Integrand size = 22, antiderivative size = 138 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {b d^5 n x^{2/3}}{4 e^5}-\frac {b d^4 n x^{4/3}}{8 e^4}+\frac {b d^3 n x^2}{12 e^3}-\frac {b d^2 n x^{8/3}}{16 e^2}+\frac {b d n x^{10/3}}{20 e}-\frac {1}{24} b n x^4-\frac {b d^6 n \log \left (d+e x^{2/3}\right )}{4 e^6}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \]
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Time = 0.07 (sec) , antiderivative size = 138, 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, 45} \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {b d^6 n \log \left (d+e x^{2/3}\right )}{4 e^6}+\frac {b d^5 n x^{2/3}}{4 e^5}-\frac {b d^4 n x^{4/3}}{8 e^4}+\frac {b d^3 n x^2}{12 e^3}-\frac {b d^2 n x^{8/3}}{16 e^2}+\frac {b d n x^{10/3}}{20 e}-\frac {1}{24} b n x^4 \]
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Rule 45
Rule 2442
Rule 2504
Rubi steps \begin{align*} \text {integral}& = \frac {3}{2} \text {Subst}\left (\int x^5 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx,x,x^{2/3}\right ) \\ & = \frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \frac {x^6}{d+e x} \, dx,x,x^{2/3}\right ) \\ & = \frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \left (-\frac {d^5}{e^6}+\frac {d^4 x}{e^5}-\frac {d^3 x^2}{e^4}+\frac {d^2 x^3}{e^3}-\frac {d x^4}{e^2}+\frac {x^5}{e}+\frac {d^6}{e^6 (d+e x)}\right ) \, dx,x,x^{2/3}\right ) \\ & = \frac {b d^5 n x^{2/3}}{4 e^5}-\frac {b d^4 n x^{4/3}}{8 e^4}+\frac {b d^3 n x^2}{12 e^3}-\frac {b d^2 n x^{8/3}}{16 e^2}+\frac {b d n x^{10/3}}{20 e}-\frac {1}{24} b n x^4-\frac {b d^6 n \log \left (d+e x^{2/3}\right )}{4 e^6}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.98 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {a x^4}{4}-\frac {1}{4} b e n \left (-\frac {d^5 x^{2/3}}{e^6}+\frac {d^4 x^{4/3}}{2 e^5}-\frac {d^3 x^2}{3 e^4}+\frac {d^2 x^{8/3}}{4 e^3}-\frac {d x^{10/3}}{5 e^2}+\frac {x^4}{6 e}+\frac {d^6 \log \left (d+e x^{2/3}\right )}{e^7}\right )+\frac {1}{4} b x^4 \log \left (c \left (d+e x^{2/3}\right )^n\right ) \]
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\[\int x^{3} \left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )d x\]
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Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {60 \, b e^{6} x^{4} \log \left (c\right ) + 20 \, b d^{3} e^{3} n x^{2} - 10 \, {\left (b e^{6} n - 6 \, a e^{6}\right )} x^{4} + 60 \, {\left (b e^{6} n x^{4} - b d^{6} n\right )} \log \left (e x^{\frac {2}{3}} + d\right ) - 15 \, {\left (b d^{2} e^{4} n x^{2} - 4 \, b d^{5} e n\right )} x^{\frac {2}{3}} + 6 \, {\left (2 \, b d e^{5} n x^{3} - 5 \, b d^{4} e^{2} n x\right )} x^{\frac {1}{3}}}{240 \, e^{6}} \]
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Timed out. \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\text {Timed out} \]
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Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + \frac {1}{4} \, a x^{4} - \frac {1}{240} \, b e n {\left (\frac {60 \, d^{6} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{7}} + \frac {10 \, e^{5} x^{4} - 12 \, d e^{4} x^{\frac {10}{3}} + 15 \, d^{2} e^{3} x^{\frac {8}{3}} - 20 \, d^{3} e^{2} x^{2} + 30 \, d^{4} e x^{\frac {4}{3}} - 60 \, d^{5} x^{\frac {2}{3}}}{e^{6}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (110) = 220\).
Time = 0.50 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.84 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left (c\right ) + \frac {1}{4} \, a x^{4} + \frac {1}{240} \, b n {\left (\frac {60 \, {\left (e x^{\frac {2}{3}} + d\right )}^{6} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{6}} - \frac {360 \, {\left (e x^{\frac {2}{3}} + d\right )}^{5} d \log \left (e x^{\frac {2}{3}} + d\right )}{e^{6}} + \frac {900 \, {\left (e x^{\frac {2}{3}} + d\right )}^{4} d^{2} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{6}} - \frac {1200 \, {\left (e x^{\frac {2}{3}} + d\right )}^{3} d^{3} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{6}} + \frac {900 \, {\left (e x^{\frac {2}{3}} + d\right )}^{2} d^{4} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{6}} - \frac {10 \, {\left (e x^{\frac {2}{3}} + d\right )}^{6}}{e^{6}} + \frac {72 \, {\left (e x^{\frac {2}{3}} + d\right )}^{5} d}{e^{6}} - \frac {225 \, {\left (e x^{\frac {2}{3}} + d\right )}^{4} d^{2}}{e^{6}} + \frac {400 \, {\left (e x^{\frac {2}{3}} + d\right )}^{3} d^{3}}{e^{6}} - \frac {450 \, {\left (e x^{\frac {2}{3}} + d\right )}^{2} d^{4}}{e^{6}} - \frac {360 \, {\left ({\left (e x^{\frac {2}{3}} + d\right )} \log \left (e x^{\frac {2}{3}} + d\right ) - e x^{\frac {2}{3}} - d\right )} d^{5}}{e^{6}}\right )} \]
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Time = 2.05 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.82 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {a\,x^4}{4}-\frac {b\,n\,x^4}{24}+\frac {b\,x^4\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{4}+\frac {b\,d\,n\,x^{10/3}}{20\,e}-\frac {b\,d^6\,n\,\ln \left (d+e\,x^{2/3}\right )}{4\,e^6}+\frac {b\,d^3\,n\,x^2}{12\,e^3}-\frac {b\,d^2\,n\,x^{8/3}}{16\,e^2}-\frac {b\,d^4\,n\,x^{4/3}}{8\,e^4}+\frac {b\,d^5\,n\,x^{2/3}}{4\,e^5} \]
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