Integrand size = 23, antiderivative size = 148 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {a e^2 x}{d^3}-\frac {b e^2 n x}{d^3}+\frac {b e n x^2}{4 d^2}-\frac {b n x^3}{9 d}+\frac {b e^2 x \log \left (c x^n\right )}{d^3}-\frac {e x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^4}-\frac {b e^3 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^4} \]
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Time = 0.13 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {269, 45, 2393, 2332, 2341, 2354, 2438} \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=-\frac {e^3 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {e x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d}+\frac {a e^2 x}{d^3}+\frac {b e^2 x \log \left (c x^n\right )}{d^3}-\frac {b e^3 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^4}-\frac {b e^2 n x}{d^3}+\frac {b e n x^2}{4 d^2}-\frac {b n x^3}{9 d} \]
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Rule 45
Rule 269
Rule 2332
Rule 2341
Rule 2354
Rule 2393
Rule 2438
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^3 (e+d x)}\right ) \, dx \\ & = \frac {\int x^2 \left (a+b \log \left (c x^n\right )\right ) \, dx}{d}-\frac {e \int x \left (a+b \log \left (c x^n\right )\right ) \, dx}{d^2}+\frac {e^2 \int \left (a+b \log \left (c x^n\right )\right ) \, dx}{d^3}-\frac {e^3 \int \frac {a+b \log \left (c x^n\right )}{e+d x} \, dx}{d^3} \\ & = \frac {a e^2 x}{d^3}+\frac {b e n x^2}{4 d^2}-\frac {b n x^3}{9 d}-\frac {e x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^4}+\frac {\left (b e^2\right ) \int \log \left (c x^n\right ) \, dx}{d^3}+\frac {\left (b e^3 n\right ) \int \frac {\log \left (1+\frac {d x}{e}\right )}{x} \, dx}{d^4} \\ & = \frac {a e^2 x}{d^3}-\frac {b e^2 n x}{d^3}+\frac {b e n x^2}{4 d^2}-\frac {b n x^3}{9 d}+\frac {b e^2 x \log \left (c x^n\right )}{d^3}-\frac {e x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^4}-\frac {b e^3 n \text {Li}_2\left (-\frac {d x}{e}\right )}{d^4} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.96 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {36 a d e^2 x-36 b d e^2 n x-18 a d^2 e x^2+9 b d^2 e n x^2+12 a d^3 x^3-4 b d^3 n x^3-36 a e^3 \log \left (1+\frac {d x}{e}\right )+6 b \log \left (c x^n\right ) \left (d x \left (6 e^2-3 d e x+2 d^2 x^2\right )-6 e^3 \log \left (1+\frac {d x}{e}\right )\right )-36 b e^3 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{36 d^4} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.18 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.84
method | result | size |
risch | \(\frac {b \ln \left (x^{n}\right ) x^{3}}{3 d}-\frac {b \ln \left (x^{n}\right ) e \,x^{2}}{2 d^{2}}+\frac {b \ln \left (x^{n}\right ) x \,e^{2}}{d^{3}}-\frac {b \ln \left (x^{n}\right ) e^{3} \ln \left (d x +e \right )}{d^{4}}-\frac {b n \,x^{3}}{9 d}+\frac {b e n \,x^{2}}{4 d^{2}}-\frac {b \,e^{2} n x}{d^{3}}-\frac {49 b n \,e^{3}}{36 d^{4}}+\frac {b n \,e^{3} \ln \left (d x +e \right ) \ln \left (-\frac {d x}{e}\right )}{d^{4}}+\frac {b n \,e^{3} \operatorname {dilog}\left (-\frac {d x}{e}\right )}{d^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{3} d^{2} x^{3}-\frac {1}{2} d e \,x^{2}+x \,e^{2}}{d^{3}}-\frac {e^{3} \ln \left (d x +e \right )}{d^{4}}\right )\) | \(272\) |
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\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{d + \frac {e}{x}} \,d x } \]
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Time = 80.12 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.80 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {a x^{3}}{3 d} - \frac {a e x^{2}}{2 d^{2}} - \frac {a e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {a e^{2} x}{d^{3}} - \frac {b n x^{3}}{9 d} + \frac {b x^{3} \log {\left (c x^{n} \right )}}{3 d} + \frac {b e n x^{2}}{4 d^{2}} - \frac {b e x^{2} \log {\left (c x^{n} \right )}}{2 d^{2}} + \frac {b e^{3} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} - \frac {b e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{3}} - \frac {b e^{2} n x}{d^{3}} + \frac {b e^{2} x \log {\left (c x^{n} \right )}}{d^{3}} \]
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\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{d + \frac {e}{x}} \,d x } \]
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\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{d + \frac {e}{x}} \,d x } \]
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Timed out. \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{d+\frac {e}{x}} \,d x \]
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