Optimal. Leaf size=232 \[ \frac{\sqrt{3} a^{2/3} (A-i B) \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}+\frac{3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac{a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac{a^{2/3} x (B+i A)}{2 \sqrt [3]{2}}+\frac{3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac{3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d} \]
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Rubi [A] time = 0.221381, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used = {3592, 3527, 3481, 55, 617, 204, 31} \[ \frac{\sqrt{3} a^{2/3} (A-i B) \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}+\frac{3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac{a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac{a^{2/3} x (B+i A)}{2 \sqrt [3]{2}}+\frac{3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac{3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d} \]
Antiderivative was successfully verified.
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Rule 3592
Rule 3527
Rule 3481
Rule 55
Rule 617
Rule 204
Rule 31
Rubi steps
\begin{align*} \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx &=-\frac{3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}+\int (a+i a \tan (c+d x))^{2/3} (-B+A \tan (c+d x)) \, dx\\ &=\frac{3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac{3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}-(i A+B) \int (a+i a \tan (c+d x))^{2/3} \, dx\\ &=\frac{3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac{3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}-\frac{(a (A-i B)) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=\frac{a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac{a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac{3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac{3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}-\frac{\left (3 a^{2/3} (A-i B)\right ) \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 )}{2 \sqrt [3]{2} d}+\frac{(3 a (A-i B)) \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 )}{2 d}\\ &=\frac{a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac{a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac{3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac{3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac{3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}-\frac{\left (3 a^{2/3} (A-i B)\right ) \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 )}{\sqrt [3]{2} d}\\ &=\frac{a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac{\sqrt{3} a^{2/3} (A-i B) \tan ^{-1}\left (\frac{1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt [3]{2} d}+\frac{a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac{3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac{3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac{3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}\\ \end{align*}
Mathematica [C] time = 1.47345, size = 115, normalized size = 0.5 \[ \frac{3 \left (e^{i d x}\right )^{2/3} (a+i a \tan (c+d x))^{2/3} \left (-5 (A-i B) \text{Hypergeometric2F1}\left (\frac{2}{3},1,\frac{5}{3},\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )+10 A+4 B \tan (c+d x)-4 i B\right )}{20 d (\cos (d x)+i \sin (d x))^{2/3}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.018, size = 321, normalized size = 1.4 \begin{align*}{\frac{-{\frac{3\,i}{5}}B}{ad} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{3}}}}+{\frac{3\,A}{2\,d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}}-{\frac{{\frac{i}{2}}{2}^{{\frac{2}{3}}}B}{d}{a}^{{\frac{2}{3}}}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }+{\frac{{2}^{{\frac{2}{3}}}A}{2\,d}{a}^{{\frac{2}{3}}}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }+{\frac{{\frac{i}{4}}{2}^{{\frac{2}{3}}}B}{d}{a}^{{\frac{2}{3}}}\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 ) }-{\frac{{2}^{{\frac{2}{3}}}A}{4\,d}{a}^{{\frac{2}{3}}}\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 ) }-{\frac{{\frac{i}{2}}\sqrt{3}{2}^{{\frac{2}{3}}}B}{d}{a}^{{\frac{2}{3}}}\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 ) }+{\frac{\sqrt{3}{2}^{{\frac{2}{3}}}A}{2\,d}{a}^{{\frac{2}{3}}}\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 ) } \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 = 1.77218, size = 1520, normalized size = 6.55 \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*} \int \left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{2}{3}} \left (A + B \tan{\left (c + d x \right )}\right ) \tan{\left (c + d x \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{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \tan \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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