Optimal. Leaf size=91 \[ \frac{4 i a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{2 a^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2}{d \sqrt{\cot (c+d x)}}-\frac{4 (-1)^{3/4} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]
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Rubi [A] time = 0.170696, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3673, 3542, 3529, 3533, 208} \[ \frac{4 i a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{2 a^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2}{d \sqrt{\cot (c+d x)}}-\frac{4 (-1)^{3/4} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3673
Rule 3542
Rule 3529
Rule 3533
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^2}{\cot ^{\frac{3}{2}}(c+d x)} \, dx &=\int \frac{(i a+a \cot (c+d x))^2}{\cot ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\int \frac{2 i a^2+2 a^2 \cot (c+d x)}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{4 i a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\int \frac{2 a^2-2 i a^2 \cot (c+d x)}{\cot ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{4 i a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2}{d \sqrt{\cot (c+d x)}}+\int \frac{-2 i a^2-2 a^2 \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 a^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{4 i a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2}{d \sqrt{\cot (c+d x)}}-\frac{\left (8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2 i a^2-2 a^2 x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{4 (-1)^{3/4} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{2 a^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{4 i a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2}{d \sqrt{\cot (c+d x)}}\\ \end{align*}
Mathematica [A] time = 3.91656, size = 152, normalized size = 1.67 \[ \frac{a^2 e^{-2 i c} (\sin (2 (c+d x))-i \cos (2 (c+d x))) \left (2 i \sec ^2(c+d x) (10 i \sin (2 (c+d x))+33 \cos (2 (c+d x))+27)-\frac{120 i \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{i \tan (c+d x)}}\right )}{30 d \sqrt{\cot (c+d x)} (\cos (d x)+i \sin (d x))^2} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.281, size = 517, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.70023, size = 217, normalized size = 2.38 \begin{align*} -\frac{4 \,{\left (3 \, a^{2} - \frac{10 i \, a^{2}}{\tan \left (d x + c\right )} - \frac{30 \, a^{2}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac{5}{2}} - 15 \,{\left (\left (2 i + 2\right ) \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + \left (2 i + 2\right ) \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + \left (i - 1\right ) \, \sqrt{2} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) - \left (i - 1\right ) \, \sqrt{2} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{2}}{30 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.45323, size = 1112, normalized size = 12.22 \begin{align*} -\frac{15 \, \sqrt{-\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{-\frac{16 i \, a^{4}}{d^{2}}}{\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - 15 \, \sqrt{-\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{-\frac{16 i \, a^{4}}{d^{2}}}{\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) -{\left (-344 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 88 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 248 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 184 i \, a^{2}\right )} \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}}}{60 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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] time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int - \frac{\tan ^{2}{\left (c + d x \right )}}{\cot ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{2 i \tan{\left (c + d x \right )}}{\cot ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{1}{\cot ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\cot \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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