Optimal. Leaf size=352 \[ -\frac{15 i b x^{4/3} \text{PolyLog}\left (2,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^2 \left (a^2+b^2\right )}+\frac{45 i b x^{2/3} \text{PolyLog}\left (4,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^4 \left (a^2+b^2\right )}+\frac{15 b x \text{PolyLog}\left (3,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}-\frac{45 b \sqrt [3]{x} \text{PolyLog}\left (5,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^5 \left (a^2+b^2\right )}-\frac{45 i b \text{PolyLog}\left (6,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 d^6 \left (a^2+b^2\right )}+\frac{3 b x^{5/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac{x^2}{2 (a+i b)} \]
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Rubi [A] time = 0.413757, antiderivative size = 352, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {3747, 3732, 2190, 2531, 6609, 2282, 6589} \[ -\frac{15 i b x^{4/3} \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^2 \left (a^2+b^2\right )}+\frac{45 i b x^{2/3} \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^4 \left (a^2+b^2\right )}+\frac{15 b x \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}-\frac{45 b \sqrt [3]{x} \text{Li}_5\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^5 \left (a^2+b^2\right )}-\frac{45 i b \text{Li}_6\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 d^6 \left (a^2+b^2\right )}+\frac{3 b x^{5/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac{x^2}{2 (a+i b)} \]
Antiderivative was successfully verified.
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Rule 3747
Rule 3732
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{x}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx &=3 \operatorname{Subst}\left (\int \frac{x^5}{a+b \tan (c+d x)} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{x^2}{2 (a+i b)}+(6 i b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x^5}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{x^2}{2 (a+i b)}+\frac{3 b x^{5/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac{(15 b) \operatorname{Subst}\left (\int x^4 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{x^2}{2 (a+i b)}+\frac{3 b x^{5/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac{15 i b x^{4/3} \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{(30 i b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d^2}\\ &=\frac{x^2}{2 (a+i b)}+\frac{3 b x^{5/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac{15 i b x^{4/3} \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{15 b x \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}-\frac{(45 b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d^3}\\ &=\frac{x^2}{2 (a+i b)}+\frac{3 b x^{5/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac{15 i b x^{4/3} \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{15 b x \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac{45 i b x^{2/3} \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac{(45 i b) \operatorname{Subst}\left (\int x \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d^4}\\ &=\frac{x^2}{2 (a+i b)}+\frac{3 b x^{5/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac{15 i b x^{4/3} \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{15 b x \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac{45 i b x^{2/3} \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac{45 b \sqrt [3]{x} \text{Li}_5\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^5}+\frac{(45 b) \operatorname{Subst}\left (\int \text{Li}_5\left (-\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{2 \left (a^2+b^2\right ) d^5}\\ &=\frac{x^2}{2 (a+i b)}+\frac{3 b x^{5/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac{15 i b x^{4/3} \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{15 b x \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac{45 i b x^{2/3} \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac{45 b \sqrt [3]{x} \text{Li}_5\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^5}-\frac{(45 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_5\left (-\frac{\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 \left (a^2+b^2\right ) d^6}\\ &=\frac{x^2}{2 (a+i b)}+\frac{3 b x^{5/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac{15 i b x^{4/3} \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{15 b x \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac{45 i b x^{2/3} \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac{45 b \sqrt [3]{x} \text{Li}_5\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^5}-\frac{45 i b \text{Li}_6\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) d^6}\\ \end{align*}
Mathematica [A] time = 1.07161, size = 310, normalized size = 0.88 \[ \frac{30 i b d^4 x^{4/3} \text{PolyLog}\left (2,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-90 i b d^2 x^{2/3} \text{PolyLog}\left (4,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+60 b d^3 x \text{PolyLog}\left (3,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-90 b d \sqrt [3]{x} \text{PolyLog}\left (5,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+45 i b \text{PolyLog}\left (6,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+12 b d^5 x^{5/3} \log \left (1+\frac{(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+2 a d^6 x^2+2 i b d^6 x^2}{4 d^6 \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+b\tan \left ( c+d\sqrt [3]{x} \right ) \right ) ^{-1}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 3.98207, size = 1094, normalized size = 3.11 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{b \tan \left (d x^{\frac{1}{3}} + c\right ) + a}, x\right ) \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{x}{a + b \tan{\left (c + d \sqrt [3]{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 \frac{x}{b \tan \left (d x^{\frac{1}{3}} + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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