Optimal. Leaf size=83 \[ \frac{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}} \]
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Rubi [A] time = 0.145958, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3545, 3544, 205} \[ \frac{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3545
Rule 3544
Rule 205
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{3}{2}}(c+d x)} \, dx &=-\frac{2 a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+(2 i a) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+\frac{\left (4 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac{(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.57436, size = 160, normalized size = 1.93 \[ -\frac{2 i \sqrt{2} a^2 e^{i (c+d x)} \sqrt{\tan (c+d x)} \left (e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-\left (-1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{d \left (-1+e^{2 i (c+d x)}\right )^{3/2} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.034, size = 318, normalized size = 3.8 \begin{align*} -{\frac{a}{2\,d}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( i\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \tan \left ( dx+c \right ) a+\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \tan \left ( dx+c \right ) a+4\,\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) \sqrt{-ia}\tan \left ( dx+c \right ) a+4\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \right ){\frac{1}{\sqrt{\tan \left ( dx+c \right ) }}}{\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{ia}}}{\frac{1}{\sqrt{-ia}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.78276, size = 748, normalized size = 9.01 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.39221, size = 1000, normalized size = 12.05 \begin{align*} \frac{\sqrt{2}{\left (-4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} -{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{\frac{8 i \, a^{3}}{d^{2}}} \log \left (\frac{{\left (2 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + i \, \sqrt{\frac{8 i \, a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a}\right ) +{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{\frac{8 i \, a^{3}}{d^{2}}} \log \left (\frac{{\left (2 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - i \, \sqrt{\frac{8 i \, a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a}\right )}{2 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \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 [A] time = 1.35031, size = 144, normalized size = 1.73 \begin{align*} -\frac{2 \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{-\left (i + 1\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} + \left (4 i + 4\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - \left (5 i + 5\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + \left (2 i + 2\right ) \, a^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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