Optimal. Leaf size=129 \[ -\frac{2 a^2 (5 A-7 i B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{4 a^2 (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i B \tan ^{\frac{3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]
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Rubi [A] time = 0.259341, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3594, 3592, 3528, 3533, 205} \[ -\frac{2 a^2 (5 A-7 i B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{4 a^2 (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i B \tan ^{\frac{3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]
Antiderivative was successfully verified.
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Rule 3594
Rule 3592
Rule 3528
Rule 3533
Rule 205
Rubi steps
\begin{align*} \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\frac{2 i B \tan ^{\frac{3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac{2}{5} \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x)) \left (\frac{1}{2} a (5 A-3 i B)+\frac{1}{2} a (5 i A+7 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{2 a^2 (5 A-7 i B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 i B \tan ^{\frac{3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac{2}{5} \int \sqrt{\tan (c+d x)} \left (5 a^2 (A-i B)+5 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac{4 a^2 (i A+B) \sqrt{\tan (c+d x)}}{d}-\frac{2 a^2 (5 A-7 i B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 i B \tan ^{\frac{3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac{2}{5} \int \frac{-5 a^2 (i A+B)+5 a^2 (A-i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{4 a^2 (i A+B) \sqrt{\tan (c+d x)}}{d}-\frac{2 a^2 (5 A-7 i B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 i B \tan ^{\frac{3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac{\left (20 a^4 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-5 a^2 (i A+B)-5 a^2 (A-i B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{4 \sqrt [4]{-1} a^2 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{4 a^2 (i A+B) \sqrt{\tan (c+d x)}}{d}-\frac{2 a^2 (5 A-7 i B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 i B \tan ^{\frac{3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\\ \end{align*}
Mathematica [B] time = 4.99916, size = 272, normalized size = 2.11 \[ \frac{\cos ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (\frac{1}{15} (\cos (2 c)-i \sin (2 c)) \sqrt{\tan (c+d x)} \sec ^2(c+d x) (-5 (A-2 i B) \sin (2 (c+d x))+(33 B+30 i A) \cos (2 (c+d x))+30 i A+27 B)-\frac{4 i e^{-2 i c} (A-i B) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.013, size = 537, normalized size = 4.2 \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 [A] time = 1.83083, size = 265, normalized size = 2.05 \begin{align*} -\frac{12 \, B a^{2} \tan \left (d x + c\right )^{\frac{5}{2}} + 4 \,{\left (5 \, A - 10 i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac{3}{2}} + 120 \,{\left (-i \, A - B\right )} a^{2} \sqrt{\tan \left (d x + c\right )} + 15 \,{\left (2 \, \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) - \sqrt{2}{\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt{2}{\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \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 = 1.95193, size = 1193, normalized size = 9.25 \begin{align*} -\frac{15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) -{\left ({\left (280 i \, A + 344 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (480 i \, A + 432 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (200 i \, A + 184 \, B\right )} 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 (4 i \, d x + 4 i \, c\right )} + 2 \, 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 = 1.2419, size = 171, normalized size = 1.33 \begin{align*} -\frac{\left (i - 1\right ) \, \sqrt{2}{\left (8 \, A a^{2} - 8 i \, B a^{2}\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{4 \, d} - \frac{6 \, B a^{2} d^{4} \tan \left (d x + c\right )^{\frac{5}{2}} + 10 \, A a^{2} d^{4} \tan \left (d x + c\right )^{\frac{3}{2}} - 20 i \, B a^{2} d^{4} \tan \left (d x + c\right )^{\frac{3}{2}} - 60 i \, A a^{2} d^{4} \sqrt{\tan \left (d x + c\right )} - 60 \, B a^{2} d^{4} \sqrt{\tan \left (d x + c\right )}}{15 \, d^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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