Optimal. Leaf size=124 \[ \frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{2 \tan ^2(c+d x)}{a^2 d}-\frac{15 i \tan (c+d x)}{4 a^2 d}-\frac{4 \log (\cos (c+d x))}{a^2 d}+\frac{15 i x}{4 a^2}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.194879, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3558, 3595, 3528, 3525, 3475} \[ \frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{2 \tan ^2(c+d x)}{a^2 d}-\frac{15 i \tan (c+d x)}{4 a^2 d}-\frac{4 \log (\cos (c+d x))}{a^2 d}+\frac{15 i x}{4 a^2}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3558
Rule 3595
Rule 3528
Rule 3525
Rule 3475
Rubi steps
\begin{align*} \int \frac{\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\tan ^3(c+d x) (-4 a+6 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan ^2(c+d x) \left (-30 i a^2-32 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{2 \tan ^2(c+d x)}{a^2 d}+\frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan (c+d x) \left (32 a^2-30 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac{15 i x}{4 a^2}-\frac{15 i \tan (c+d x)}{4 a^2 d}-\frac{2 \tan ^2(c+d x)}{a^2 d}+\frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{4 \int \tan (c+d x) \, dx}{a^2}\\ &=\frac{15 i x}{4 a^2}-\frac{4 \log (\cos (c+d x))}{a^2 d}-\frac{15 i \tan (c+d x)}{4 a^2 d}-\frac{2 \tan ^2(c+d x)}{a^2 d}+\frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}
Mathematica [B] time = 1.61437, size = 300, normalized size = 2.42 \[ \frac{\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-128 i d x \sin ^2(c)+60 d x \sin (2 c)-\sin (2 c) \sin (4 d x)-64 d x \tan (c)-i \sin (2 c) \cos (4 d x)-16 \sec (c) \cos (2 c-d x) \sec (c+d x)+16 \sec (c) \cos (2 c+d x) \sec (c+d x)+8 i \sin (2 c) \sec ^2(c+d x)-16 i \sec (c) \sin (2 c-d x) \sec (c+d x)+16 i \sec (c) \sin (2 c+d x) \sec (c+d x)+32 i \sin (2 c) \log \left (\cos ^2(c+d x)\right )+64 (\sin (2 c)-i \cos (2 c)) \tan ^{-1}(\tan (d x))+\cos (2 c) \left (-64 d x \tan (c)+8 \sec ^2(c+d x)+32 \log \left (\cos ^2(c+d x)\right )-60 i d x-i \sin (4 d x)+\cos (4 d x)\right )+64 i d x+16 i \sin (2 d x)-16 \cos (2 d x)\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.024, size = 108, normalized size = 0.9 \begin{align*}{\frac{-2\,i\tan \left ( dx+c \right ) }{{a}^{2}d}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,{a}^{2}d}}-{\frac{{\frac{9\,i}{4}}}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{1}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{31\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{8\,{a}^{2}d}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}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 = 2.24751, size = 451, normalized size = 3.64 \begin{align*} \frac{124 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 8 \,{\left (-31 i \, d x - 6\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (124 i \, d x + 95\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 64 \,{\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, e^{\left (6 i \, d x + 6 i \, c\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 ) + 14 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1}{16 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, 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.
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Sympy [A] time = 1.98362, size = 219, normalized size = 1.77 \begin{align*} \frac{\frac{2 e^{- 2 i c} e^{2 i d x}}{a^{2} d} + \frac{4 e^{- 4 i c}}{a^{2} d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \begin{cases} \frac{\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} - a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{16 a^{4} d^{2}} & \text{for}\: 16 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac{\left (31 i e^{4 i c} - 8 i e^{2 i c} + i\right ) e^{- 4 i c}}{4 a^{2}} - \frac{31 i}{4 a^{2}}\right ) & \text{otherwise} \end{cases} + \frac{31 i x}{4 a^{2}} - \frac{4 \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.
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Giac [A] time = 3.62507, size = 132, normalized size = 1.06 \begin{align*} \frac{\frac{2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{62 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac{8 \,{\left (a^{2} \tan \left (d x + c\right )^{2} + 4 i \, a^{2} \tan \left (d x + c\right )\right )}}{a^{4}} - \frac{93 \, \tan \left (d x + c\right )^{2} - 150 i \, \tan \left (d x + c\right ) - 61}{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.
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