Integrand size = 22, antiderivative size = 143 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {b e^5 n x^{2/3}}{4 d^5}-\frac {b e^4 n x^{4/3}}{8 d^4}+\frac {b e^3 n x^2}{12 d^3}-\frac {b e^2 n x^{8/3}}{16 d^2}+\frac {b e n x^{10/3}}{20 d}-\frac {b e^6 n \log \left (d+\frac {e}{x^{2/3}}\right )}{4 d^6}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )-\frac {b e^6 n \log (x)}{6 d^6} \]
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Time = 0.07 (sec) , antiderivative size = 143, 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} \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )-\frac {b e^6 n \log \left (d+\frac {e}{x^{2/3}}\right )}{4 d^6}-\frac {b e^6 n \log (x)}{6 d^6}+\frac {b e^5 n x^{2/3}}{4 d^5}-\frac {b e^4 n x^{4/3}}{8 d^4}+\frac {b e^3 n x^2}{12 d^3}-\frac {b e^2 n x^{8/3}}{16 d^2}+\frac {b e n x^{10/3}}{20 d} \]
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Rule 46
Rule 2442
Rule 2504
Rubi steps \begin{align*} \text {integral}& = -\left (\frac {3}{2} \text {Subst}\left (\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^7} \, dx,x,\frac {1}{x^{2/3}}\right )\right ) \\ & = \frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \frac {1}{x^6 (d+e x)} \, dx,x,\frac {1}{x^{2/3}}\right ) \\ & = \frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \left (\frac {1}{d x^6}-\frac {e}{d^2 x^5}+\frac {e^2}{d^3 x^4}-\frac {e^3}{d^4 x^3}+\frac {e^4}{d^5 x^2}-\frac {e^5}{d^6 x}+\frac {e^6}{d^6 (d+e x)}\right ) \, dx,x,\frac {1}{x^{2/3}}\right ) \\ & = \frac {b e^5 n x^{2/3}}{4 d^5}-\frac {b e^4 n x^{4/3}}{8 d^4}+\frac {b e^3 n x^2}{12 d^3}-\frac {b e^2 n x^{8/3}}{16 d^2}+\frac {b e n x^{10/3}}{20 d}-\frac {b e^6 n \log \left (d+\frac {e}{x^{2/3}}\right )}{4 d^6}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )-\frac {b e^6 n \log (x)}{6 d^6} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {a x^4}{4}+\frac {1}{4} b x^4 \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )+\frac {1}{6} b e n \left (\frac {3 e^4 x^{2/3}}{2 d^5}-\frac {3 e^3 x^{4/3}}{4 d^4}+\frac {e^2 x^2}{2 d^3}-\frac {3 e x^{8/3}}{8 d^2}+\frac {3 x^{10/3}}{10 d}-\frac {3 e^5 \log \left (d+\frac {e}{x^{2/3}}\right )}{2 d^6}-\frac {e^5 \log (x)}{d^6}\right ) \]
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\[\int x^{3} \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )d x\]
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Time = 0.37 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.12 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {60 \, b d^{6} x^{4} \log \left (c\right ) + 60 \, a d^{6} x^{4} + 20 \, b d^{3} e^{3} n x^{2} - 120 \, b d^{6} n \log \left (x^{\frac {1}{3}}\right ) + 60 \, {\left (b d^{6} - b e^{6}\right )} n \log \left (d x^{\frac {2}{3}} + e\right ) + 60 \, {\left (b d^{6} n x^{4} - b d^{6} n\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right ) - 15 \, {\left (b d^{4} e^{2} n x^{2} - 4 \, b d e^{5} n\right )} x^{\frac {2}{3}} + 6 \, {\left (2 \, b d^{5} e n x^{3} - 5 \, b d^{2} e^{4} n x\right )} x^{\frac {1}{3}}}{240 \, d^{6}} \]
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Timed out. \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\text {Timed out} \]
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Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.69 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + \frac {1}{4} \, a x^{4} - \frac {1}{240} \, b e n {\left (\frac {60 \, e^{5} \log \left (d x^{\frac {2}{3}} + e\right )}{d^{6}} - \frac {12 \, d^{4} x^{\frac {10}{3}} - 15 \, d^{3} e x^{\frac {8}{3}} + 20 \, d^{2} e^{2} x^{2} - 30 \, d e^{3} x^{\frac {4}{3}} + 60 \, e^{4} x^{\frac {2}{3}}}{d^{5}}\right )} \]
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Time = 0.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.73 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left (c\right ) + \frac {1}{4} \, a x^{4} + \frac {1}{240} \, {\left (60 \, x^{4} \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right ) - e {\left (\frac {60 \, e^{5} \log \left ({\left | d x^{\frac {2}{3}} + e \right |}\right )}{d^{6}} - \frac {12 \, d^{4} x^{\frac {10}{3}} - 15 \, d^{3} e x^{\frac {8}{3}} + 20 \, d^{2} e^{2} x^{2} - 30 \, d e^{3} x^{\frac {4}{3}} + 60 \, e^{4} x^{\frac {2}{3}}}{d^{5}}\right )}\right )} b n \]
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Time = 1.93 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.78 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {x^{10/3}\,\left (\frac {b\,e\,n}{5\,d}-\frac {b\,e^2\,n}{4\,d^2\,x^{2/3}}-\frac {b\,e^4\,n}{2\,d^4\,x^2}+\frac {b\,e^3\,n}{3\,d^3\,x^{4/3}}+\frac {b\,e^5\,n}{d^5\,x^{8/3}}\right )}{4}+\frac {a\,x^4}{4}+\frac {b\,x^4\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{4}-\frac {b\,e^6\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{2/3}}+1\right )}{2\,d^6} \]
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