Optimal. Leaf size=357 \[ \frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{49 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{72 a^2 d}+\frac{5 i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}-\frac{49 \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}+\frac{49 \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{5 i \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{18 a^2 d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.56227, antiderivative size = 357, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 15, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.577, Rules used = {3558, 3595, 3528, 3538, 3476, 329, 209, 634, 618, 204, 628, 203, 275, 292, 31} \[ \frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{49 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{72 a^2 d}+\frac{5 i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}-\frac{49 \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}+\frac{49 \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{5 i \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{18 a^2 d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3558
Rule 3595
Rule 3528
Rule 3538
Rule 3476
Rule 329
Rule 209
Rule 634
Rule 618
Rule 204
Rule 628
Rule 203
Rule 275
Rule 292
Rule 31
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\tan ^{\frac{4}{3}}(c+d x) \left (-\frac{7 a}{3}+\frac{13}{3} i a \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \sqrt [3]{\tan (c+d x)} \left (-\frac{80 i a^2}{9}-\frac{98}{9} a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\frac{98 a^2}{9}-\frac{80}{9} i a^2 \tan (c+d x)}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{8 a^4}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(10 i) \int \sqrt [3]{\tan (c+d x)} \, dx}{9 a^2}+\frac{49 \int \frac{1}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{36 a^2}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(10 i) \operatorname{Subst}\left (\int \frac{\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{9 a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{36 a^2 d}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(10 i) \operatorname{Subst}\left (\int \frac{x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{x}{1+x^3} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{3 a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{1-\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{49 \operatorname{Subst}\left (\int \frac{1+\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{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{1+x}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}+\frac{49 \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{49 \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{49 \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{49 \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{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{5 i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}-\frac{49 \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{49 \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{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(5 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{(5 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{49 \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{49 \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{49 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{5 i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}-\frac{49 \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{49 \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{5 i \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{18 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{(5 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{49 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{5 i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{5 i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}-\frac{49 \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{49 \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{5 i \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{18 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 1.62348, size = 192, normalized size = 0.54 \[ -\frac{\sqrt [3]{\tan (c+d x)} \sec ^2(c+d x) \left (9\ 2^{2/3} e^{2 i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{3},\frac{1}{3};\frac{4}{3};\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+2 \left (89 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};-\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-88 i \sin (2 (c+d x))-85 \cos (2 (c+d x))-13\right )\right )}{48 a^2 d (\tan (c+d x)-i)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.048, size = 373, normalized size = 1. \begin{align*} -3\,{\frac{\sqrt [3]{\tan \left ( dx+c \right ) }}{{a}^{2}d}}+{\frac{{\frac{i}{18}}}{{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{1}{18\,{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{5\,\tan \left ( dx+c \right ) }{12\,{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}{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{89\,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{89\,\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{i}{36}}}{{a}^{2}d} \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) ^{-2}}+{\frac{{\frac{89\,i}{72}}}{{a}^{2}d}\ln \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) }-{\frac{5}{12\,{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 = 2.7649, size = 1573, normalized size = 4.41 \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(-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 = 1.18894, size = 329, normalized size = 0.92 \begin{align*} -\frac{89 \, \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{89 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{89 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{3 \, \tan \left (d x + c\right )^{\frac{1}{3}}}{a^{2} d} - \frac{-16 i \, \tan \left (d x + c\right )^{\frac{4}{3}} - 13 \, \tan \left (d x + c\right )^{\frac{1}{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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