Integrand size = 25, antiderivative size = 335 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=-\frac {b e n}{2 d f^2 x}-\frac {b e^2 n \log (x)}{2 d^2 f^2}-\frac {2 b e g n \log (x)}{d f^3}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}-\frac {3 b g^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^4} \]
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Time = 0.23 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {46, 2463, 2442, 36, 29, 31, 2441, 2352, 2440, 2438} \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac {3 g^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}-\frac {b e^2 n \log (x)}{2 d^2 f^2}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}-\frac {3 b g^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^4}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {2 b e g n \log (x)}{d f^3}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e n}{2 d f^2 x} \]
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Rule 29
Rule 31
Rule 36
Rule 46
Rule 2352
Rule 2438
Rule 2440
Rule 2441
Rule 2442
Rule 2463
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x^3}-\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x^2}+\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 x}-\frac {g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)^2}-\frac {3 g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 (f+g x)}\right ) \, dx \\ & = \frac {\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3} \, dx}{f^2}-\frac {(2 g) \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2} \, dx}{f^3}+\frac {\left (3 g^2\right ) \int \frac {a+b \log \left (c (d+e x)^n\right )}{x} \, dx}{f^4}-\frac {\left (3 g^3\right ) \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{f^4}-\frac {g^3 \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx}{f^3} \\ & = -\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}+\frac {(b e n) \int \frac {1}{x^2 (d+e x)} \, dx}{2 f^2}-\frac {(2 b e g n) \int \frac {1}{x (d+e x)} \, dx}{f^3}-\frac {\left (3 b e g^2 n\right ) \int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx}{f^4}+\frac {\left (3 b e g^2 n\right ) \int \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right )}{d+e x} \, dx}{f^4}-\frac {\left (b e g^2 n\right ) \int \frac {1}{(d+e x) (f+g x)} \, dx}{f^3} \\ & = -\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^4}+\frac {(b e n) \int \left (\frac {1}{d x^2}-\frac {e}{d^2 x}+\frac {e^2}{d^2 (d+e x)}\right ) \, dx}{2 f^2}-\frac {(2 b e g n) \int \frac {1}{x} \, dx}{d f^3}+\frac {\left (2 b e^2 g n\right ) \int \frac {1}{d+e x} \, dx}{d f^3}+\frac {\left (3 b g^2 n\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{f^4}-\frac {\left (b e^2 g^2 n\right ) \int \frac {1}{d+e x} \, dx}{f^3 (e f-d g)}+\frac {\left (b e g^3 n\right ) \int \frac {1}{f+g x} \, dx}{f^3 (e f-d g)} \\ & = -\frac {b e n}{2 d f^2 x}-\frac {b e^2 n \log (x)}{2 d^2 f^2}-\frac {2 b e g n \log (x)}{d f^3}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}-\frac {3 b g^2 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^4} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=-\frac {\frac {4 b e f g n (\log (x)-\log (d+e x))}{d}+\frac {b e f^2 n (d+e x \log (x)-e x \log (d+e x))}{d^2 x}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^2}-\frac {4 f g \left (a+b \log \left (c (d+e x)^n\right )\right )}{x}-\frac {2 f g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-6 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {2 b e f g^2 n (\log (d+e x)-\log (f+g x))}{e f-d g}+6 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )+6 b g^2 n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )-6 b g^2 n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{2 f^4} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.66 (sec) , antiderivative size = 549, normalized size of antiderivative = 1.64
method | result | size |
risch | \(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{2 f^{2} x^{2}}+\frac {3 b \ln \left (\left (e x +d \right )^{n}\right ) g^{2} \ln \left (x \right )}{f^{4}}+\frac {2 b \ln \left (\left (e x +d \right )^{n}\right ) g}{f^{3} x}-\frac {3 b \ln \left (\left (e x +d \right )^{n}\right ) g^{2} \ln \left (g x +f \right )}{f^{4}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g^{2}}{f^{3} \left (g x +f \right )}+\frac {3 b e n \ln \left (e x +d \right ) g^{2}}{f^{3} \left (d g -e f \right )}-\frac {3 b \,e^{2} n \ln \left (e x +d \right ) g}{2 f^{2} \left (d g -e f \right ) d}-\frac {b \,e^{3} n \ln \left (e x +d \right )}{2 f \left (d g -e f \right ) d^{2}}-\frac {2 b e g n \ln \left (x \right )}{d \,f^{3}}-\frac {b \,e^{2} n \ln \left (x \right )}{2 d^{2} f^{2}}-\frac {b e n}{2 d \,f^{2} x}-\frac {b e n \,g^{2} \ln \left (g x +f \right )}{f^{3} \left (d g -e f \right )}-\frac {3 b n \,g^{2} \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{4}}-\frac {3 b n \,g^{2} \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{4}}+\frac {3 b n \,g^{2} \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{4}}+\frac {3 b n \,g^{2} \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {1}{2 f^{2} x^{2}}+\frac {3 g^{2} \ln \left (x \right )}{f^{4}}+\frac {2 g}{f^{3} x}-\frac {3 g^{2} \ln \left (g x +f \right )}{f^{4}}+\frac {g^{2}}{f^{3} \left (g x +f \right )}\right )\) | \(549\) |
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\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}} \,d x } \]
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\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int \frac {a + b \log {\left (c \left (d + e x\right )^{n} \right )}}{x^{3} \left (f + g x\right )^{2}}\, dx \]
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\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}} \,d x } \]
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\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}} \,d x } \]
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Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^3\,{\left (f+g\,x\right )}^2} \,d x \]
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