Optimal. Leaf size=163 \[ \frac{(A+i B) \sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]
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Rubi [A] time = 0.495813, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4241, 3596, 3598, 12, 3544, 205} \[ \frac{(A+i B) \sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]
Antiderivative was successfully verified.
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Rule 4241
Rule 3596
Rule 3598
Rule 12
Rule 3544
Rule 205
Rubi steps
\begin{align*} \int \frac{\cot ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{A+B \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{(A+i B) \sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (3 A+i B)-a (i A-B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{(A+i B) \sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{a^2 (i A+B) \sqrt{a+i a \tan (c+d x)}}{4 \sqrt{\tan (c+d x)}} \, dx}{a^3}\\ &=\frac{(A+i B) \sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\left ((i A+B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=\frac{(A+i B) \sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{\left (i a (i A+B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\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{\left (\frac{1}{2}-\frac{i}{2}\right ) (i A+B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{\sqrt{a} d}+\frac{(A+i B) \sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}\\ \end{align*}
Mathematica [A] time = 3.08865, size = 165, normalized size = 1.01 \[ \frac{e^{-2 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{\cot (c+d x)} \left ((A-i B) e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-5 A e^{2 i (c+d x)}+A-i B \left (-1+e^{2 i (c+d x)}\right )\right )}{\sqrt{2} a d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.676, size = 484, normalized size = 3. \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 [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 [B] time = 1.59125, size = 1176, normalized size = 7.21 \begin{align*} \frac{{\left (a d \sqrt{\frac{2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{{\left (i \, a d \sqrt{\frac{2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A - B\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 )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right ) - a d \sqrt{\frac{2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{{\left (-i \, a d \sqrt{\frac{2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A - B\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 )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right ) - 2 \, \sqrt{2}{\left ({\left (5 \, A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - i \, B\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 )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{\frac{3}{2}}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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