Optimal. Leaf size=113 \[ \frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 i a B \sqrt{a+i a \tan (c+d x)}}{d} \]
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Rubi [A] time = 0.374943, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used = {3594, 3600, 3480, 206, 3599, 63, 208} \[ \frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 i a B \sqrt{a+i a \tan (c+d x)}}{d} \]
Antiderivative was successfully verified.
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Rule 3594
Rule 3600
Rule 3480
Rule 206
Rule 3599
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac{2 i a B \sqrt{a+i a \tan (c+d x)}}{d}+2 \int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{a A}{2}+\frac{1}{2} a (i A+2 B) \tan (c+d x)\right ) \, dx\\ &=\frac{2 i a B \sqrt{a+i a \tan (c+d x)}}{d}+A \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx+(2 a (i A+B)) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{2 i a B \sqrt{a+i a \tan (c+d x)}}{d}+\frac{\left (a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (4 a^2 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 i a B \sqrt{a+i a \tan (c+d x)}}{d}-\frac{(2 i a A) \operatorname{Subst}\left (\int \frac{1}{i-\frac{i x^2}{a}} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac{2 a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 i a B \sqrt{a+i a \tan (c+d x)}}{d}\\ \end{align*}
Mathematica [A] time = 1.92784, size = 157, normalized size = 1.39 \[ \frac{\sqrt{2} a e^{-i (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (\sqrt{2} (A-i B) \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-A \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{1+e^{2 i (c+d x)}}}\right )+i \sqrt{2} B e^{i (c+d x)}\right )}{d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.345, size = 467, normalized size = 4.1 \begin{align*} -{\frac{a}{d \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 2\,iA\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sin \left ( dx+c \right ) +2\,iB\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sin \left ( dx+c \right ) -2\,A\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sin \left ( dx+c \right ) +iA\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) \sin \left ( dx+c \right ) +2\,B\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \sin \left ( dx+c \right ) -A\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\ln \left ( -{\frac{1}{\sin \left ( dx+c \right ) } \left ( -\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) } \right ) \sin \left ( dx+c \right ) -2\,iB\cos \left ( dx+c \right ) +2\,iB+2\,B\sin \left ( dx+c \right ) \right ) } \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: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.86401, size = 1354, normalized size = 11.98 \begin{align*} \frac{4 i \, \sqrt{2} B a \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 2 \, \sqrt{\frac{A^{2} a^{3}}{d^{2}}} d \log \left (\frac{{\left (\sqrt{2}{\left (A a e^{\left (2 i \, d x + 2 i \, c\right )} + A a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 2 \, \sqrt{\frac{A^{2} 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 )}}{A a}\right ) + 2 \, \sqrt{\frac{A^{2} a^{3}}{d^{2}}} d \log \left (\frac{{\left (\sqrt{2}{\left (A a e^{\left (2 i \, d x + 2 i \, c\right )} + A a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 2 \, \sqrt{\frac{A^{2} 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 )}}{A a}\right ) + \sqrt{\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a^{3}}{d^{2}}} d \log \left (\frac{{\left (\sqrt{2}{\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (2 i \, A + 2 \, B\right )} a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + i \, \sqrt{\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} 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 )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) - \sqrt{\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a^{3}}{d^{2}}} d \log \left (\frac{{\left (\sqrt{2}{\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (2 i \, A + 2 \, B\right )} a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - i \, \sqrt{\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} 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 )}}{{\left (2 i \, A + 2 \, B\right )} a}\right )}{2 \, 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{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cot \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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