Integrand size = 22, antiderivative size = 190 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=-\frac {b e^8 n \sqrt [3]{x}}{3 d^8}+\frac {b e^7 n x^{2/3}}{6 d^7}-\frac {b e^6 n x}{9 d^6}+\frac {b e^5 n x^{4/3}}{12 d^5}-\frac {b e^4 n x^{5/3}}{15 d^4}+\frac {b e^3 n x^2}{18 d^3}-\frac {b e^2 n x^{7/3}}{21 d^2}+\frac {b e n x^{8/3}}{24 d}+\frac {b e^9 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{3 d^9}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )+\frac {b e^9 n \log (x)}{9 d^9} \]
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Time = 0.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2504, 2442, 46} \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )+\frac {b e^9 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{3 d^9}+\frac {b e^9 n \log (x)}{9 d^9}-\frac {b e^8 n \sqrt [3]{x}}{3 d^8}+\frac {b e^7 n x^{2/3}}{6 d^7}-\frac {b e^6 n x}{9 d^6}+\frac {b e^5 n x^{4/3}}{12 d^5}-\frac {b e^4 n x^{5/3}}{15 d^4}+\frac {b e^3 n x^2}{18 d^3}-\frac {b e^2 n x^{7/3}}{21 d^2}+\frac {b e n x^{8/3}}{24 d} \]
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Rule 46
Rule 2442
Rule 2504
Rubi steps \begin{align*} \text {integral}& = -\left (3 \text {Subst}\left (\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^{10}} \, dx,x,\frac {1}{\sqrt [3]{x}}\right )\right ) \\ & = \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {1}{3} (b e n) \text {Subst}\left (\int \frac {1}{x^9 (d+e x)} \, dx,x,\frac {1}{\sqrt [3]{x}}\right ) \\ & = \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {1}{3} (b e n) \text {Subst}\left (\int \left (\frac {1}{d x^9}-\frac {e}{d^2 x^8}+\frac {e^2}{d^3 x^7}-\frac {e^3}{d^4 x^6}+\frac {e^4}{d^5 x^5}-\frac {e^5}{d^6 x^4}+\frac {e^6}{d^7 x^3}-\frac {e^7}{d^8 x^2}+\frac {e^8}{d^9 x}-\frac {e^9}{d^9 (d+e x)}\right ) \, dx,x,\frac {1}{\sqrt [3]{x}}\right ) \\ & = -\frac {b e^8 n \sqrt [3]{x}}{3 d^8}+\frac {b e^7 n x^{2/3}}{6 d^7}-\frac {b e^6 n x}{9 d^6}+\frac {b e^5 n x^{4/3}}{12 d^5}-\frac {b e^4 n x^{5/3}}{15 d^4}+\frac {b e^3 n x^2}{18 d^3}-\frac {b e^2 n x^{7/3}}{21 d^2}+\frac {b e n x^{8/3}}{24 d}+\frac {b e^9 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{3 d^9}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )+\frac {b e^9 n \log (x)}{9 d^9} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.90 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {a x^3}{3}+\frac {1}{3} b x^3 \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )+\frac {1}{9} b e n \left (-\frac {3 e^7 \sqrt [3]{x}}{d^8}+\frac {3 e^6 x^{2/3}}{2 d^7}-\frac {e^5 x}{d^6}+\frac {3 e^4 x^{4/3}}{4 d^5}-\frac {3 e^3 x^{5/3}}{5 d^4}+\frac {e^2 x^2}{2 d^3}-\frac {3 e x^{7/3}}{7 d^2}+\frac {3 x^{8/3}}{8 d}+\frac {3 e^8 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{d^9}+\frac {e^8 \log (x)}{d^9}\right ) \]
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\[\int x^{2} \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )\right )d x\]
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Time = 0.34 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.01 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {840 \, b d^{9} x^{3} \log \left (c\right ) + 140 \, b d^{6} e^{3} n x^{2} + 840 \, a d^{9} x^{3} - 280 \, b d^{3} e^{6} n x - 840 \, b d^{9} n \log \left (x^{\frac {1}{3}}\right ) + 840 \, {\left (b d^{9} + b e^{9}\right )} n \log \left (d x^{\frac {1}{3}} + e\right ) + 840 \, {\left (b d^{9} n x^{3} - b d^{9} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right ) + 21 \, {\left (5 \, b d^{8} e n x^{2} - 8 \, b d^{5} e^{4} n x + 20 \, b d^{2} e^{7} n\right )} x^{\frac {2}{3}} - 30 \, {\left (4 \, b d^{7} e^{2} n x^{2} - 7 \, b d^{4} e^{5} n x + 28 \, b d e^{8} n\right )} x^{\frac {1}{3}}}{2520 \, d^{9}} \]
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Time = 57.67 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.85 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {a x^{3}}{3} + b \left (\frac {e n \left (\frac {3 x^{\frac {8}{3}}}{8 d} - \frac {3 e x^{\frac {7}{3}}}{7 d^{2}} + \frac {e^{2} x^{2}}{2 d^{3}} - \frac {3 e^{3} x^{\frac {5}{3}}}{5 d^{4}} + \frac {3 e^{4} x^{\frac {4}{3}}}{4 d^{5}} - \frac {e^{5} x}{d^{6}} + \frac {3 e^{6} x^{\frac {2}{3}}}{2 d^{7}} + \frac {3 e^{8} \left (\begin {cases} \frac {\sqrt [3]{x}}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d \sqrt [3]{x} + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{8}} - \frac {3 e^{7} \sqrt [3]{x}}{d^{8}}\right )}{9} + \frac {x^{3} \log {\left (c \left (d + \frac {e}{\sqrt [3]{x}}\right )^{n} \right )}}{3}\right ) \]
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Time = 0.19 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.67 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{3} \, b x^{3} \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right ) + \frac {1}{3} \, a x^{3} + \frac {1}{2520} \, b e n {\left (\frac {840 \, e^{8} \log \left (d x^{\frac {1}{3}} + e\right )}{d^{9}} + \frac {105 \, d^{7} x^{\frac {8}{3}} - 120 \, d^{6} e x^{\frac {7}{3}} + 140 \, d^{5} e^{2} x^{2} - 168 \, d^{4} e^{3} x^{\frac {5}{3}} + 210 \, d^{3} e^{4} x^{\frac {4}{3}} - 280 \, d^{2} e^{5} x + 420 \, d e^{6} x^{\frac {2}{3}} - 840 \, e^{7} x^{\frac {1}{3}}}{d^{8}}\right )} \]
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Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.71 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{3} \, b x^{3} \log \left (c\right ) + \frac {1}{3} \, a x^{3} + \frac {1}{2520} \, {\left (840 \, x^{3} \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right ) + e {\left (\frac {840 \, e^{8} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{d^{9}} + \frac {105 \, d^{7} x^{\frac {8}{3}} - 120 \, d^{6} e x^{\frac {7}{3}} + 140 \, d^{5} e^{2} x^{2} - 168 \, d^{4} e^{3} x^{\frac {5}{3}} + 210 \, d^{3} e^{4} x^{\frac {4}{3}} - 280 \, d^{2} e^{5} x + 420 \, d e^{6} x^{\frac {2}{3}} - 840 \, e^{7} x^{\frac {1}{3}}}{d^{8}}\right )}\right )} b n \]
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Time = 1.91 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.81 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {840\,a\,d^9\,x^3+1680\,b\,e^9\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{1/3}}+1\right )+840\,b\,d^9\,x^3\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )-280\,b\,d^3\,e^6\,n\,x-840\,b\,d\,e^8\,n\,x^{1/3}+105\,b\,d^8\,e\,n\,x^{8/3}+140\,b\,d^6\,e^3\,n\,x^2+420\,b\,d^2\,e^7\,n\,x^{2/3}+210\,b\,d^4\,e^5\,n\,x^{4/3}-168\,b\,d^5\,e^4\,n\,x^{5/3}-120\,b\,d^7\,e^2\,n\,x^{7/3}}{2520\,d^9} \]
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