \cos (c)+i A \sin (c)-2 B \sin (c)) \left (\tan ^{-1}(\tan (d x)) \sin (c)-i \tan ^{-1}(\tan (d x)) \cos (c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{\sec (c+d x) (A \cos (c)+2 i B \cos (c)+i A \sin (c)-2 B \sin (c)) \left (\frac{1}{2} \cos (c) \log \left (\cos ^2(c+d x)\right )+\frac{1}{2} i \sin (c) \log \left (\cos ^2(c+d x)\right )\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{(A+i B) \cos (4 d x) \sec (c+d x) \left (\frac{1}{16} \cos (2 c)-\frac{1}{16} i \sin (2 c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{(3 B-i A) \sec (c+d x) \left (\frac{3}{4} d x \cos (2 c)+\frac{3}{4} i d x \sin (2 c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{i (2 A+3 i B) \sec (c+d x) \sin (2 d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{4 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{(B-i A) \sec (c+d x) \left (\frac{1}{16} \cos (2 c)-\frac{1}{16} i \sin (2 c)\right ) \sin (4 d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.033, size = 177, normalized size = 1.3 \begin{align*} -{\frac{B\tan \left ( dx+c \right ) }{{a}^{2}d}}+{\frac{{\frac{5\,i}{4}}A}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{7\,B}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{A}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{4}}B}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{7\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{8\,{a}^{2}d}}-{\frac{{\frac{17\,i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{{a}^{2}d}}-{\frac{A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}}+{\frac{{\frac{i}{8}}B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{2}d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 1.46559, size = 424, normalized size = 2.99 \begin{align*} \frac{{\left (-28 i \, A + 68 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} +{\left ({\left (-28 i \, A + 68 \, B\right )} d x - 8 \, A - 44 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (7 \, A + 11 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 16 \,{\left ({\left (A + 2 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (A + 2 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + A + i \, B}{16 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [A] time = 14.4321, size = 223, normalized size = 1.57 \begin{align*} - \frac{2 i B e^{- 2 i c}}{a^{2} d \left (e^{2 i d x} + e^{- 2 i c}\right )} - \frac{\left (\begin{cases} 7 i A x e^{4 i c} + \frac{2 A e^{2 i c} e^{- 2 i d x}}{d} - \frac{A e^{- 4 i d x}}{4 d} - 17 B x e^{4 i c} + \frac{3 i B e^{2 i c} e^{- 2 i d x}}{d} - \frac{i B e^{- 4 i d x}}{4 d} & \text{for}\: d \neq 0 \\x \left (7 i A e^{4 i c} - 4 i A e^{2 i c} + i A - 17 B e^{4 i c} + 6 B e^{2 i c} - B\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i c}}{4 a^{2}} + \frac{\left (A + 2 i B\right ) \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.85702, size = 162, normalized size = 1.14 \begin{align*} -\frac{\frac{2 \,{\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{2 \,{\left (7 \, A + 17 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac{16 \, B \tan \left (d x + c\right )}{a^{2}} - \frac{21 \, A \tan \left (d x + c\right )^{2} + 51 i \, B \tan \left (d x + c\right )^{2} - 22 i \, A \tan \left (d x + c\right ) + 74 \, B \tan \left (d x + c\right ) - 5 \, A - 27 i \, B}{a^{2}{\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]