Integrand size = 23, antiderivative size = 135 \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+\frac {e}{x}\right ) x^4} \, dx=-\frac {b n}{4 e x^2}+\frac {b d n}{e^2 x}-\frac {a+b \log \left (c x^n\right )}{2 e x^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^3 n}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{e^3}-\frac {b d^2 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^3} \]
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Time = 0.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {269, 46, 2393, 2341, 2338, 2354, 2438} \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^3 n}-\frac {d^2 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x}-\frac {a+b \log \left (c x^n\right )}{2 e x^2}-\frac {b d^2 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^3}+\frac {b d n}{e^2 x}-\frac {b n}{4 e x^2} \]
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Rule 46
Rule 269
Rule 2338
Rule 2341
Rule 2354
Rule 2393
Rule 2438
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a+b \log \left (c x^n\right )}{e x^3}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x^2}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 x}-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^3 (e+d x)}\right ) \, dx \\ & = \frac {d^2 \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{e^3}-\frac {d^3 \int \frac {a+b \log \left (c x^n\right )}{e+d x} \, dx}{e^3}-\frac {d \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx}{e^2}+\frac {\int \frac {a+b \log \left (c x^n\right )}{x^3} \, dx}{e} \\ & = -\frac {b n}{4 e x^2}+\frac {b d n}{e^2 x}-\frac {a+b \log \left (c x^n\right )}{2 e x^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^3 n}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{e^3}+\frac {\left (b d^2 n\right ) \int \frac {\log \left (1+\frac {d x}{e}\right )}{x} \, dx}{e^3} \\ & = -\frac {b n}{4 e x^2}+\frac {b d n}{e^2 x}-\frac {a+b \log \left (c x^n\right )}{2 e x^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^3 n}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{e^3}-\frac {b d^2 n \text {Li}_2\left (-\frac {d x}{e}\right )}{e^3} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+\frac {e}{x}\right ) x^4} \, dx=-\frac {\frac {b e^2 n}{x^2}-\frac {4 b d e n}{x}+\frac {2 e^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {4 d e \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+4 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )+4 b d^2 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{4 e^3} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.18 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.96
method | result | size |
risch | \(-\frac {b \ln \left (x^{n}\right ) d^{2} \ln \left (d x +e \right )}{e^{3}}-\frac {b \ln \left (x^{n}\right )}{2 e \,x^{2}}+\frac {b \ln \left (x^{n}\right ) d^{2} \ln \left (x \right )}{e^{3}}+\frac {b \ln \left (x^{n}\right ) d}{e^{2} x}+\frac {b d n}{e^{2} x}-\frac {b n}{4 e \,x^{2}}-\frac {b n \,d^{2} \ln \left (x \right )^{2}}{2 e^{3}}+\frac {b n \,d^{2} \ln \left (d x +e \right ) \ln \left (-\frac {d x}{e}\right )}{e^{3}}+\frac {b n \,d^{2} \operatorname {dilog}\left (-\frac {d x}{e}\right )}{e^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {d^{2} \ln \left (d x +e \right )}{e^{3}}-\frac {1}{2 e \,x^{2}}+\frac {d^{2} \ln \left (x \right )}{e^{3}}+\frac {d}{e^{2} x}\right )\) | \(264\) |
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\[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{4}} \,d x } \]
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Time = 42.97 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.96 \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+\frac {e}{x}\right ) x^4} \, dx=- \frac {a d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {a d^{2} \log {\left (x \right )}}{e^{3}} + \frac {a d}{e^{2} x} - \frac {a}{2 e x^{2}} + \frac {b d^{3} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {b d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} - \frac {b d^{2} n \log {\left (x \right )}^{2}}{2 e^{3}} + \frac {b d^{2} \log {\left (x \right )} \log {\left (c x^{n} \right )}}{e^{3}} + \frac {b d n}{e^{2} x} + \frac {b d \log {\left (c x^{n} \right )}}{e^{2} x} - \frac {b n}{4 e x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 e x^{2}} \]
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\[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{4}} \,d x } \]
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\[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{4}} \,d x } \]
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Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,\left (d+\frac {e}{x}\right )} \,d x \]
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