Integrand size = 21, antiderivative size = 141 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}-\frac {x \left (6 a+5 b n+6 b \log \left (c x^n\right )\right )}{6 e^3 (d+e x)}+\frac {\left (6 a+11 b n+6 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{6 e^4}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4} \]
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Time = 0.14 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2384, 2354, 2438} \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\frac {\log \left (\frac {e x}{d}+1\right ) \left (6 a+6 b \log \left (c x^n\right )+11 b n\right )}{6 e^4}-\frac {x \left (6 a+6 b \log \left (c x^n\right )+5 b n\right )}{6 e^3 (d+e x)}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{6 e^2 (d+e x)^2}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4} \]
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Rule 2354
Rule 2384
Rule 2438
Rubi steps \begin{align*} \text {integral}& = -\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}+\frac {\int \frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx}{3 e} \\ & = -\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}+\frac {\int \frac {x \left (3 b n+2 (3 a+b n)+6 b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx}{6 e^2} \\ & = -\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}-\frac {x \left (6 a+5 b n+6 b \log \left (c x^n\right )\right )}{6 e^3 (d+e x)}+\frac {\int \frac {9 b n+2 (3 a+b n)+6 b \log \left (c x^n\right )}{d+e x} \, dx}{6 e^3} \\ & = -\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}-\frac {x \left (6 a+5 b n+6 b \log \left (c x^n\right )\right )}{6 e^3 (d+e x)}+\frac {\left (6 a+11 b n+6 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{6 e^4}-\frac {(b n) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^4} \\ & = -\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}-\frac {x \left (6 a+5 b n+6 b \log \left (c x^n\right )\right )}{6 e^3 (d+e x)}+\frac {\left (6 a+11 b n+6 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{6 e^4}+\frac {b n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\frac {\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}-\frac {9 d^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {18 d \left (a+b \log \left (c x^n\right )\right )}{d+e x}-b n \left (\frac {d (3 d+2 e x)}{(d+e x)^2}+2 \log (x)-2 \log (d+e x)\right )-18 b n (\log (x)-\log (d+e x))+9 b n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )+6 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+6 b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{6 e^4} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.54 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.20
method | result | size |
risch | \(\frac {b \ln \left (x^{n}\right ) d^{3}}{3 e^{4} \left (e x +d \right )^{3}}+\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{e^{4}}+\frac {3 b \ln \left (x^{n}\right ) d}{e^{4} \left (e x +d \right )}-\frac {3 b \ln \left (x^{n}\right ) d^{2}}{2 e^{4} \left (e x +d \right )^{2}}+\frac {7 b n d}{6 e^{4} \left (e x +d \right )}+\frac {11 b n \ln \left (e x +d \right )}{6 e^{4}}-\frac {b n \,d^{2}}{6 e^{4} \left (e x +d \right )^{2}}-\frac {11 b n \ln \left (e x \right )}{6 e^{4}}-\frac {b n \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{4}}-\frac {b n \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {d^{3}}{3 e^{4} \left (e x +d \right )^{3}}+\frac {\ln \left (e x +d \right )}{e^{4}}+\frac {3 d}{e^{4} \left (e x +d \right )}-\frac {3 d^{2}}{2 e^{4} \left (e x +d \right )^{2}}\right )\) | \(310\) |
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\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{4}} \,d x } \]
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Time = 30.94 (sec) , antiderivative size = 518, normalized size of antiderivative = 3.67 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\text {Too large to display} \]
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\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{4}} \,d x } \]
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\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{4}} \,d x } \]
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Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^4} \,d x \]
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