Optimal. Leaf size=213 \[ -\frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{x}{8 \sqrt [3]{2} a^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}} \]
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Rubi [A] time = 0.202164, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {3540, 3526, 3481, 55, 617, 204, 31} \[ -\frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{x}{8 \sqrt [3]{2} a^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}} \]
Antiderivative was successfully verified.
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Rule 3540
Rule 3526
Rule 3481
Rule 55
Rule 617
Rule 204
Rule 31
Rubi steps
\begin{align*} \int \frac{\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx &=-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{\int \frac{a-2 i a \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{\int (a+i a \tan (c+d x))^{2/3} \, dx}{4 a^2}\\ &=-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{i \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{4 a d}\\ &=\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 a d}\\ &=\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}\\ &=\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{i \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac{9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 0.662838, size = 128, normalized size = 0.6 \[ \frac{3 \sec ^2(c+d x) \left (\, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+6 i \sin (2 (c+d x))+5 \cos (2 (c+d x))+5\right )}{16 a d (\tan (c+d x)-i) \sqrt [3]{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.02, size = 181, normalized size = 0.9 \begin{align*}{\frac{-{\frac{i}{8}}{2}^{{\frac{2}{3}}}}{d}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ){a}^{-{\frac{4}{3}}}}+{\frac{{\frac{i}{16}}{2}^{{\frac{2}{3}}}}{d}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ){a}^{-{\frac{4}{3}}}}-{\frac{{\frac{i}{8}}\sqrt{3}{2}^{{\frac{2}{3}}}}{d}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \right ) } \right ){a}^{-{\frac{4}{3}}}}+{\frac{{\frac{9\,i}{4}}}{ad}{\frac{1}{\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{{\frac{3\,i}{8}}}{d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}} \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 = 1.80004, size = 1040, normalized size = 4.88 \begin{align*} \frac{{\left (32 \, a^{2} d \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (32 \, a^{3} d^{2} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right ) + 2^{\frac{2}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{2}{3}}{\left (33 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 30 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\right )} e^{\left (\frac{4}{3} i \, d x + \frac{4}{3} i \, c\right )} +{\left (-16 i \, \sqrt{3} a^{2} d - 16 \, a^{2} d\right )} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left (16 i \, \sqrt{3} a^{3} d^{2} - 16 \, a^{3} d^{2}\right )} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right ) +{\left (16 i \, \sqrt{3} a^{2} d - 16 \, a^{2} d\right )} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left (-16 i \, \sqrt{3} a^{3} d^{2} - 16 \, a^{3} d^{2}\right )} \left (\frac{i}{128 \, a^{4} d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right )\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \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{\tan ^{2}{\left (c + d x \right )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{4}{3}}}\, 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{\tan \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{4}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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