Optimal. Leaf size=343 \[ -\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}+\frac{7 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac{7 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{12 a d}-\frac{5 i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{2 \sqrt{3} a d}-\frac{7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{6 a d}+\frac{7 \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt{3} a d}-\frac{7 \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt{3} a d}+\frac{5 i \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{12 a d} \]
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Rubi [A] time = 0.457812, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 14, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3550, 3528, 3538, 3476, 329, 209, 634, 618, 204, 628, 203, 275, 292, 31} \[ -\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}+\frac{7 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac{7 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{12 a d}-\frac{5 i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{2 \sqrt{3} a d}-\frac{7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{6 a d}+\frac{7 \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt{3} a d}-\frac{7 \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt{3} a d}+\frac{5 i \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{12 a d} \]
Antiderivative was successfully verified.
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Rule 3550
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)} \, dx &=-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \tan ^{\frac{4}{3}}(c+d x) \left (\frac{7 a}{3}-\frac{10}{3} i a \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \sqrt [3]{\tan (c+d x)} \left (\frac{10 i a}{3}+\frac{7}{3} a \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \frac{-\frac{7 a}{3}+\frac{10}{3} i a \tan (c+d x)}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{2 a^2}\\ &=\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{(5 i) \int \sqrt [3]{\tan (c+d x)} \, dx}{3 a}-\frac{7 \int \frac{1}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{6 a}\\ &=\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{3 a d}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 a d}\\ &=\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{a d}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a d}\\ &=\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{x}{1+x^3} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{2 a d}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac{7 \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 )}{6 a d}-\frac{7 \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 )}{6 a d}\\ &=-\frac{7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{6 a 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 )}{6 a d}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}+\frac{7 \operatorname{Subst}\left (\int \frac{-\sqrt{3}+2 x}{1-\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt{3} a d}-\frac{7 \operatorname{Subst}\left (\int \frac{\sqrt{3}+2 x}{1+\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt{3} a d}\\ &=-\frac{7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac{5 i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{6 a d}+\frac{7 \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{8 \sqrt{3} a d}-\frac{7 \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{8 \sqrt{3} a d}+\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\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 )}{12 a d}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{4 a d}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,-\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}\\ &=\frac{7 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac{7 \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac{7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac{5 i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{6 a d}+\frac{7 \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{8 \sqrt{3} a d}-\frac{7 \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{8 \sqrt{3} a d}+\frac{5 i \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{12 a d}+\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac{2}{3}}(c+d x)\right )}{2 a d}\\ &=\frac{7 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac{7 \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac{5 i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{2 \sqrt{3} a d}-\frac{7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac{5 i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{6 a d}+\frac{7 \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{8 \sqrt{3} a d}-\frac{7 \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{8 \sqrt{3} a d}+\frac{5 i \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{12 a d}+\frac{7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{4 a d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end{align*}
Mathematica [C] time = 1.64355, size = 181, normalized size = 0.53 \[ -\frac{i \sqrt [3]{\tan (c+d x)} \sec ^2(c+d x) \left (3\ 2^{2/3} \left (1+e^{2 i (c+d x)}\right )^{4/3} \, _2F_1\left (\frac{1}{3},\frac{1}{3};\frac{4}{3};\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )-34 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};-\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (i \sin (2 (c+d x))+\cos (2 (c+d x))+1)+18 i \sin (2 (c+d x))+22 \cos (2 (c+d x))+34\right )}{16 a d (\tan (c+d x)-i)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.058, size = 291, normalized size = 0.9 \begin{align*} 3\,{\frac{\sqrt [3]{\tan \left ( dx+c \right ) }}{ad}}-{\frac{{\frac{3\,i}{4}}}{ad} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}}}-{\frac{1}{6\,ad}\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 ) ^{-1}}-{\frac{{\frac{i}{6}}}{ad} \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) ^{-1}}+{\frac{{\frac{17\,i}{24}}}{ad}\ln \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) }-{\frac{17\,\sqrt{3}}{12\,ad}{\it Artanh} \left ({\frac{\sqrt{3}}{3} \left ( -i+2\,\sqrt [3]{\tan \left ( dx+c \right ) } \right ) } \right ) }+{\frac{{\frac{i}{8}}}{ad}\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}}{4\,ad}{\it Artanh} \left ({\frac{\sqrt{3}}{3} \left ( i+2\,\sqrt [3]{\tan \left ( dx+c \right ) } \right ) } \right ) }-{\frac{{\frac{i}{4}}}{ad}\ln \left ( \sqrt [3]{\tan \left ( dx+c \right ) }-i \right ) }-{\frac{{\frac{17\,i}{12}}}{ad}\ln \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) }+{\frac{1}{6\,ad} \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 [B] time = 2.81985, size = 1948, normalized size = 5.68 \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 = 1.19589, size = 346, normalized size = 1.01 \begin{align*} \frac{17 \, \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 )}{24 \, a 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 )}{8 \, a 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 )}{8 \, a d} + \frac{17 i \, \log \left (\tan \left (d x + c\right )^{\frac{2}{3}} - i \, \tan \left (d x + c\right )^{\frac{1}{3}} - 1\right )}{24 \, a d} - \frac{17 i \, \log \left (\tan \left (d x + c\right )^{\frac{1}{3}} + i\right )}{12 \, a d} - \frac{i \, \log \left (\tan \left (d x + c\right )^{\frac{1}{3}} - i\right )}{4 \, a d} - \frac{i \, \tan \left (d x + c\right )^{\frac{1}{3}}}{2 \, a d{\left (\tan \left (d x + c\right ) - i\right )}} - \frac{3 \,{\left (i \, a^{3} d^{3} \tan \left (d x + c\right )^{\frac{4}{3}} - 4 \, a^{3} d^{3} \tan \left (d x + c\right )^{\frac{1}{3}}\right )}}{4 \, a^{4} d^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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