Optimal. Leaf size=337 \[ \frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{\tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{72 a^2 d}+\frac{i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}+\frac{\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{i \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac{\log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{\log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}+\frac{i \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{18 a^2 d}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.588446, antiderivative size = 337, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 14, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3559, 3595, 3538, 3476, 329, 275, 200, 31, 634, 618, 204, 628, 295, 203} \[ \frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{\tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{72 a^2 d}+\frac{i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}+\frac{\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{i \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac{\log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{\log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}+\frac{i \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{18 a^2 d}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3559
Rule 3595
Rule 3538
Rule 3476
Rule 329
Rule 275
Rule 200
Rule 31
Rule 634
Rule 618
Rule 204
Rule 628
Rule 295
Rule 203
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{2}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\tan ^{\frac{2}{3}}(c+d x) \left (\frac{7 a}{3}-\frac{1}{3} i a \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\frac{16 i a^2}{9}-\frac{2}{9} a^2 \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}} \, dx}{8 a^4}\\ &=\frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(2 i) \int \frac{1}{\sqrt [3]{\tan (c+d x)}} \, dx}{9 a^2}+\frac{\int \tan ^{\frac{2}{3}}(c+d x) \, dx}{36 a^2}\\ &=\frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{9 a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{x^{2/3}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{36 a^2 d}\\ &=\frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{x}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{x^4}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}\\ &=\frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{i \operatorname{Subst}\left (\int \frac{1}{1+x^3} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{3 a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2}+\frac{\sqrt{3} x}{2}}{1-\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2}-\frac{\sqrt{3} x}{2}}{1+\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}\\ &=\frac{\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{i \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}-\frac{i \operatorname{Subst}\left (\int \frac{2-x}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{-\sqrt{3}+2 x}{1-\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt{3} a^2 d}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{3}+2 x}{1+\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt{3} a^2 d}\\ &=\frac{\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}+\frac{\log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{\log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}+\frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{i \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{18 a^2 d}-\frac{i \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{6 a^2 d}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,-\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}\\ &=-\frac{\tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{\tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}+\frac{\log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{\log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}+\frac{i \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{18 a^2 d}+\frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{i \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac{2}{3}}(c+d x)\right )}{3 a^2 d}\\ &=-\frac{\tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{\tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}+\frac{\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}+\frac{\log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{\log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}+\frac{i \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{18 a^2 d}+\frac{i \tan ^{\frac{2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac{\tan ^{\frac{5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 1.37696, size = 194, normalized size = 0.58 \[ \frac{i \tan ^{\frac{2}{3}}(c+d x) \sec ^2(c+d x) \left (9 \sqrt [3]{2} e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac{2}{3},\frac{2}{3};\frac{5}{3};\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+14 \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};-\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-4 (i \sin (2 (c+d x))+4 \cos (2 (c+d x))+4)\right )}{96 a^2 d (\tan (c+d x)-i)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.053, size = 357, normalized size = 1.1 \begin{align*}{\frac{-{\frac{i}{9}}}{{a}^{2}d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}} \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) ^{-2}}-{\frac{7}{36\,{a}^{2}d}\sqrt [3]{\tan \left ( dx+c \right ) } \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) ^{-2}}+{\frac{\tan \left ( dx+c \right ) }{18\,{a}^{2}d} \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) ^{-2}}+{\frac{{\frac{i}{18}}}{{a}^{2}d} \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) ^{-2}}+{\frac{{\frac{7\,i}{144}}}{{a}^{2}d}\ln \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) }+{\frac{7\,\sqrt{3}}{72\,{a}^{2}d}{\it Artanh} \left ({\frac{\sqrt{3}}{3} \left ( -i+2\,\sqrt [3]{\tan \left ( dx+c \right ) } \right ) } \right ) }+{\frac{{\frac{i}{16}}}{{a}^{2}d}\ln \left ( i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) }-{\frac{\sqrt{3}}{8\,{a}^{2}d}{\it Artanh} \left ({\frac{\sqrt{3}}{3} \left ( i+2\,\sqrt [3]{\tan \left ( dx+c \right ) } \right ) } \right ) }-{\frac{{\frac{i}{8}}}{{a}^{2}d}\ln \left ( \sqrt [3]{\tan \left ( dx+c \right ) }-i \right ) }-{\frac{{\frac{7\,i}{72}}}{{a}^{2}d}\ln \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) }+{\frac{{\frac{i}{36}}}{{a}^{2}d} \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) ^{-2}}+{\frac{1}{36\,{a}^{2}d} \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) ^{-1}} \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: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.81643, size = 1571, normalized size = 4.66 \begin{align*} \text{result too large to display} \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*} \frac{\int \frac{\tan ^{\frac{2}{3}}{\left (c + d x \right )}}{- \tan ^{2}{\left (c + d x \right )} + 2 i \tan{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.19647, size = 305, normalized size = 0.91 \begin{align*} -\frac{7 \, \sqrt{3} \log \left (-\frac{\sqrt{3} - 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} + i}{\sqrt{3} + 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} - i}\right )}{144 \, a^{2} d} + \frac{\sqrt{3} \log \left (-\frac{\sqrt{3} - 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} - i}{\sqrt{3} + 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} + i}\right )}{16 \, a^{2} d} + \frac{i \, \log \left (\tan \left (d x + c\right )^{\frac{2}{3}} + i \, \tan \left (d x + c\right )^{\frac{1}{3}} - 1\right )}{16 \, a^{2} d} + \frac{7 i \, \log \left (\tan \left (d x + c\right )^{\frac{2}{3}} - i \, \tan \left (d x + c\right )^{\frac{1}{3}} - 1\right )}{144 \, a^{2} d} - \frac{7 i \, \log \left (\tan \left (d x + c\right )^{\frac{1}{3}} + i\right )}{72 \, a^{2} d} - \frac{i \, \log \left (\tan \left (d x + c\right )^{\frac{1}{3}} - i\right )}{8 \, a^{2} d} + \frac{\tan \left (d x + c\right )^{\frac{5}{3}} - 4 i \, \tan \left (d x + c\right )^{\frac{2}{3}}}{12 \, a^{2} d{\left (\tan \left (d x + c\right ) - i\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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