Optimal. Leaf size=205 \[ -\frac{(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}+\frac{(-B+i A) \tan ^3(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(5 A+7 i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 a d}+\frac{4 (5 A+7 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]
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Rubi [A] time = 0.524372, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3595, 3597, 3592, 3527, 3480, 206} \[ -\frac{(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}+\frac{(-B+i A) \tan ^3(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(5 A+7 i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 a d}+\frac{4 (5 A+7 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d} \]
Antiderivative was successfully verified.
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Rule 3595
Rule 3597
Rule 3592
Rule 3527
Rule 3480
Rule 206
Rubi steps
\begin{align*} \int \frac{\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{(i A-B) \tan ^3(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (3 a (i A-B)+\frac{1}{2} a (5 A+7 i B) \tan (c+d x)\right ) \, dx}{a^2}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(5 A+7 i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 a d}-\frac{2 \int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-a^2 (5 A+7 i B)+\frac{1}{4} a^2 (25 i A-23 B) \tan (c+d x)\right ) \, dx}{5 a^3}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{(5 A+7 i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 a d}-\frac{(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}-\frac{2 \int \sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (25 i A-23 B)-a^2 (5 A+7 i B) \tan (c+d x)\right ) \, dx}{5 a^3}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{4 (5 A+7 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d}-\frac{(5 A+7 i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 a d}-\frac{(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}+\frac{(i A+B) \int \sqrt{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=\frac{(i A-B) \tan ^3(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{4 (5 A+7 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d}-\frac{(5 A+7 i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 a d}-\frac{(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}+\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} \sqrt{a} d}+\frac{(i A-B) \tan ^3(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{4 (5 A+7 i B) \sqrt{a+i a \tan (c+d x)}}{5 a d}-\frac{(5 A+7 i B) \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 a d}-\frac{(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}\\ \end{align*}
Mathematica [A] time = 3.33997, size = 176, normalized size = 0.86 \[ \frac{(A+B \tan (c+d x)) \left ((A-i B) \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+\frac{1}{30} \sec ^2(c+d x) (5 (23 A+37 i B) \cos (c+d x)+(25 A+59 i B) \cos (3 (c+d x))+4 i \sin (c+d x) ((5 A+22 i B) \cos (2 (c+d x))+5 A+16 i B))\right )}{2 d \sqrt{a+i a \tan (c+d x)} (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.064, size = 168, normalized size = 0.8 \begin{align*} -2\,{\frac{1}{{a}^{3}d} \left ( -i/5B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}+2/3\,iB \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}a+1/3\,A \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}a-2\,iB\sqrt{a+ia\tan \left ( dx+c \right ) }{a}^{2}-{a}^{2}A\sqrt{a+ia\tan \left ( dx+c \right ) }-1/2\,{\frac{{a}^{3} \left ( A+iB \right ) }{\sqrt{a+ia\tan \left ( dx+c \right ) }}}-1/4\,{a}^{5/2} \left ( A-iB \right ) \sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\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 = 2.09766, size = 1266, normalized size = 6.18 \begin{align*} \frac{2 \, \sqrt{2}{\left ({\left (35 \, A + 103 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \,{\left (25 \, A + 41 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 15 \,{\left (7 \, A + 11 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \, A + 15 i \, B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 15 \,{\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} \log \left (\frac{{\left (i \, a d \sqrt{\frac{2 \, A^{2} - 4 i \, A B - 2 \, 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \,{\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} \log \left (\frac{{\left (-i \, a d \sqrt{\frac{2 \, A^{2} - 4 i \, A B - 2 \, 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right )}{60 \,{\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\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*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}\, dx \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 )^{3}}{\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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