Optimal. Leaf size=167 \[ \frac{(-7 B+i A) \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(-B+i A) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{-11 B+5 i A}{6 a d \sqrt{a+i a \tan (c+d x)}} \]
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Rubi [A] time = 0.353451, antiderivative size = 167, 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 = {3595, 3592, 3526, 3480, 206} \[ \frac{(-7 B+i A) \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(-B+i A) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{-11 B+5 i A}{6 a d \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3595
Rule 3592
Rule 3526
Rule 3480
Rule 206
Rubi steps
\begin{align*} \int \frac{\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{(i A-B) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\tan (c+d x) \left (2 a (i A-B)+\frac{1}{2} a (A+7 i B) \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(i A-7 B) \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}-\frac{\int \frac{-\frac{1}{2} a (A+7 i B)+2 a (i A-B) \tan (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i A-11 B}{6 a d \sqrt{a+i a \tan (c+d x)}}+\frac{(i A-7 B) \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}-\frac{(A-i B) \int \sqrt{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i A-11 B}{6 a d \sqrt{a+i a \tan (c+d x)}}+\frac{(i A-7 B) \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}+\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{2 a d}\\ &=\frac{(i A+B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(i A-B) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i A-11 B}{6 a d \sqrt{a+i a \tan (c+d x)}}+\frac{(i A-7 B) \sqrt{a+i a \tan (c+d x)}}{3 a^2 d}\\ \end{align*}
Mathematica [A] time = 2.7572, size = 167, normalized size = 1. \[ \frac{3 (A-i B) e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+A \left (7 e^{2 i (c+d x)}+8 e^{4 i (c+d x)}-1\right )+i B \left (13 e^{2 i (c+d x)}+38 e^{4 i (c+d x)}-1\right )}{3 a d \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.032, size = 116, normalized size = 0.7 \begin{align*}{\frac{-2\,i}{{a}^{2}d} \left ( -iB\sqrt{a+ia\tan \left ( dx+c \right ) }-{\frac{a \left ( 3\,A+5\,iB \right ) }{4}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}+{\frac{{a}^{2} \left ( A+iB \right ) }{6} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{ \left ( A-iB \right ) \sqrt{2}}{8}\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \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 = 1.9881, size = 1031, normalized size = 6.17 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{-\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{-\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{-\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{-\frac{A^{2} - 2 i \, A B - B^{2}}{a^{3} 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt{2}{\left ({\left (8 i \, A - 38 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (7 i \, A - 13 \, 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{12 \, a^{2} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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 )} \tan \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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