Optimal. Leaf size=203 \[ \frac{15 i b x^{4/3} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac{45 i b x^{2/3} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}-\frac{15 b x \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{45 b \sqrt [3]{x} \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac{45 i b \text{PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}+\frac{a x^2}{2}-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{1}{2} i b x^2 \]
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Rubi [A] time = 0.271423, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {14, 3747, 3719, 2190, 2531, 6609, 2282, 6589} \[ \frac{a x^2}{2}+\frac{15 i b x^{4/3} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac{45 i b x^{2/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}-\frac{15 b x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{45 b \sqrt [3]{x} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac{45 i b \text{Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{1}{2} i b x^2 \]
Antiderivative was successfully verified.
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Rule 14
Rule 3747
Rule 3719
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx &=\int \left (a x+b x \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx\\ &=\frac{a x^2}{2}+b \int x \tan \left (c+d \sqrt [3]{x}\right ) \, dx\\ &=\frac{a x^2}{2}+(3 b) \operatorname{Subst}\left (\int x^5 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{a x^2}{2}+\frac{1}{2} i b x^2-(6 i b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x^5}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{a x^2}{2}+\frac{1}{2} i b x^2-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{(15 b) \operatorname{Subst}\left (\int x^4 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=\frac{a x^2}{2}+\frac{1}{2} i b x^2-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{15 i b x^{4/3} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac{(30 i b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}\\ &=\frac{a x^2}{2}+\frac{1}{2} i b x^2-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{15 i b x^{4/3} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac{15 b x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{(45 b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^3}\\ &=\frac{a x^2}{2}+\frac{1}{2} i b x^2-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{15 i b x^{4/3} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac{15 b x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{45 i b x^{2/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac{(45 i b) \operatorname{Subst}\left (\int x \text{Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^4}\\ &=\frac{a x^2}{2}+\frac{1}{2} i b x^2-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{15 i b x^{4/3} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac{15 b x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{45 i b x^{2/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac{45 b \sqrt [3]{x} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}-\frac{(45 b) \operatorname{Subst}\left (\int \text{Li}_5\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{2 d^5}\\ &=\frac{a x^2}{2}+\frac{1}{2} i b x^2-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{15 i b x^{4/3} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac{15 b x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{45 i b x^{2/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac{45 b \sqrt [3]{x} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac{(45 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}\\ &=\frac{a x^2}{2}+\frac{1}{2} i b x^2-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{15 i b x^{4/3} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac{15 b x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{45 i b x^{2/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac{45 b \sqrt [3]{x} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac{45 i b \text{Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}\\ \end{align*}
Mathematica [A] time = 0.0410655, size = 203, normalized size = 1. \[ \frac{15 i b x^{4/3} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac{45 i b x^{2/3} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}-\frac{15 b x \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{45 b \sqrt [3]{x} \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac{45 i b \text{PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}+\frac{a x^2}{2}-\frac{3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{1}{2} i b x^2 \]
Antiderivative was successfully verified.
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Maple [F] time = 0.077, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\tan \left ( c+d\sqrt [3]{x} \right ) \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.06424, size = 834, normalized size = 4.11 \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 (b x \tan \left (d x^{\frac{1}{3}} + c\right ) + a x, 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 x \left (a + b \tan{\left (c + d \sqrt [3]{x} \right )}\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^{\frac{1}{3}} + c\right ) + a\right )} x\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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