Integrand size = 21, antiderivative size = 98 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=-\frac {b n x}{e^2}+\frac {2 x \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}-\frac {d \left (2 a+b n+2 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}-\frac {2 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3} \]
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Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2384, 45, 2393, 2332, 2354, 2438} \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=-\frac {d \log \left (\frac {e x}{d}+1\right ) \left (2 a+2 b \log \left (c x^n\right )+b n\right )}{e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}+\frac {x (2 a+b n)}{e^2}+\frac {2 b x \log \left (c x^n\right )}{e^2}-\frac {2 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3}-\frac {2 b n x}{e^2} \]
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Rule 45
Rule 2332
Rule 2354
Rule 2384
Rule 2393
Rule 2438
Rubi steps \begin{align*} \text {integral}& = -\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}+\frac {\int \frac {x \left (2 a+b n+2 b \log \left (c x^n\right )\right )}{d+e x} \, dx}{e} \\ & = -\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}+\frac {\int \left (\frac {2 a+b n+2 b \log \left (c x^n\right )}{e}-\frac {d \left (2 a+b n+2 b \log \left (c x^n\right )\right )}{e (d+e x)}\right ) \, dx}{e} \\ & = -\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}+\frac {\int \left (2 a+b n+2 b \log \left (c x^n\right )\right ) \, dx}{e^2}-\frac {d \int \frac {2 a+b n+2 b \log \left (c x^n\right )}{d+e x} \, dx}{e^2} \\ & = \frac {(2 a+b n) x}{e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}-\frac {d \left (2 a+b n+2 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {(2 b) \int \log \left (c x^n\right ) \, dx}{e^2}+\frac {(2 b d n) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^3} \\ & = -\frac {2 b n x}{e^2}+\frac {(2 a+b n) x}{e^2}+\frac {2 b x \log \left (c x^n\right )}{e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}-\frac {d \left (2 a+b n+2 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}-\frac {2 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\frac {a e x-b e n x+b e x \log \left (c x^n\right )-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}+b d n (\log (x)-\log (d+e x))-2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-2 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.41 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.55
method | result | size |
risch | \(\frac {b \ln \left (x^{n}\right ) x}{e^{2}}-\frac {2 b \ln \left (x^{n}\right ) d \ln \left (e x +d \right )}{e^{3}}-\frac {b \ln \left (x^{n}\right ) d^{2}}{e^{3} \left (e x +d \right )}-\frac {b n d \ln \left (e x +d \right )}{e^{3}}+\frac {b n d \ln \left (e x \right )}{e^{3}}-\frac {b n x}{e^{2}}-\frac {b n d}{e^{3}}+\frac {2 b n d \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{3}}+\frac {2 b n d \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x}{e^{2}}-\frac {2 d \ln \left (e x +d \right )}{e^{3}}-\frac {d^{2}}{e^{3} \left (e x +d \right )}\right )\) | \(250\) |
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\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{{\left (e x + d\right )}^{2}} \,d x } \]
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Time = 12.08 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.74 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\frac {a d^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {2 a d \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {a x}{e^{2}} - \frac {b d^{2} n \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {\log {\left (x \right )}}{d e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{d e} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{2}} + \frac {2 b d n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\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 (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {2 b d \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{2}} - \frac {b n x}{e^{2}} + \frac {b x \log {\left (c x^{n} \right )}}{e^{2}} \]
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\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{{\left (e x + d\right )}^{2}} \,d x } \]
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\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{{\left (e x + d\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^2} \,d x \]
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